User:Egm6321.f10.team5/hw5

= Problem 1 - Higher order deriv.= From Meeting 26, p. 26-2

Find
Find $$y_{xxxxx}$$ in terms of derivative of $$y$$ with respect to $$t$$.

Author and Proof-Reader
[Author] Jo

[Proof-Reader] Oh, Sang Min

= Problem 2 - Solve using Method 2= From Meeting 26, p. 26-3

Find
Solve and plot $$x^2 y'' - 2 x y' + 2 y = 0$$ using the trial solution method where $$y = x^r$$ and boundary conditions are $$y(1) = 3$$ & $$y(2) = 4$$ or $$y(1) = -2$$ & $$y(2) = 5$$.

Solve
For $$y(1) = 3$$ & $$y(2) = 4$$ case, we get the following solution. For $$y(1) = -2$$ & $$y(2) = 5$$ case, we get the following solution.

Author and Proof-Reader
[Author] Jo

[Proof-Reader] Oh, Sang Min

= Problem 3 - Consider Char. eq.= From Meeting 27, p. 27-1

Given
The Euler L2-ODE-VC: $$ a_2x^2y''+a_1xy'+a_0y=0 \,$$ $$\displaystyle (Eq. 3.1) $$

The Euler L2-ODE-CC: $$ b_2y''+b_1y'+b_0y=0 \,$$ $$\displaystyle (Eq. 3.2) $$

The characteristic equation: $$ (r-\lambda)^2 = r^2-2\lambda r + \lambda^2 = 0 \,$$ $$\displaystyle (Eq. 3.3) $$

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: $$ y=x^r \,$$ $$\displaystyle (Eq. 3.1.1) $$

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

$$ y'=rx^(r-1) \,$$ $$\displaystyle (Eq. 3.1.2) $$

$$ y''=r(r-1)x^(r-2) \,$$ $$\displaystyle (Eq. 3.1.3) $$

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

$$ a_2x^2r(r-1)x^(r-2)+a_1xrx^(r-1)+a_0x^r=0 \,$$ $$\displaystyle (Eq. 3.1.4) $$

Then simplify,

$$ x^r[a_2r(r-1)+a_1r+a_0]=0 \,$$ $$\displaystyle (Eq. 3.1.5) $$

$$ a_2r^2-a_2r+a_1r+a_0=0 \,$$ $$\displaystyle (Eq. 3.1.6) $$

$$ a_2r^2+(a_1-a_2)r+a_0=0 \,$$ $$\displaystyle (Eq. 3.1.7) $$

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,

$$ a_2r^2+(a_1-a_2)r+a_0=r^2-2\lambda r+\lambda^2 \,$$ $$\displaystyle (Eq. 3.1.8) $$

It is then clear that,

$$ a_2 = 1 \,$$ $$\displaystyle (Eq. 3.1.9) $$

$$ (a_1-a_2)= -2\lambda ; a_1 = a_2-2\lambda = 1-2\lambda \,$$ $$\displaystyle (Eq. 3.1.10) $$

$$ a_0=\lambda^2 \,$$ $$\displaystyle (Eq. 3.1.11) $$

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,

$$ y_1=x^\lambda \,$$ $$\displaystyle (Eq. 3.1.12) $$

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

$$ y_2(x)=U(x)y_1(x) \,$$ $$\displaystyle (Eq. 3.1.13) $$

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;">$$\displaystyle (Eq. 3.2.29) $$

Author and Proof-Reader
[Author] Rob Carroll

[Proof-Reader] Oh, Sang Min

= Problem 4 - Find particular solution= From Meeting 27, p 27-2.

Given
We are given the differential equation,
 * $$\displaystyle y'+P(x)y=Q(x)$$

where the homogeneous solution has been previously found to be,
 * $$\displaystyle y_h=Ae^{-\int{P(x)dx}}$$

Find
We are to find a particular solution, $$y_p$$, by variation of parameters.

