Question

Solve the given boundary-value problem.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{2}\mathit{x}\mathbf{-}\mathbf{2}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{5}\mathbf{,}\mathit{y}\mathbf{\left(}\mathit{\pi }\mathbf{\right)}\mathbf{=}\mathit{\pi }$

Short answer

$y=k_2\sin x{e}^x+x$

Step-by-step solution

Obtain the general solution of the homogeneous differential equation  

First, we need to solve the associated homogeneous equation $y\text{'}\text{'}-2y\text{'}+2y=0$

Corresponding auxiliary equation is $m^2-2m+2=0$

Solving,

$m_1=1+i$, $m_2=1-i$, which are complex and conjugate.

So the solution of corresponding homogeneous equation is

$y_h=c_1{e}^{(1+i)x}+c_2{e}^{(1-i)x}=e^x(k_1\cos x+k_2\sin x)$$y_h=c_1{e}^{(1+i)x}+c_2{e}^{(1-i)x}=e^x(k_1\cos x+k_2\sin x)$

Finding the particular solution

Given non homogeneous differential equation is$y\text{'}\text{'}-2y\text{'}+2y=2x-2$

Assume that $y_p=Ax+B$ is a solution of the non-homogeneous differential equation

Differentiate with respect to x

$y_p\text{'}=A$

$y_p\text{'}\text{'}=0$

Substituting in given differential equation

$0-2A+2Ax+2B=2x-2$

comparing the coefficients,

$$\begin{gathered} 2A=2\Rightarrow A=1 \\ 2B-2A=-2\Rightarrow B=0 \\ 2A+C=1\Rightarrow C=-1 \end{gathered}$$

Hence the particular solution is

$y_p=x$

Finding the solution

Hence the solution of given differential equation is

y = y_h + y_p

$y=K_1\cos x+K_2\sin x)e^x+x$

Applying the given conditions

At the point (0,0)

(4) becomes$0=[K_1\cos 0+K_2\sin 0]e^0+\left(0\right)$

Or $K_1=0$

At the point$(\pi ,\pi )$

$\pi =[K_1\cos \pi +K_2\sin \pi ]e^{\pi }+\left(\pi \right)$$\pi =[K_1\cos \pi +K_2\sin \pi ]e^{\pi }+\left(\pi \right)$

Since , we cant find value of k₂

Hence the solution is localid="1664170514310" $\mathit{y}\mathbf{=}{\mathit{k}}_{\mathbf{2}}\mathit{s}\mathit{i}\mathit{n}\mathbf{}\mathit{x}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathit{x}$