User talk:Egm6341.f10.team5.robcarroll

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Problem 4
Find $$F(x,y,y') = \frac {d\phi} {dx} (x,y)$$ where $$\phi(x,y) = x^2 y^{\frac 3 2} + \log (x^3 y^2) = k$$ and verify that $$F$$ is exact N1_ODE and invent 3 more.

Solution 4
First, differentiate $$\frac {d\phi} {dx} (x,y)$$ To do so, we use the product rule and the chain rule for the first term in $${phi}(x,y)$$ and a Logarithmic derivative for the second term. So then $$F(x,y,y')$$ comes out to be: $$F(x,y,y') = 2 x y^{\frac 3 2} + \frac 3 {x} + [{\frac 3 2} x^2 y^{\frac 1 2} + \frac 2 y] y'=0$$

where the constant $$k$$ goes to zero after differentiation.

Verification
For $$F(x,y,y')$$ to meet the criteria of an exact N1_ODE it must meet three conditions: 1) Have an integral $$\frac {d\phi} {dx} (x,y)$$ that exists, which it clearly does.

2) Fit the form $$M(x,y) + N(x,y) y' = 0$$, which it does if we make $$M(x,y) = 2 x y^{\frac 3 2} + \frac 3 {x}$$and $$N(x,y) = [{\frac 3 2} x^2 y^{\frac 1 2} + \frac 2 y]$$

3) If $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$$, which we can see that it does as well.

Therefore it can be said that $$F(x,y,y')$$ is an exact non-linear first order ordinary differential equation.

Three other examples
Here are three other examples of exact N1_ODE

1) $$F(x,y,y') = 5 x^4 ln(y) + x^5 \frac 1 y y'$$

2) $$F(x,y,y') = 5 x^{\frac 3 2} y^2 + 4 x^{\frac 5 2} y y' $$

3) $$F(x,y,y') = 3 \frac 1 x y^3 + 9 ln(x) y^2 y' $$

Problem 1 - Derive EOM (SC-N1-ODE)
From Meeting 13, p. 13-1 ~ p. 13-2

Given
A projectile of mass (m) is being shot through the air at an angle (α) to the horizontal with velocity (v). The projectile experiences a drag force due to air resistance, $$F_{air} = k v^n \,$$ Where $$k \,$$ is the coefficient of air resistance and is a constant in this problem. The force equations are therefore given as such, $$m(\frac {dv_x} {dt}) = -k v^{n} {\cos {\alpha}}$$ $$\displaystyle (Eq. 1.1) $$ $$m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$ $$\displaystyle (Eq. 1.2) $$

Find
For the given situation: $$m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$
 * 1) Derive the equations of motion for the projectile
 * 2) For the particular case $$k = 0 \,$$ verify that $$y(x)\,$$ is a parabola
 * 3) Consider $$k \ne 0 \,$$ and $$v_{xo}=0 \,$$, therefore

becomes

$$m(\frac {dv_y} {dt}) = -k v_y^{n} {\sin {\alpha}} - mg$$


 * 3.1 Find $$v_y (t)\,$$ and $$y(t) \,$$ for $$m = constant \,$$
 * 3.2 Find $$v_y (t)\,$$ and $$y(t) \,$$ for $$m = m(t) \,$$


 * 4. Is $$m(\frac {dv_y} {dt}) = -k v_y^{n} {\sin {\alpha}} - mg$$ exact or can it be made exact?

Part 1
The projectile is given an initial velocity $$ v $$ in a direction that is at an angle $$ \alpha $$ to the horizontal. The Equations of motion (Eq. 1.1 and Eq. 1.2) given in the problem statement are acquired through summing the forces on the projectile and decomposing those forces into their vector quantities acting on the projectile. We sum the forces by looking at which ones are having an effect on the motion of the object.

1) inertia of the projectile after being fired from its initial position:

$$F_{inertia} = m(\frac {dv} {dt}) $$

which must be decomposed into its x and y components:

$$F_{x-inertia} = m(\frac {dv_x} {dt}) $$

$$F_{y-inertia} = m(\frac {dv_y} {dt}) $$

2) The effects of gravity on the y-component of the projectile's inertia:

$$F_{g} = -m g \,$$

$$g =  \,$$ acceleration  due  to  gravity

3) The force of air resistance on the projectile's motion decomposed into its x and y directional components:

$$F_{x-air} = -k v^{n} {\cos {\alpha}} \,$$

$$F_{y-air} = -k v^{n} {\sin {\alpha}} \,$$

Now if we sum the forces in the x and y directions and consider the force equilibrium in those directions we obtain the Equations of Motion given in the problem statement:

$$\sum {F_x} = m(\frac {dv_x} {dt}) = -k v^{n} {\cos {\alpha}}$$

$$\sum {F_y} = m(\frac {dv_y} {dt}) = -k v^{n} {\sin {\alpha}} - mg$$

Part 2
For the particular case where $$ k = 0 $$ equations 1.1 and 1.2 reduce to:

$$ m(\frac {dv_x} {dt}) = 0$$

which reduces to

$$ (\frac {dv_x} {dt}) = 0$$$$\displaystyle (Eq. 1.3) $$

and

$$m(\frac {dv_y} {dt}) = -mg $$

which reduces further to

$$(\frac {dv_y} {dt}) = -g $$$$\displaystyle (Eq. 1.4) $$

Now we can integrate Eqs. 1.3 and 1.4 with respect to time in order to get expressions for velocity and position in the x and y directions. We will start with the x-direction equations:

x-component:

$$ \frac {dv_x} {dt} = 0$$

Integrating with respect to time we obtain,

$$ v_x(t) = c_1 \,$$$$\displaystyle (Eq. 1.5) $$

Using the initial condition $$ v_x(t=0) = v_{xo} \,$$ the constant of integration $$ c_1 \,$$ can be found,

$$ c_1 = v_{xo} \,$$

So Eq. 1.5 becomes,

$$ v_x(t) = v_{xo} \,$$ $$\displaystyle (Eq. 1.6) $$

Now we repeat the same procedure for Eq. 1.4 to obtain the y component of the velocity vector. Once we integrate Eq. 1.4 with respect to time we get,

$$ v_y(t) = -gt + c_2 \,$$$$\displaystyle (Eq. 1.7) $$

Then using the initial condition $$ v_y(t=0) = v_{yo} \,$$ we can determine that $$ c_2 = v_{yo} \,$$ so then Eq. 1.7 becomes,