Solution
We first let
 * $$\displaystyle y(x)=u(x)y_h$$

By taking the derivative we find,
 * $$\displaystyle y'(x)=e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)]$$

We substitution this into the given equation,
 * $$\displaystyle e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)]+P(x)u(x)e^{-\int{P(x)dx}}=Q(x)$$

Rearranging yields,
 * $$\displaystyle e^{-\int{P(x)dx}}[u'(x)-u(x)P(x)+u(x)P(x)]=Q(x)$$

Which then simplifies to,
 * $$\displaystyle e^{-\int{P(x)dx}}u'(x)=Q(x)$$

We solve for $$u'(x)$$,
 * $$\displaystyle u'(x)=e^{\int{P(x)dx}}Q(x)$$

Integration yields,
 * $$\displaystyle u(x)=\int{e^{\int{P(x)dx}}Q(x)dx}$$

Keeping in mind that $$y(x)=u(x)y_h$$ and $$y(x)=y_p(x)+y_h(x)$$, we use $$u(x)$$ to find the particular solution,
 * $$\displaystyle y_p(x)=e^{-\int{P(x)dx}}\int{e^{\int{P(x)dx}}Q(x)dx}$$

Author and Proof-Reader
[Author]
 * Raul Riveros

[Proof-Reader]

= Problem 5 - Non-homog. L2-ODE-CC = From Meeting 28, p 28-2 to Meeting 29, p 29-1.

Given
Non-homogeneous L2-ODE-CC

Find
1)  Find PDEs for integrating factor $$ h(t,y) \! $$

2.1) Find $$ \bar{a}_1 $$ and $$ \bar{a}_0 $$ in terms of $$ a_2, a_1, a_0 \! $$

2.2) Find quadratic equation for $$ \alpha \!$$

2.3) Show reduced order equation using integrating factor

2.4) Use IFM to find $$ y \! $$ for general excitation $$ f(t) \! $$

2.5) Show that $$ \alpha \beta = \frac{a_0}{a_2} $$ and $$ \alpha + \beta = \frac{a_1}{a_2} $$

2.6) Deduce expression for particular solution for general excitation $$ f(t) \!$$

2.7) Verify with table of particular solutions considering $$ f(t) = t^2 \! $$

2.8) Solve with $$ f(t) = exp(-t^2) \! $$. Find coefficients $$ a_2, a_1, a_0 \! $$ such that (5.1) has the following characteristic equations:

2.8.1) $$ (r+1)(r-2) = 0 \! $$

2.8.2) $$ (r-4)^2 = 0 \! $$

Part 1
We can rearrange Eqn. 5.1 into the following form:

When in this form, it is apparent that the equation is exact since it can be represented as

where

The 2nd condition of exactness involves two relations which are as follows:

Computing the partial derivatives for (5.1.5) and (5.1.6),

$$ f_{tt} = f_{ty} = f_{yy} = f_{yp} = f_{y} = 0 \! $$

$$ g_{tp} = g_{yp} = g_{pp} = 0 \! $$

$$ g_{y} = a_0 \! $$

In order for (5.1.5) to be satisfied, $$ a_0 \! $$ must be zero.

$$ (5.1.3) \Rightarrow \phi(t,y,p) = h(t,y) + \int a_2 dp \! $$

$$ \phi_t = h_t \! $$

$$ \phi_y = h_y \! $$

Substituting these results into (5.1.4) yields

From (5.1.7) we can solve for $$ h \! $$ to arrive at

Part 2
Let $$ h(t) = exp(\alpha t), \alpha \in \Re \! $$.

Multiplying both sides of Eqn. 5.1 and integrating yields

Part 2.1
Assume that the LHS of (5.2.1) will be of the following form

We set $$ \bar{a}_2 = 0 $$ to reduce the order, differentiate (5.2.2) and set it equal to the integrand of (5.2.1) as follows

$$ \frac{d}{dt}\Big[exp(\alpha t)(\bar{a}_2y+\bar{a}_1y'+\bar{a}_0y) \Big] = exp(\alpha t)\Big(\bar{a}_1 y + (\alpha \bar{a}_1+\bar{a}_0)y' + \alpha \bar{a}_0 y \Big) = exp(\alpha t)\big[a_2y''+a_1y'+a_0y\big] \! $$

The remaining coefficients of (5.2.2) are thus given by

$$ \bar{a}_1 = a_2 \! $$

$$ \bar{a}_0 = a_1 - \alpha a_2 = a_0/\alpha \! $$

Part 2.2
With

$$ a_1 - \alpha a_2 = a_0/\alpha \! $$

we can rewrite it in quadratic form by multiplying both sides by $$ \alpha \! $$ to result in the following

