User:Egm6321.f10.team3.cook/HW5

=Problem 7: Find $$ {u}_{1},  {u}_{2} $$ using trial solution and variation of parameters=

Given
Give the reverse engineered equation on [|Mtg 31-1] of the lecture notes


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$$ \left(x+1 \right) {y}^{''} - \left(2x+3 \right) {y}^{'} + 2y=0 $$ $$
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 * $$\displaystyle (Eq. 7.1)
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Find
Find $$ {u}_{1},  {u}_{2} $$ 2 Homogeneous solutions of $$\displaystyle (Eq 7.1) $$ using the trial solution:


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y= {e}^{rx} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.2)
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Solution
To solve we first find the roots to the characteristic equation by plugging $$\displaystyle (Eq 7.2) $$ into $$\displaystyle (Eq 7.1) $$ to obtain


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\left(x+1 \right) {r}^{2} - \left(2x+3 \right) r + 2=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.3)
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Find the roots by using the quadratic formula
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r= \frac{\left(2x+3 \right) \pm \sqrt{{\left(2x+3 \right)}^{2}-8\left(x+1 \right)}} {2 \left( x+1\right)}= \frac{\left(2x+3 \right) \pm \left(2x+1 \right) } {2 \left( x+1\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.4)
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Solving we get the two roots to the characteristic equation:


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{r}_{1} = 2 ; {r}_{2}=  \frac{1}{ \left( x+1\right)} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.5)
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Now we use the trial solution $$\displaystyle y= {e}^{rx} $$ and the constant root to obtain:


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{u}_{1} = exp(2x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.6)
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To find $$\displaystyle {u}_{2}(x) $$ we will use the variation of parameters method.


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y(x)=U(x) {u}_{1}(x) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.7)
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Plugging $$\displaystyle (Eq. 7.7)$$ into $$\displaystyle (Eq. 7.7)$$ gives us


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{U}^{'}(x) \left[ -\left(2x+3 \right) {u}_{1}(x)+ \left( x+1\right) {u}^{'}_{1}(x) \right] + \left( x+1\right) {u}_{1}(x) {U}^{''}(x)=0 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.8)
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Now use the reduction of order method 0 to find the expression for $$\displaystyle {u}_{2}$$. Using equation 1 from [|Mtg 31-1] we can now solve for $$\displaystyle {u}_{2}(x)$$


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{u}_{2}(x)= {u}_{1}(x) \int \frac{1}{ {u}^{2}_{1}}exp \left[ - \int {a}_{1}dx\right]dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.9)
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where $$\displaystyle {a}_{1}:= \frac{2x+3}{x+1} $$. Solving the integral for $$\displaystyle {a}_{1}$$ first we have
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\int {a}_{1}dx= \int \frac{2x+3}{x+1} dx = \left[2x+ log(x+1)\right] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.10)
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Now plug $$\displaystyle (Eq. 7.10)$$ into $$\displaystyle (Eq. 7.9) $$ to obtain


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{u}_{2}(x)= {u}_{1}(x) \int \frac{x+1}{ {u}^{2}_{1}}exp \left[ -2x\right]dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.11)
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Substituting in our expression for $$\displaystyle {u}_{1}(x)=exp\left(2x\right)$$ we have


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{u}_{2}(x)= exp \left(2x \right) \int \left(x+1\right) {exp \left(-2x \right)}^{2}exp \left[ -2x\right]dx= exp \left(2x \right) \int \left(x+1\right) exp \left[ -6x\right]dx $$ $$ Solving the above equation by parts yields the solution for $$\displaystyle {u}_{2}(x)$$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.11)
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{u}_{2}(x)= exp \left(2x \right) \left[ - \frac{1}{6}x exp \left(-6x \right)- \frac{1}{36}exp \left(-6x \right)-\frac{1}{6}exp \left(-6x \right)\right]= - \frac{1}{36} \left( 6x+7\right)exp \left( -4x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 7.12)
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The two homogeneous solutions are:


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{u}_{1} = exp(2x); {u}_{2}(x)= - \frac{1}{36} \left( 6x+7\right)exp \left( -4x\right) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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