$$ v_y(t) = -gt + v_{yo} \,$$$$\displaystyle (Eq. 1.8) $$

Now we integrate Eq. 1.7 and 1.8 with respect to time once more to obtain the equations for position as a function of time in the x and y directions. After integrating with respect to time Eq. 1.7 and 1.8 become,

$$ x(t) = v_{xo}t + c_3 \,$$$$\displaystyle (Eq. 1.9) $$

and

$$ y(t) = - \frac 1 2 gt^2 + v_{yo}t + c_4 \,$$$$\displaystyle (Eq. 1.10) $$

Using initial conditions for position $$ x(t=0) = x_{o}\,$$ and $$ y(t=0) = y_{o}\,$$ then $$c_3\,$$ and $$c_4\,$$ can be solved for,

$$ c_3 = x_{o}\,$$

and

$$ c_4 = y_{o}\,$$

Plugging these values back into Eq. 1.9 and 1.10 yields the position equations for the x and y directions,

$$ x(t) = v_{xo}t + x_o \,$$$$\displaystyle (Eq. 1.11) $$

$$ y(t) = - \frac 1 2 gt^2 + v_{yo}t + y_o \,$$$$\displaystyle (Eq. 1.12) $$

To express y as a function of x we will manipulate eq. 1.11 to get t in terms of x. This yeilds the expression,

$$ t = \frac {x-x_o} {v_{xo}} \,$$$$\displaystyle (Eq. 1.13) $$

We then plug this value in for t in Eq. 1.12 to get,

$$ y(x) = - \frac 1 2 g(\frac {x-x_o} {v_{xo}})^2 + v_{yo}(\frac {x-x_o} {v_{xo}}) + y_o \,$$$$\displaystyle (Eq. 1.14) $$

to confirm that Eq. 1.14 is a parabola we can simply compare the general form of a parabolic function to Eq. 1.14,

$$ y(x) = ax^2 + bx + c \,$$

$$ y(x) = - \frac 1 2 g(\frac {x-x_o} {v_{xo}})^2 + v_{yo}(\frac {x-x_o} {v_{xo}}) + y_o \,$$

where $$a$$, $$b$$ and $$c$$ are all constants. It is apparent that Eq. 1.14 fits the form of a parabolic function so it can be concluded that our expression for $$y(x)$$ is a parabola.

Part 3
For the case where $$v_{xo}=0 \,$$, and $$k \ne 0$$ we can see from Eq. 1.1 that as long as $$v^n \ne 0 \,$$ then $$\alpha = 90^\circ \,$$ so that there are no net forces acting on the projectile in the x direction which leaves Eq. 1.2 as the net force acting on the projectile. However, since $$\alpha = 90^\circ \,$$ Eq. 1.2 can be re-written as,

$$m(\frac {dv_y} {dt}) = -k v_y^{n}- mg \,$$ $$\displaystyle (Eq. 1.15) $$

Part 3.1
Solving Eq. 1.15 for $$v_y(t)$$ and $$y(t)$$ when $$m =$$ constant. Start with the equation of motion (Eq. 1.15),

$$m(\frac {dv_y} {dt}) = -k (v_y)^{n}- mg \,$$

In order to solve this we must first show exactness. So we first put Eq 1.15 into a form that shows the first condition of exactness is met

$$m(\frac {dv_y} {dt})+ k v_y^{n}+ mg = 0\,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.16) $$

So that we can see that the two parts, $$M(v_y,t)$$ and $$N(v_y,t)$$ are,

$$M(v_y,t)=k v_y^{n}+ mg $$

$$N(v_y,t)=m \,$$

Now that the first condition of exactness has been shown we must see if the second condition of exactness is met,

$$ M(v_y,t)_{v_y} = N(v_y,t)_t$$<p style="text-align:right;">$$\displaystyle (Eq. 1.17) $$

Which is clearly not met if $$m$$ is a constant

$$M(v_y,t)_{v_y} = knv_y^{n-1} \ne N(v_y,t)_{t} = \frac{dm(t)}{dt}$$<p style="text-align:right;">$$\displaystyle (Eq. 1.18) $$

Therefore, the equation is non-exact.

However, the equation can be made exact through the integrating factor method

Then expressing as a total derivative,

Keeping in mind that Eq. 1.20 is a sum of the forces so we can introduce the $$\phi$$ notation that will simplify the equation a bit.

$$ \sum F = \frac {\partial \phi} {\partial t} dt + \frac {\partial \phi} {\partial v_y} dv_y =0 $$

Therefore the $$\phi$$ derivatives are as follows,

$$\phi_t = h(v_y,t)(kv_y^n+mg)$$

$$\phi_{v_y} = h(v_y,t)m)$$

$$\phi_{t,v_y} = \phi_{v_y,t}$$

Now, we test for exactness. In order for Eq. 1.20 to be exact it must meet the condition,

$$h_{v_y}N+hN_{v_y}=h_tM+hM_t \,$$<p style="text-align:right;">$$\displaystyle (Eq. 1.21) $$

So we substitute the problem parameters into Eq. 1.21 and get,

$$\frac{\partial h(v_y,t)}{\partial v_y}(kv_y^n+mg)+h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m+h(v_y,t)\frac{\partial m}{\partial t}$$<p style="text-align:right;">$$\displaystyle (Eq. 1.22) $$

Letting $$\partial h / \partial v_y = 0$$ and since $$m$$ is a constant $$\partial m / \partial t = 0$$. So Eq. 1.22 reduces to,

$$h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m$$<p style="text-align:right;">$$\displaystyle (Eq. 1.23) $$

Now, we manipulate Eq 1.23 in order to obtain an expression for $$h(v_y,t)$$,

$$h(v_y,t)\frac{\partial(kv_y^n+mg)}{\partial v_y} = \frac{\partial h(v_y,t)}{\partial t}m$$<p style="text-align:right;">$$\displaystyle (Eq. 1.24) $$

$$h(v_y,t)\frac{nkv_y^n}{m} = \frac{\partial h(v_y,t)}{\partial t}$$<p style="text-align:right;">$$\displaystyle (Eq. 1.25) $$

$$h(v_y,t) = \exp(\frac{kv_y^n}{m}t)$$<p style="text-align:right;">$$\displaystyle (Eq. 1.26) $$