Part 2.3
Returning now to (5.2.1), we substitute (5.2.2) as the result of the LHS integration yielding the following reduced order equation

Dividing through by $$ exp(\alpha t) \! $$ and then by $$ \bar{a}_1 $$ results in the following

Part 2.4
With the L1-ODE-CC found in (5.2.5), we choose the integrating factor as

$$ h(t) = exp \Big[ \int \frac{\bar{a}_0}{\bar{a}_1} dt \Big] = e^{\beta t} \! $$

And since we have (5.2.5) in the correct form (see (2) p. 10-3), we can solve for y using (6) on p. 10-3

Part 2.5
$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_0/\alpha}{a_2} \Rightarrow \alpha\beta = \frac{a_0}{a_2} $$

$$ \beta = \frac{\bar{a}_0}{\bar{a}_1} = \frac{a_1 - \alpha a_2}{a_2} = \frac{a_1}{a_2} - \alpha \Rightarrow \alpha + \beta = \frac{a_1}{a_2} $$

Part 2.6
Eqn. (5.2.6) is actually the particular solution of the L2-ODE-CC given. Since $$ \alpha $$ and $$ \beta $$ are the roots of the characteristic equation, the complementary solution (or homogeneous solution) is found to be

Because the general solution is the linear combination of the homogeneous solution and particular solution, it is as follows

Part 2.7
If we use the trial solution lookup table for a particular solution if $$ f(t) = t^2 $$, we come up with a trial solution of

where $$ C_2 $$,$$ C_1 $$ and $$ C_0 $$ are undetermined constant coefficients.

To compare this particular solution to the one found in (5.2.6), we substitute $$ f(t) = t^2 $$ into (5.2.6) and find that

The result shown in (5.2.10) is clearly the same form as the trial solution in (5.2.9).

Part 2.8
We will now solve (5.1) for the forcing function $$ f(t) = exp(-t^2) $$.

Part 2.8.1
To begin with, since the characteristic equation is $$ (r+1)(r-2) $$, we know that the homogeneous solution is

and we know that the coefficients in (5.1) are

$$a_2 = 1 \!$$

$$a_1 = -1 \!$$

$$a_0 = -2 \!$$

Our integrating factor (p. 29-3) becomes

Now we turn to (2) p. 30-1 to find the particular solution

Part 2.8.2
To begin with, since the characteristic equation is $$ (r-4)^2 $$, we know that the homogeneous solution is

and we know that the coefficients in (5.1) are

$$a_2 = 1 \!$$

$$a_1 = -8 \!$$

$$a_0 = 16 \!$$

Our integrating factor (p. 29-3) becomes

Now using (2) p. 30-1 again to find the particular solution

Author and Proof-Reader
[Author] Michael Steinberg

[Proof-Reader]

= Problem 6 - Show that lecture agree with the book = From Meeting 30, p 30-1.

Given
Particular Solution from meeting 30.

$$\displaystyle y_p(x) = u_1(x)\int{\frac{1}{h(x)}\left(\int{h(x)f(x)\,dx}\right)\,dx} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.1) $$

Particular solution from King p. 8 (1.6)

$$\displaystyle y_p(x) = \oint{f(s)\left[\frac{u_1(s)u_2(x)-u_1(x)u_2(s)}{W(s)}\right]\,ds} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.2) $$

Where $$\displaystyle W(x)$$ is the Wronskian defined as

$$\displaystyle W(x)=u_1(x)\,u_2(x)'-u_2(x)\,u_1(x)' $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.3) $$

Hint from meeting 30.

$$\displaystyle \frac{1}{h(x)}=\left( \frac{u_2(x)}{u_1(x)} \right)' $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.4) $$

Find
Show that 6.1 agrees with 6.2

Solve
Using quotient rule on equation 6.4 to solve for $$\displaystyle h(x)$$.

$$\displaystyle d\left(\frac{u}{v}\right)=\frac{v\,du-u\,dv}{v^2} $$

$$\displaystyle d\left(\frac{u_2(x)}{u_1(x)}\right)=\frac{u_1(x)\,u_2(x)'-u_2(x)\,u_1(x)'}{u_1(x)^2} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.5) $$

Inverse of equation 6.5 solves for $$\displaystyle h(x)$$.

$$\displaystyle h(x)=\frac{u_1(x)^2}{u_1(x)\,u_2(x)'-u_2(x)\,u_1(x)'} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.6) $$

