User talk:Egm6321.f10.team5.oh/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] Soo W. Jo

[Proof-Reader]

= 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] Soo W. Jo

[Proof-Reader]

= 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]

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

Author and Proof-Reader
[Author]

[Proof-Reader]

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

Author and Proof-Reader
[Author]

[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

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

Author and Proof-Reader
[Author]

[Proof-Reader]

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

Given

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$$\left( x+1 \right)y''-\left( 2x+3 \right)y'+2y=0$$
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 * <p style="text-align:right">$$\displaystyle (7.1)$$
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Trial solution :
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$$\displaystyle y= {e}^{rx} $$
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 * <p style="text-align:right">$$\displaystyle (7.2)$$
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Find
$$ u_1\ $$ and $$ u_2\ $$ of (7.1) using trial solution (7.2)

Solve
(7.2) ->
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$$\displaystyle y'=r{{e}^{rx}}$$
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 * <p style="text-align:right">$$\displaystyle (7.3)$$
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$$\displaystyle y''={{r}^{2}}{{e}^{rx}}$$
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 * <p style="text-align:right">$$\displaystyle (7.4)$$
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(7.1) substitute with (7.3) and (7.4)
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$$\displaystyle (x+1)r^2{e}^{rx} - (2x+3)r{e}^{rx} + 2{e}^{rx} = 0$$
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 * <p style="text-align:right">$$\displaystyle (7.5)$$
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->
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$$\displaystyle {e}^{rx} ((x+1)r^2 - (2x+3)r + 2) = 0$$
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 * <p style="text-align:right">$$\displaystyle (7.6)$$
 * }

Using basic knowledge of Quadratic equation
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$$\displaystyle r= \frac{\left(2x+3 \right) \pm \sqrt{{\left(2x+3 \right)}^{2}-8\left(x+1 \right)}} {2 \left( x+1\right)}= \frac{\left(2x+3 \right) \pm \left(2x+1 \right) } {2 \left( x+1\right)} $$
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 * <p style="text-align:right">$$\displaystyle (7.7)$$
 * }

We can get two solutions about r


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

We choose $$r_1 $$ because it is constant
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$$\displaystyle {u}_{1}(x) = {e}^{2x} $$
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 * <p style="text-align:right">$$\displaystyle (7.9)$$
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Using (1) p.30-1


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$$\displaystyle {u}_{2}(x)= \underbrace{{u}_{1}(x)}_{{e}^{2x}} \int \frac{1}{ u_1^2(x)}exp[-\int \underbrace{a_1(x)}_{\frac{-(2x+3)}{x+1}} \ dx] \ dx $$
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$$\displaystyle \ = {e}^{2x} \int \frac{1}{ {e}^{4x}}exp[\underbrace{\int \frac{2x+3}{x+1} \ dx}_{2x+log(x+1)}] \ dx $$
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$$\displaystyle \ = {e}^{2x} \int \frac{x+1}{{e}^{2x}} \ dx $$
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Using wolfram mathematica


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$$\displaystyle {u}_{2}(x) = - \frac{1}{4} (2x+3) $$
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 * <p style="text-align:right">$$\displaystyle (7.10)$$
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Therefore
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$$ u_1 = {e}^{2x}, \ u_2 = - \frac{1}{4} (2x+3) $$
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Author and Proof-Reader
[Author] Oh, Sang Min

[Proof-Reader]

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

Given
Trial solution


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

Character equation


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$$\displaystyle r^2 + 3 = 0 $$ <p style="text-align:right;">$$\displaystyle (8-2) $$
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Find
L2-ODE-VC

Solve

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


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


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$$\displaystyle \frac{{e}^{rx}}{x^2} [a_2(r^2 - \frac{4}{x}r + \frac{6}{x^2}) + a_1(r - \frac{2}{x}) + a_0] = \frac{{e}^{rx}}{x^2} (r^2 + 3) = 0 $$ <p style="text-align:right;">$$\displaystyle (8-6) $$
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$$\displaystyle a_2 r^2 + (a_1 - \frac{4a_2}{x})r + (\frac{6a_2}{x^2} - \frac{2a_1}{x} + a_0) = r^2 + 3 $$ <p style="text-align:right;">$$\displaystyle (8-7) $$
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Therefore


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$$\displaystyle a_2 = 1, \ a_1 = \frac{4}{x}, \ a_0 = 3 + \frac{2}{x^2} $$ <p style="text-align:right;">$$\displaystyle (8-8) $$
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$$ y'' + \frac{4}{x} y' + (3 + \frac{2}{x^2})y = 0 $$
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Author and Proof-Reader
[Author] Oh, Sang min

[Proof-Reader]

= References =

= Contributing members =

Soo W. Jo : Authored Problem 1 and Problem 2.

Rob Carroll : Authored Problem 3