Therefore the $$\phi$$ derivative equations can be solved to get $$\phi$$,

$$ \phi_t = \exp(\frac{kv_y^n}{m}t)(kv_y^n+mg) $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.27) $$

$$ \phi = \int{\phi_t dt} = \int{ \exp(\frac{kv_y^n}{m}t)(kv_y^n+mg)dt} $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.28) $$

$$ \phi =(kv_y^n+mg)\frac{m}{kv_y^n}\exp(\frac{kv_y^n}{m}t)+f(v_y) = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.29) $$

$$ \phi_v = m\exp(\frac{kv_y^n}{m}t) $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.30) $$

$$ \phi = \int{\phi_{v_y} dv_y} = \int{ m\exp(\frac{kv_y^n}{m}t)dv_y} = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.31) $$

$$ \phi = m\int{\exp(\frac{kv_y^n}{m}t)dv_y} + f(t) = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.32) $$

We can now combine these $$\phi$$ equations to come up with an expression of the sum of the forces that is exact,

$$ \sum F =\frac {\partial \phi} {\partial t} dt + \frac {\partial \phi} {\partial v_y} dv_y =\phi = m\int{\exp(\frac{kv_y^n}{m}t)dv_y} + (kv_y^n+mg)\frac{m}{kv_y^n}\exp(\frac{kv_y^n}{m}t) = c_5 $$ <p style="text-align:right;">$$\displaystyle (Eq. 1.33) $$

This resulting equation is not integrable and therefore no explicit general form of $$v_y$$ was found. However, through the integrating factor method an exact non-linear expression was found. Numerical solutions can be found given a value of n, and initial conditions.

Part 3.2
For the condition where $$m=m(t) \ne 0$$, the integrating factor becomes even more complicated because in Eq 1.22 the initial simplification that occurs when $$ \frac {\partial m} {\partial t} = 0$$ does not happen. This will result in another implicit solution for $$v_y$$ when the integrating factor method is applied.

Part 3.2
If $$m(t)\ne0$$ the integrating factor method is complicated in equation 1.22 as the partial of $$m$$ with respect to time remains, complicating the expression for $$h$$

Author and Proof-reader
[Author]

[Proof-reader]

=HW 5=

Given
The Euler L2-ODE-VC: $$ a_2x^2y''+a_1xy'+a_0y=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1) $$

The Euler L2-ODE-CC: $$ b_2y''+b_1y'+b_0y=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2) $$

The characteristic equation: $$ (r-\lambda)^2 = r^2-2\lambda r + \lambda^2 = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.3) $$

Find
1.1) Find $$a_2$$, $$a_1$$ , $$a_0$$ such that Eqn. 3.3 is the characteristic equation of Eqn. 3.1

1.2) Find the first homogeneous solution, $$y_1$$

1.3) Find the second homogeneous solution, $$y_2$$

1.4) Find the general homogeneous solution

2.1) Find $$b_2$$, $$b_1$$ , $$b_0$$ such that Eqn. 3.3 is the characteristic equation of Eqn. 3.2

2.2) Find the first homogeneous solution, $$y_1$$

2.3) Find the second homogeneous solution, $$y_2$$

2.4) Find the general homogeneous solution

Part 1.1
The coefficients of Eqn. 3.1 (a terms) can be found if we use the trial solution method. The trial solution for an Euler L2-ODE-VC is: $$ y=x^r \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.1) $$

Therefore we can calculate the first and second derivatives of the trial solution:

$$ y'=rx^(r-1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.2) $$

$$ y''=r(r-1)x^(r-2) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.3) $$

Then substitute back into Egn. 3.1 so that it becomes,

$$ a_2x^2r(r-1)x^(r-2)+a_1xrx^(r-1)+a_0x^r=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.4) $$

Then simplify,

$$ x^r[a_2r(r-1)+a_1r+a_0]=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.5) $$

$$ a_2r^2-a_2r+a_1r+a_0=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.6) $$

$$ a_2r^2+(a_1-a_2)r+a_0=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.7) $$

We then set Eqn. 3.1.7 equal to the expanded out version of the characteristic equation (Eqn. 3.3) to obtain the values of the coefficients,

$$ a_2r^2+(a_1-a_2)r+a_0=r^2-2\lambda r+\lambda^2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.8) $$

It is then clear that,

$$ a_2 = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.9) $$

$$ (a_1-a_2)= -2\lambda ; a_1 = a_2-2\lambda = 1-2\lambda \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.10) $$

$$ a_0=\lambda^2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.11) $$

Part 1.2
To find the first homogeneous solution use the trial solution (Eqn. 3.1.1) and the characteristic equation (Eqn. 3.3). We find the roots of Eqn. 3.3 and substitute the first (real) one in for $$r$$ in the trial solution (Egn. 3.1.1). In this particular case the two roots are the same so $$r_1=r_2=\lambda$$. Substituting this into Eqn. 3.1.1 we get,

$$ y_1=x^\lambda \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.12) $$

Part 1.3
Since $$\lambda$$ has two identical roots we can find the second homogeneous solution using the form,

$$ y_2(x)=U(x)y_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.13) $$

We then calculate the first and second derivatives of Eqn 3.1.13. The $$(x)$$ argument will be ignored to simplify the equations.

$$ y_2'=U'y_1+Uy_1' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.14) $$

$$ y_2=Uy_1+2U'y_1'+Uy_1'' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.15) $$

Plug Eqns. 3.1.13, 3.1.14, 3.1.15 into Eqn 3.1 and simplify,

$$ a_2x^2(Uy_1+2U'y_1'+Uy_1)+a_1x(U'y_1+Uy_1')+a_0Uy_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.16) $$

$$ U(a_2x^2y_1+a_1xy_1'+a_0y_1)+U'(2a_2x^2y_1'+a_1xy_1)+Ua_2x^2y_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.17) $$