Using integration by parts on equation 6.1

$$\displaystyle \int{u\,dv}=u\,v-\int{v\,du} $$

$$\displaystyle dv=\frac{1}{h(x)}=\left(\frac{u_2(x)}{u_1(x)}\right)' $$

$$\displaystyle u=\left(\int{h(x)f(x)\,dx}\right) $$

$$\displaystyle y_p(x) = u_1(x)\left\{\frac{u_2(x)}{u_1(x)} \left(\int{h(s)f(s)\,ds}\right)-\int{h(s)f(s)\frac{u_2(s)}{u_1(s)}\,ds}             \right\} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.7) $$

Subsitute for h(s) as defined by equation 6.6

$$\displaystyle y_p(x) = u_1(x)\left\{\frac{u_2(x)}{u_1(x)} \left(\int{\frac{u_1(s)^2}{u_1(s)\,u_2(s)'-u_2(s)\,u_1(s)'}f(s)\,ds}\right)-\int{\frac{u_1(s)^2}{u_1(s)\,u_2(s)'-u_2(s)\,u_1(s)'}f(s)\frac{u_2(s)}{u_1(s)}\,ds}             \right\} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.8) $$

Simplify equation 6.8

$$\displaystyle y_p(x) = \int{f(x)\frac{u_2(x)u_1(s)^2}{W(s)} \,ds}-\int{f(x)\frac{u_1(x)u_1(s)u_2(s)}{W(s)} \,ds}=\int{\frac{u_2(x)u_1(s)^2-u_1(x)u_1(s)u_2(s)}{W(s)}\,ds} $$ <p style="text-align:right;">$$\displaystyle (Eq. 6.9) $$

Equation 6.9 does not equal equation 6.2 from King. This suggest there is either misprint in King or course notes.

If equation 6.1 was $$\displaystyle  y_p(x) = u_1(x)\int{\frac{1}{h(x)}\left(\int{\frac{h(x)f(x)\,dx}{u_1(x)}}\right)\,dx}$$ and solved similarly as shown above would be equal to equation 6.2.

Author and Proof-Reader
[Author] Mike Faraone

[Proof-Reader]

= Problem 7 - Finding two homog. solutions = From Meeting 31, p 31-1.

Given

 * {| style="width:100%" border="0"

$$\left( x+1 \right)y''-\left( 2x+3 \right)y'+2y=0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.1)$$
 * }

Trial solution :
 * {| style="width:100%" border="0"

$$\displaystyle y= {e}^{rx} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.2)$$
 * }

Find
$$ u_1\ $$ and $$ u_2\ $$ of (7.1) using trial solution (7.2)

Solve
(7.2) ->
 * {| style="width:100%" border="0"

$$\displaystyle y'=r{{e}^{rx}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.3)$$
 * }


 * {| style="width:100%" border="0"

$$\displaystyle y''={{r}^{2}}{{e}^{rx}}$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.4)$$
 * }

(7.1) substitute with (7.3) and (7.4)
 * {| style="width:100%" border="0"

$$\displaystyle (x+1)r^2{e}^{rx} - (2x+3)r{e}^{rx} + 2{e}^{rx} = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.5)$$
 * }

->
 * {| style="width:100%" border="0"

$$\displaystyle {e}^{rx} ((x+1)r^2 - (2x+3)r + 2) = 0$$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.6)$$
 * }

Using basic knowledge of Quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle r= \frac{\left(2x+3 \right) \pm \sqrt{{\left(2x+3 \right)}^{2}-8\left(x+1 \right)}} {2 \left( x+1\right)}= \frac{\left(2x+3 \right) \pm \left(2x+1 \right) } {2 \left( x+1\right)} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.7)$$
 * }

We can get two solutions about r


 * {| style="width:100%" border="0"

$$\displaystyle {r}_{1} = 2 \, \  {r}_{2}=  \frac{1}{ \left( x+1\right)} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.8)$$
 * }

We choose $$r_1 $$ because it is constant
 * {| style="width:100%" border="0"

$$\displaystyle {u}_{1}(x) = {e}^{2x} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.9)$$
 * }