This can be simplified further noting that $$U$$ is multiplied by Eqn 3.1, which equals $$0$$. So that completely eliminates one term in Egn. 3.1.17. We then substitute in the calculated values of the $$a$$ coefficients and the values for $$y_1$$ from part 1.1 and simplify,

$$ U'(2a_2x^2y_1'+a_1xy_1)+U''a_2x^2y_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.18) $$

$$ U'[2x^2\lambda x^{\lambda -1}+(1-2\lambda )xx^\lambda ]+U''x^2x^\lambda =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.19) $$

$$ U'[2x^2\lambda x^{\lambda -1}+(1-2\lambda )xx^\lambda ]+U''x^2x^\lambda =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.19) $$

$$ U'[2\lambda x^{\lambda +1}+x^{\lambda +1}-2\lambda x^{\lambda +1}]+U''x^{\lambda +2} =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.20) $$

$$ U'x^{\lambda +1}+U''x^{\lambda +2} =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.21) $$

$$ (U'+ U''x)x^{\lambda +1} =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.22) $$

$$ U'+ U''x =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.23) $$

We then use variation of parameters to reduce the order of Eqn. 3.1.23,

$$ m=U' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.24) $$

So we substitute in the change of parameters into Egn. 3.1.24 and use separation of variables to solve for $$m$$,

$$ m+ m'x =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.25) $$

$$ m + x \frac {dm} {dx}=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.26) $$

$$ m = -x \frac {dm} {dx} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.27) $$

$$ \frac {m} {dm} = - \frac {x} {dx} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.28) $$

Then invert to get,

$$ \frac {dm} {m} = - \frac {dx} {x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.29) $$

Then integrate each side and simplify,

$$ ln (m) = - (ln (x) + k_1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.30) $$

$$ m = \frac 1 x e^{- k_1} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.31) $$

$$ k_2 = e^{-k_1} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.32) $$

$$ m = \frac 1 x k_2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.33) $$

Then substitute the identity $$m = U'$$ into Eqn. 3.1.33 and integrate to solve for $$U$$,

$$ U = \int U' dx = \int \frac 1 x k_2 dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.34) $$

$$ U = k_2 [ln(x) + k_3] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.35) $$

Substitute $$U$$ back into Eqn. 3.1.13 along with the known value of $$y_1$$ to obtain the 2nd homogeneous solution,

$$ y_2 = k_2 [ln(x) + k_3]x^{\lambda} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.36) $$

Part 1.4
The general homogeneous solution of the Euler L2-ODE-VC will have the form,

$$ y(x) = C_1 y_1(x)+C_2 y_2(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.37) $$

Plugging Eqns. 3.1.36 and 3.1.12 into Eqn. 3.1.37 and simplifying results in the general homogeneous solution,

$$ y(x) = C_1 x^{\lambda} + C_2 [k_2 [ln(x) + k_3]x^{\lambda}] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.38) $$

$$ y(x) = x^{\lambda}[C_1 + C_2 k_2 ln(x) + C_2 k_3] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.39) $$

$$ C_3 = C_1 + C_2 k_3 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.40) $$

$$ C_4 = C_2 k_2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.41) $$

Therefore the general homogeneous solution simplifies to,

$$ y(x) = x^{\lambda}[C_4 + C_5 ln(x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.1.42) $$

Part 2.1
The coefficients of Eqn. 3.2 (b terms) can be found if we use the trial solution method. The trial solution for an Euler L2-ODE-CC is: $$ y=e^{rx} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.1) $$

Therefore we can calculate the first and second derivatives of the trial solution:

$$ y'=re^{rx} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.2) $$

$$ y''=r^2e^{rx} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.3) $$

Then substitute back into Egn. 3.2 so that it becomes,

$$ b_2r^2e^{rx} + b_1re^{rx} + b_0e^{rx}=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.4) $$

Then simplify,

$$ e^{rx}[b_2r^2+b_1r+b_0]=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.5) $$

$$ b_2r^2+b_1r+b_0=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.6) $$

We then set Eqn. 3.2.6 equal to the expanded out version of the characteristic equation (Eqn. 3.3) to obtain the values of the coefficients,

$$ b_2r^2+b_1r+b_0=r^2-2\lambda r+\lambda^2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.7) $$

It is then clear that,

$$ b_2 = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.8) $$

$$ b_1= -2\lambda \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.9) $$

$$ b_0=\lambda^2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.10) $$

Part 2.2
To find the first homogeneous solution use the trial solution (Eqn. 3.2.1) and the characteristic equation (Eqn. 3.3). We find the roots of Eqn. 3.3 and substitute the first (real) one in for $$r$$ in the trial solution (Egn. 3.2.1). In this particular case the two roots are the same so $$r_1=r_2=\lambda$$. Substituting this into Eqn. 3.2.1 we get,

$$ y_1=e^{\lambda x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.11) $$

Part 2.3
Since $$\lambda$$ has two identical roots we can find the second homogeneous solution using the form,

$$ y_2(x)=U(x)y_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.12) $$

We then calculate the first and second derivatives of Eqn 3.2.12. The $$(x)$$ argument will be ignored to simplify the equations.

$$ y_2'=U'y_1+Uy_1' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.13) $$

$$ y_2=Uy_1+2U'y_1'+Uy_1'' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.14) $$

Plug Eqns. 3.2.12, 3.2.13, 3.2.14 into Eqn 3.2 and simplify,

$$ b_2(Uy_1+2U'y_1'+Uy_1)+b_1(U'y_1+Uy_1')+b_0Uy_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.15) $$

$$ U(b_2y_1+b_1y_1'+b_0y_1)+U'(2b_2y_1'+b_1y_1)+Ub_2y_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.16) $$

This can be simplified further noting that $$U$$ is multiplied by Eqn 3.2, which equals $$0$$. So that completely eliminates one term in Egn. 3.2.16. We then substitute in the calculated values of the $$a$$ coefficients and the values for $$y_1$$ from part 2.1 and simplify,

$$ U'(2b_2y_1'+b_1y_1)+U''b_2y_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.17) $$

$$ U'[2\lambda e^{\lambda x}+(-2\lambda )e^{\lambda x}]+U''e^{\lambda x} =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.18) $$

$$ U''e^{\lambda x} =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.19) $$

Therefore

$$ U'' =0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.20) $$

Then Integrate Eqn. 3.2.20 twice with respect to $$x$$

$$ U' = k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.21) $$

$$ U = x k_1 + k_2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.22) $$

Then plug Eqn. 3.2.22 and 3.2.1 into Eqn. 3.2.12 to get,

$$ y_2(x) = e^{\lambda x} (x k_1 + k_2) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.23) $$

Part 2.4
The general homogeneous solution of the Euler L2-ODE-CC will have the form,

$$ y(x) = C_1 y_1(x)+C_2 y_2(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.24) $$