Using (1) p.30-1


 * {| style="width:100%" border="0"

$$\displaystyle {u}_{2}(x)= \underbrace{{u}_{1}(x)}_{{e}^{2x}} \int \frac{1}{ u_1^2(x)}exp[-\int \underbrace{a_1(x)}_{\frac{-(2x+3)}{x+1}} \ dx] \ dx $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \ = {e}^{2x} \int \frac{1}{ {e}^{4x}}exp[\underbrace{\int \frac{2x+3}{x+1} \ dx}_{2x+log(x+1)}] \ dx $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }


 * {| style="width:100%" border="0"

$$\displaystyle \ = {e}^{2x} \int \frac{x+1}{{e}^{2x}} \ dx $$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Using wolfram mathematica


 * {| style="width:100%" border="0"

$$\displaystyle {u}_{2}(x) = - \frac{1}{4} (2x+3) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\displaystyle (7.10)$$
 * }

Therefore
 * {| style="width:100%" border="0" align="left"

$$ u_1 = {e}^{2x}, \ u_2 = - \frac{1}{4} (2x+3) $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * }
 * }
 * }

Author and Proof-Reader
[Author] Oh, Sang Min

[Proof-Reader] Mike Faraone

= Problem 8 - Finding L2-ODE-VC(homog.) = From Meeting 31, p 31-3.

Given
Trial solution


 * {| style="width:100%" border="0" align="left"

$$\displaystyle y = \frac{{e}^{rx}}{x^2} $$ <p style="text-align:right;">$$\displaystyle (8-1) $$
 * }
 * }

Character equation


 * {| style="width:100%" border="0" align="left"

$$\displaystyle r^2 + 3 = 0 $$ <p style="text-align:right;">$$\displaystyle (8-2) $$
 * }
 * }

Find
L2-ODE-VC

Solve

 * {| style="width:100%" border="0" align="left"

$$\displaystyle y = \frac{{e}^{rx}}{x^2} $$ <p style="text-align:right;">$$\displaystyle (8-3) $$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$\displaystyle y' = \frac{r{e}^{rx}}{x^2} - \frac{2{e}^{rx}}{x^3} = \frac{{e}^{rx}}{x^2} (r - \frac{2}{x}) $$ <p style="text-align:right;">$$\displaystyle (8-4) $$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$\displaystyle y'' = \frac{{e}^{rx}}{x^2} {(r - \frac{2}{x})(r - \frac{2}{x}) + \frac{2}{x^2}} = \frac{{e}^{rx}}{x^2} (r^2 - \frac{4}{x}r + \frac{6}{x^2}) $$ <p style="text-align:right;">$$\displaystyle (8-5) $$
 * }
 * }

 Using Reverse Eng.


 * {| style="width:100%" border="0" align="left"

$$\displaystyle \frac{{e}^{rx}}{x^2} [a_2(r^2 - \frac{4}{x}r + \frac{6}{x^2}) + a_1(r - \frac{2}{x}) + a_0] = \frac{{e}^{rx}}{x^2} (r^2 + 3) = 0 $$ <p style="text-align:right;">$$\displaystyle (8-6) $$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$\displaystyle a_2 r^2 + (a_1 - \frac{4a_2}{x})r + (\frac{6a_2}{x^2} - \frac{2a_1}{x} + a_0) = r^2 + 3 $$ <p style="text-align:right;">$$\displaystyle (8-7) $$
 * }
 * }

Therefore


 * {| style="width:100%" border="0" align="left"

$$\displaystyle a_2 = 1, \ a_1 = \frac{4}{x}, \ a_0 = 3 + \frac{2}{x^2} $$ <p style="text-align:right;">$$\displaystyle (8-8) $$
 * }
 * }


 * {| style="width:100%" border="0" align="left"

$$ y'' + \frac{4}{x} y' + (3 + \frac{2}{x^2})y = 0 $$
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * style="width:92%; padding:10px; border:2px solid #8888aa" align="center"|
 * }
 * }
 * }

Author and Proof-Reader
[Author] Oh, Sang min

[Proof-Reader] Mike Faraone

= References =

= Contributing members =

Jo : Authored Problem 1 and Problem 2.

Rob Carroll : Authored Problem 3

Oh, Sang Min : : Authored Problem 7 and Problem 8, proof-read problem 1, 2 and 3

Michael Faraone : Authored Problem 6, Proofed problems 7 and 8.

Michael Steinberg : Authored Problem 5