Plugging Eqns. 3.2.23 and 3.2.12 into Eqn. 3.2.24 and simplifying results in the general homogeneous solution,

$$ y(x) = C_1 e^{\lambda x} + C_2 e^{\lambda x} (x k_1 + k_2) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.25) $$

$$ y(x) = e^{\lambda x}[C_1 + C_2 + k_2] + C_2 k_1 x e^{\lambda x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.26) $$

$$ C_3 = C_1 + C_2 + k_3 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.27) $$

$$ C_4 = C_2 k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.28) $$

Therefore the general homogeneous solution simplifies to,

$$ y(x) = C_3 e^{\lambda x} + C_4 x e^{\lambda x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 3.2.29) $$

=HW 6=

Given
$$ P_2(x) = \frac 1 2 (3x^2-1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.1) $$

$$ Q_2(x) = \frac 1 4 (3x^2-1) log(\frac {1+x} {1-x}) - \frac 3 2 x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.2) $$

Find
Assuming Eq. 1.1 is the first homogeneous solution to the Legendre function show that Eq. 1.2 is a second homogeneous solution using the variation of parameters method.

Solution
First we must identify the Equation that Eq. 1.1 and 1.2 are solutions to. The Legendre function is written as

$$ 0 = (1-x^2)y'' - 2xy' + n(n+1)y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.3) $$

For a second order equation $$n=2$$ so the Legendre function becomes

$$ 0 = (1-x^2)y'' - 2xy' + 6y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.4) $$

To obtain the second homogeneous solution the general form of a homogeneous L2-ODE-VC will be needed,

$$ 0 = a_2y'' + a_1 y' + a_0 y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.5) $$

So we use the general solution form,

$$ y(x)=U(x) u_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.6) $$

Where $$U(x)$$ is unknown, $$u_1(x) = P_2(x)$$ and $$u_2(x)$$ is the second homogeneous solution in this case. So we now solve for $$u_2(x)$$ in a general form,

$$ u_2(x)=U(x)u_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.7) $$

$$ u_2'=U'u_1+Uu_1' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.8) $$

$$ u_2=Uu_1+2U'u_1'+Uu_1'' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.9) $$

Plug Eqs. 1.7, 1.8, 1.9 into Eq. 1.5 and simplify,

$$ a_2(Uu_1+2U'u_1'+Uu_1)+a_1(U'u_1+Uu_1')+a_0Uu_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.10) $$

$$ U(a_2u_1+a_1u_1'+a_0u_1)+U'(2a_2u_1'+a_1u_1)+Ua_2u_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.11) $$

This can be simplified further noting that $$U$$ is multiplied by the original general form of the L2-ODE-VC, which equals $$0$$. So that completely eliminates one term in Eg. 1.11. We then get the resulting equation,

$$ U'(2a_2u_1'+a_1u_1)+U''a_2u_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.12) $$

Then we force $$a_2 = 1$$, use the identity $$Z=U'$$ to reduce the order of Eq. 1.12, and divide through by $$u_1$$ to get,

$$ Z(\frac{2u_1'} {u_1} + a_1)+Z'=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.13) $$

This now makes the equation of first order. We can then use the integrating factor method to make the equation exact and thus easier to solve for $$U(x)$$. According to Eq. (1) on p.10-3 of the class notes the integrating factor for a linear first order ODE is of the form,

$$ h(x) = exp[\int a_0(s) ds] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.14) $$

In this case $$a_0 = (\frac{2u_1'} {u_1} + a_1)$$ so Eq. 1.14 becomes,

$$ h(x) = exp[\int (\frac{2u_1'} {u_1} + a_1) dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.15) $$

Simplifying Eq. 1.15 results in,

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.16) $$

Then multiply Eq. 1.13 through by Eq. 1.16 and simplify we find that

$$ hZ' + h'Z = (hZ)' = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.17) $$

Solving for $$Z$$ results in

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.18) $$

Invoking the identity $$Z = U'$$ to solve for $$ U $$.

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.19) $$

Then plug Eq. 1.19 back into Eq. 1.6 to get the expression,

$$ y(x) = u_1 [\int \frac {k_2} {h} dx + k_1] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.20) $$

Then multiply through by the $$u_1$$ term,

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.21) $$

Knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.22) $$

and $$ k_1 = 1 $$

Now we make the Legendre function, Eq. 1.4, fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to sole for $$h(x)$$

$$ 0 = y'' - \frac {2x} {(1-x^2)} y' + \frac {6} {(1-x^2)}y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.23) $$

Therefore,

$$ a_1 = \frac {-2x} {(1-x^2)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.24) $$

$$ a_0 = \frac {6} {(1-x^2)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.25) $$

Now plug Eq. 1.22 and Eq. 1.1 back into Eq. 1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = (\frac 3 2 x^2 - \frac 1 2)^2 (x^2 -1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.26) $$

Then plug Eq. 1.26 and Eq. 1.1 into Eq. 1.22 to get the expression for the 2nd homogeneous solution. We will also assume $$k_2=-1$$ for reasons that will become clear later. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = [\frac 1 2 (3 x^2 - 1)] (\int \frac{-1} {(\frac 3 2 x^2 - \frac 1 2)^2 (x^2 - 1)}) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.27) $$

This simplifies to

$$ u_2(x) = \frac 1 2 (3 x^2 - 1) \frac 1 2 [\frac{-6x} {3x^2 - 1} - log (1-x) + log (x+1)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.28) $$

This equation then simplifies to the form of Eq. 1.2

$$ u_2(x) = Q_2(x) = \frac 1 4 (3x^2-1) log(\frac {1+x} {1-x}) - \frac 3 2 x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 1.29) $$

Given
From p.28 problem 1.1 of King we are given

Part 1 A) $$ (x-1)y'' - xy'+y = f(x) = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.1) $$

with solution

$$ u_1(x) = e^x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.2) $$

and

B) $$ xy'' + 2y'+ xy = f(x) = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.1) $$

with solution

$$ u_1(x) = x^{-1}sin(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.2) $$

Part 2

Instead of homogeneous functions, make $$f(x) = sin(x)$$ for Eqs. 2.1.1 and 2.2.1 and find the general solutions.

Find
Show that $$u_1$$ in parts 1 and 2 are homogeneous solutions to their respective ODE's. Then use variation of parameters to find $$u_2$$ for Eq. 2.1.1 and Eq. 2.2.1

Part 1
Part A

First we must show that Eq. 2.1.2 is a solution to Eq. 2.1.1. To do this we simply plug Eq. 2.1.2 in for $$y$$ in Eq. 2.1.1. To do this we first must find the 1st and 2nd derivatives of Eq. 2.1.2

$$ u_1' = e^x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.3) $$

$$ u_1'' = e^x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.4) $$

Then plug Eq. 2.1.2, 2.1.3 and 2.1.4 into Eq. 2.1.1

$$ (x-1)e^x - x e^x + e^x = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.5) $$

It is then clear that the RHS reduces to zero, which matches the LHS, so we have confirmed that Eq. 2.1.2 is a solution of Eq. 2.1.1.

Next, we wil find the 2nd homogeneous solution to Eq. 2.1.1 using variation of parameters. to do this we use the general form of the homogeneous L2-ODE-VC,

$$ a_2y'' + a_1y' + a_0y = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.6) $$

To solve the general form we use the general form of the solution,

$$ y(x)=U(x) u_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.7) $$

Where $$U(x)$$ is unknown, $$u_1(x)$$ is Eq. 2.1.2 and $$u_2(x)$$ is the second homogeneous solution in this case. So we now solve Eq. 2.1.6 for $$y(x)$$ in a general form. To do this we must have the 1st and 2nd derivatives of $$y(x)$$,

$$ y =U(x)u_1(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.7) $$

$$ y'=U'u_1+Uu_1' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.8) $$

$$ y=Uu_1+2U'u_1'+Uu_1'' \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.9) $$

Plug Eqs. 2.1.7, 2.1.8, 2.1.9 into Eq. 2.1.6 and simplify,

$$ a_2(Uu_1+2U'u_1'+Uu_1)+a_1(U'u_1+Uu_1')+a_0Uu_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.10) $$

$$ U(a_2u_1+a_1u_1'+a_0u_1)+U'(2a_2u_1'+a_1u_1)+Ua_2u_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.11) $$

This can be simplified further noting that $$U$$ is multiplied by the original general form of the L2-ODE-VC, which equals $$0$$. So that completely eliminates one term in Eg. 2.1.11. We then get the resulting equation,

$$ U'(2a_2u_1'+a_1u_1)+U''a_2u_1=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.12) $$

Then we force $$a_2 = 1$$, use the identity $$Z=U'$$ to reduce the order of Eq. 2.1.12, and divide through by $$u_1$$ to get,

$$ Z(\frac{2u_1'} {u_1} + a_1)+Z'=0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.13) $$

This now makes the equation of first order. We can then use the integrating factor method to make the equation exact and thus easier to solve for $$U(x)$$. According to Eq. (1) on p.10-3 of the class notes the integrating factor for a linear first order ODE is of the form,

$$ h(x) = exp[\int a_0(s) ds] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.14) $$

In this case $$a_0 = (\frac{2u_1'} {u_1} + a_1)$$ so Eq. 2.1.14 becomes,

$$ h(x) = exp[\int (\frac{2u_1'} {u_1} + a_1) dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.15) $$

Simplifying Eq. 2.1.15 results in,

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.16) $$

Then multiply Eq. 2.1.13 through by Eq. 2.1.16 and simplify we find that

$$ hZ' + h'Z = (hZ)' = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.17) $$

Solving for $$Z$$ results in

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.18) $$

Invoking the identity $$Z = U'$$ to solve for $$ U $$.

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.19) $$

Then plug Eq. 2.1.19 back into Eq. 2.1.7 to get the expression,

$$ y(x) = u_1 [\int \frac {k_2} {h} dx + k_1] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.20) $$

Then multiply through by the $$u_1$$ term,

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.21) $$

Knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.22) $$

and $$ k_1 = 1 $$

Now we make the Eq. 2.1.1 fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to solve for $$h(x)$$

$$ 0 = y'' - \frac {x} {(x-1)} y' + \frac {1} {(x-1)}y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.23) $$

Therefore,

$$ a_1 = \frac {-x} {(x-1)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.24) $$

$$ a_0 = \frac {1} {(x-1)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.25) $$

Now plug Eq. 2.1.22 and Eq. 2.1.2 back into Eq. 2.1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = (\frac {e^x} {x-1} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.26) $$

Then plug Eq. 2.1.26 and Eq. 2.1.2 into Eq. 2.1.22 to get the expression for the 2nd homogeneous solution. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = e^x (k_2 \int \frac {(x-1)} {e^x}) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.27) $$

This simplifies to

$$ u_2(x) = e^x \frac {-k_2x} {e^x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.28) $$

After the negative sign is absorbed into the constant $$k_2$$, the 2nd homogeneous solution simplifies to the form

$$ u_2(x) = k_2 x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.29) $$

Part B

First we must show that Eq. 2.2.2 is a solution to Eq. 2.2.1. To do this we simply plug Eq. 2.2.2 in for $$y$$ in Eq. 2.2.1. To do this we first must find the 1st and 2nd derivatives of Eq. 2.2.2

$$ u_1' = -x^{-2} sin(x) + x^{-1} cos(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.3) $$

$$ u_1'' = 2x^{-3} sin(x) - 2x^{-2} cos(x) - x^{-1} sin(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.4) $$

Then plug Eq. 2.2.2, 2.2.3 and 2.2.4 into Eq. 2.2.1

$$ [2x^{-2} sin(x) - 2x^{-1} cos(x) - sin(x)] + [2x^{-1} cos(x) - 2x^{-2} sin(x)] + sin(x) = 0 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.5) $$

It is then clear that the RHS reduces to zero, which matches the LHS, so we have confirmed that Eq. 2.2.2 is a solution of Eq. 2.2.1.

Next, we wil find the 2nd homogeneous solution to Eq. 2.2.1 using variation of parameters. to do this we use the same techniques to find the general solutions to $$h(x)$$, $$Z$$ , and $$U$$ as we did in Part 1. So we can use Eqs. 2.1.16, 2.1.18 , 2.1.19 , and 2.1.21 shown below.

$$ h(x) = u_1^2 exp[\int a_1 dx] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.16) $$

$$ Z = \frac {1} {h(x)} \int 0 dx = \frac {k_2} {h(x)} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.18) $$

$$ U = \int \frac {k_2} {h} dx +k_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.19) $$

$$ y(x) = u_1 \int \frac {k_2} {h} dx + k_1u_1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.21) $$

Again, knowing that the complete homogeneous solution to the 2nd order equation is a linear combination of all homogeneous solutions means,

$$ u_2 = u_1 \int \frac {k_2} {h} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.22) $$

and $$ k_1 = 1 $$

Now we make the Eq. 2.2.1 fit the homogeneous form of our general L2-ODE-VC so that $$a_2 = 1$$ so that we can find $$a_1$$ in order to solve for $$h(x)$$

$$ 0 = y'' - \frac {2} {x} y' + y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.6) $$

Therefore,

$$ a_1 = \frac {2} {x} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.7) $$

$$ a_0 = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.8) $$

Now plug Eq. 2.2.7 and Eq. 2.2.2 into Eq. 2.1.16 and simplify the expression for $$h(x)$$ to obtain

$$ h(x) = sin^2(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.9) $$

Then plug Eq. 2.2.9 and Eq. 2.2.2 into Eq. 2.1.22 to get the expression for the 2nd homogeneous solution. So now the expression for $$u_2$$ becomes,

$$ u_2(x) = [\frac {sin(x)} {x}] [k_2 \int \frac {1} {sin^2(x)}] dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.10) $$

This simplifies to

$$ u_2(x) = [\frac {sin(x)} {x}] [-k_2cot(x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.11) $$

The 2nd homogeneous solution then simplifies to the form

$$ u_2(x) = k_2 [\frac {-cos(x)} {x}] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.12) $$

Part 2
Part A

To find the general solution of

$$ sin(x) = (x-1)y'' - xy' + y \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.30) $$

we assume the general solution is of the form

$$ y(x) = A(x) u_1 + B(x) u_2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.31) $$

Since we already know $$u_1$$ and $$u_2$$ from part 1 A we need to find $$A(x)$$ and $$B(x)$$. To do this we utilize the equations,

$$ A(x) = - \int \frac 1 W u_2(x) f(x) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.32) $$

and

$$ B(x) = \int \frac 1 W u_1(x) f(x) dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.33) $$

where W is the Wronskian. The wronskian is the determinant of a simple 2x2 matrix in this case. It will simplify to,

$$ W =u_1 u_2' - u_2 u_1' = e^x -xe^x = e^x (1-x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.34) $$

Now we plug Eq. 2.1.34 and 2.1.29 into Eq. 2.1.32, where the constant will be neglected in this case for $$u_2$$ and $$f(x) = sin(x)$$ so we get,

$$ A(x) = - \int \frac {x sin(x)} {e^x (1-x)} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.35) $$

For simplicity we will leave Eq. 2.1.35 in its integral form. Now we calculate $$B(x)$$. Plug Eq. 2.1.2 and Eq. 2.1.34 into Eq. 2.1.33 to get

$$ B(x) = \int \frac {sin(x)} {(1-x)} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.36) $$

Again, for simplicity we will leave Eq. 2.1.36 in its integral form. So we then plug Eqs. 2.1.35, 2.1.36 into Eq. 2.1.31 to get our general solution for Part A,

$$ y(x) = -e^x \int \frac {x sin(x)} {e^x (1-x)} dx + x \int \frac {sin(x)} {(1-x)} dx \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.1.37) $$

Part B We will now use the same procedure to find the general solution to the nonhomogeneous equation,

$$ sin(x) = xy'' + 2y' + xy \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.13) $$

Our Wronskian is now,

$$ W = u_1 u_2' - u_2 u_1' = (x^{-1} sin(x)) (x^{-2}cos(x) + x^{-1}sin(x)) - (-x^{-1}cos(x)) (-x^{-2}sin(x) + x^{-1}cos(x)) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.14) $$

which simplifies all the way down to,

$$ W = x^{-2} \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.15) $$

So the coefficients of our general solution $$A(x)$$ and $$B(x)$$ can be abtained by pluging Eq. 2.2.15, 2.2.12, and 2.2.2 into Eq. 2.1.32 and 2.1.33 as they are needed,

$$ A(x) = - \int x^2 \frac {-cos(x)} {x} sin(x) dx = \frac 1 8 [sin(2x)-2xcos(2x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.16) $$

$$ B(x) = \int x^2 \frac {sin(x)} {x} sin(x) dx = \frac {x^2} {4} - \frac 1 4 x sin(2x)- \frac 1 8 cos(2x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.17) $$

Now we plug Eq. 2.2.17, 2.2.16, 2.2.2 and 2.2.12 into Eq. 2.1.31 to obtain the general solution for Eq. 2.2.13.

$$ y(x) = \frac {sin(x)} {8x} [sin(2x)-2xcos(2x)] - \frac {cos(x)} {x} [\frac {x^2} {4} - \frac 1 4 x sin(2x)- \frac 1 8 cos(2x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 2.2.18) $$

=HW 7=

Given
$$ Q_n(x) = P_n(x) tanh^{-1}(x) - 2 \sum_{j=1,3, 5,...}^J \frac {2n-2j+1} {(2n-j+1)j} P_{n-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.1) $$

$$ J = 1+2[\frac {n-1} {2}] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.2) $$

Find
Use Eq. 7.1 to show $$Q_n(x)$$ is even or odd depending on $$n$$

Solution
In order to get a preliminary solution we must review when a function is even or odd. A function is said to be odd if $$-f(x) = f(-x)$$ and even if $$f(x) = f(-x)$$.

Since Eq. 7.1 is a combination of functions we must also review the properties of what happens when even/odd functions are combined either through multiplication or summation.

Even/Odd function Basic Properties

1) The sum of a number of odd functions is odd

2) The sum of a number of even functions is even

3) The sum of an even and an odd function is not even or odd, unless one of the functions is equal to zero

4) The product of an even (or odd) function and a constant multiple is even (odd)

5) The product of two even functions is even

6) The product of two odd functions is even

7) the product of an even and odd function is odd

Reference: Even and Odd functions

Now, to use these properties in eq. 7.1 we must determine whether $$P_n$$ is even or odd based on $$n$$. To do that we simply recall that $$P_n$$ is odd if $$n$$ is odd and $$P_n$$ is even if $$n$$ is even.

Reference: Legendre Polynomials

So, we can now look at Eq. 7.1 in terms of evenness and oddness. For the first case we assume $$ n = even $$

$$ Q_n(x) = (even) tanh^{-1}(x) - 2 \sum_{j=1,3,5,..}^J \frac {2n-2j+1} {(2n-j+1)j} (odd) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.3) $$

Eq. 7.3 must be further simplified in terms of it evenness/oddness. By simple visual analysis of a graph of $$tanh^{-1}(x)$$ or by plugging in random numbers and then their negative counterparts we can see that

$$ tanh^{-1}(x) = odd \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.4) $$

Also, we can apply Basic Properties (1) and (4) above to see that the summation term is the sum of a number of odd functions after each has been multiplied by a constant. Therefore the entire second term of Eq. 7.3 is an odd function. This allows us to reduce Eq. 7.3 to

$$ Q_n(x) = (even) (odd) - (odd) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.5) $$

Then using Basic Properties (1) and (7) we see that when

$$ n = even \,$$

$$ Q_n(x) = odd \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.6) $$

Now we use the same process to determine if $$Q_n(x)$$ is even or odd if $$n$$ is odd

If $$n = odd$$, then $$P_n(x) = odd$$ and $$P_{n-j}(x) = even$$. Also keeping in mind that $$tanh^{-1}(x)$$ is always odd. So Eq. 7.1 reduces to

$$ Q_n(x) = (odd) (odd) - 2 \sum_{j=1,3,5,..}^J (constant) (even) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.7) $$

Using Basic Properties (2), (4) and (5) it is clear that when

$$ n = odd \,$$

Eq. 7.7 reduces to

$$ Q_n(x) = even \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.8) $$

To prove that $$Q_n(x)$$ is even when $$n$$ is odd and vice versa we will look at the specific examples when $$n = 1, 2, 3, 4$$

To do this we must first get the Legendre Polynomials for $$n = 0, 1, 2, 3, 4$$. The quickest way to do this is to use the Rodriquez formula.

$$ P_n(x) = \frac {1} {2^n n!} \frac {d^n} {dx^n} (x^2-1)^n \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.9) $$

Reference: Rodrigues Formula

From Eq. 7.9 we can get

$$ P_0(x) = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.10) $$

$$ P_1(x) = x \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.11) $$

$$ P_2(x) = \frac {1} {2} (3x^2-1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.12) $$

$$ P_3(x) = \frac {1} {2} (5x^3-3x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.13) $$

$$ P_4(x) = \frac {1} {8} (35x^4 - 30x^2 + 3) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.14) $$

For $$n=1$$

First, we see for $$n=1$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {1-1} {2}] = 1 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.15) $$

Then plug Eq. 7.15, 7.11 into Eq 7.1 to get

$$ Q_1(x) = (x) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^1 \frac {2(1)-2(j)+1} {(2(1)-(j)+1)(j)} P_{1-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.16) $$

It is then clear that $$j=1$$ so once we plug that in and Eq. 7.10 for $$P_0(x)$$ Eq. 7.16 reduces to

$$ Q_1(x) = (x) tanh^{-1}(x) - (1) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.17) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_1(\frac 1 2) = -0.725347... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.18) $$

and

$$ Q_1(\frac {-1} {2}) = -0.725347... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.19) $$

Therefore we can see that $$Q_1(x)$$ is an even function

Used Wolfram Alpha for numerical computation

For $$n=2$$

First, we see for $$n=2$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {2-1} {2}] = 2 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.20) $$

Then plug Eq. 7.12, and 7.20 into Eq 7.1 to get

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^2 \frac {2(2)-2(j)+1} {(2(2)-(j)+1)(j)} P_{2-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.21) $$

It is then clear that $$j=1$$ so once we plug that in and Eq. 7.11 for $$P_1(x)$$ Eq. 7.21 reduces to

$$ Q_2(x) = (\frac 1 2 (3x^2-1)) tanh^{-1}(x) - (\frac 3 2) (x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.22) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_2(\frac 1 2) = -0.818663... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.23) $$

and

$$ Q_2(\frac {-1} {2}) = 0.818663... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.24) $$

Therefore we can see that $$Q_2(x)$$ is an odd function

Used Wolfram Alpha for numerical computation

For $$n=3$$

First, we see for $$n=3$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {3-1} {2}] = 3 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.25) $$

Then plug Eq. 7.13, and 7.25 into Eq 7.1 to get

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^3 \frac {2(3)-2(j)+1} {(2(3)-(j)+1)(j)} P_{3-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.26) $$

It is then clear that $$j=1, 3$$ so once we plug that in we see that Eq. 7.26 reduces to

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - 2 [\frac 5 6 (\frac 1 2 (3x^2-1)) + \frac 1 12 (1)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.27) $$

Eq. 7.27 can then be further simplified to get

$$ Q_3(x) = (\frac 1 2 (5x^3-3x)) tanh^{-1}(x) - \frac 5 6 (3x^2-1) - \frac 1 6 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.28) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_3(\frac 1 2) = -0.198655... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.29) $$

and

$$ Q_3(\frac {-1} {2}) = -0.198655... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.30) $$

Therefore we can see that $$Q_3(x)$$ is an even function

Used Wolfram Alpha for numerical computation

For $$n=4$$

First, we see for $$n=4$$ Eq. 7.2 reduces to

$$ J = 1+2[\frac {4-1} {2}] = 4 \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.31) $$

Then plug Eq. 7.14, and 7.31 into Eq 7.1 to get

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 \sum_{j=1,3,5,...}^4 \frac {2(4)-2(j)+1} {(2(4)-(j)+1)(j)} P_{4-j}(x) \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.32) $$

It is then clear that $$j=1, 3$$ so once we plug that in we see that Eq. 7.32 reduces to

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - 2 [\frac 7 8 (\frac 1 2 (5x^3-3x)) + \frac 3 18 (x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.33) $$

Eq. 7.33 can then be further simplified to get

$$ Q_4(x) = [\frac 1 8 (35x^4 - 30x^2 + 3)] tanh^{-1}(x) - \frac 7 8 (5x^3-3x)) - \frac 1 3 (x)] \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.34) $$

To test whether it is even or odd we can plug in $$x= \frac 1 2$$ and $$x= \frac {-1} {2}$$

$$ Q_4 (frac 1 2) = 0.440175... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.35) $$

and

$$ Q_4(\frac {-1} {2}) = -0.440175... \,$$ <p style="text-align:right;">$$\displaystyle (Eq. 7.36) $$

Therefore we can see that $$Q_4(x)$$ is an odd function

Used Wolfram Alpha for numerical computation

From the generalization of evenness and oddness made in the first part of the problem along with the examples provided it has been proven that Eq. 7.1 is an odd function when "n" is an even integer and even when "n" is an odd integer.