Question

In Problems 39–44 solve the given differential equation subject to the indicated conditions.

$\mathit{y}\mathbf{"}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{(}\mathbf{\pi }\mathbf{/}\mathbf{2}\mathbf{)}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{-}\mathbf{1}$

Short answer

The solution for the given differential equation at the given conditions is,

$y=e^{\left(x-\mathrm{\pi }\right)}\cos x$

Step-by-step solution

Given

We have given the second order homogeneous differential equation,

$y"+2y\text{'}+y=0$with the conditions$y\left(\frac{\mathrm{\pi }}{2}\right)=0$and $y\left(\mathrm{\pi }\right)=-1$.

Now, we will solve it by assuming that$y=e^{mx}$is a solution of the given D.E.

Differentiation and then substitution

When we differentiate our equation of assumption with respect to x, then we get:

$$\begin{gathered} y\text{'}=m{e}^{mx}\left(1\right) \\ y"=m^2{e}^{mx}\left(2\right) \end{gathered}$$

After that, substitute with y’ and y’’ shown in equations (1) and (2) and our assumption $y=e^{mx}$into given equation, then we have :

$$\begin{gathered} m^2{e}^{mx}-2m{e}^{mx}+2e^{mx}=0 \\ \left(m^2-2m+2\right)e^{mx}=0 \end{gathered}$$

As $e^{mx}$can not be 0 , then we have the auxiliary equation as,$m^2-2m+2=0$.

Then the roots of auxiliary equation :

$$\begin{gathered} m_{1,2}=\frac{2\pm \sqrt{4-4\times 1\times 2}}{2} \\ =\frac{2\pm \sqrt{-4}}{2} \\ =1\pm i \end{gathered}$$

Applying the given conditions

So, the solution of the given differential equation $y"-2y\text{'}+2y=0$as,

$$\begin{gathered} y=c_1{e}^{\left(1+i\right)x}+c_2{e}^{\left(1+i\right)x} \\ =c_1{e}^x{e}^{ix}+c_2{e}^x{e}^{-ix} \\ =c_1{e}^x\left(\cos x+\sin x\right)+c_2{e}^x\left(\cos x-\sin x\right) \\ =c_1+c_2{e}^x\cos x+\left(c_1-c_2\right)e^x\sin x \\ =k_1{e}^x\cos x+k_2{e}^x\sin x\left(3\right) \end{gathered}$$

Here, $k_1=c_1+c_2$and $k_2=c_1-c_2$are an arbitrary constants.

Now, we have to apply the given conditions into the solution of the given differential equation to find the constants $k_1$and$k_2$:

Apply the point $\left(x,y\right)=\left(\frac{\mathrm{\pi }}{2},0\right)$in the equation (3), then :

$$\begin{gathered} 0=k_1{e}^{\frac{\mathrm{\pi }}{2}}\cos \frac{\mathrm{\pi }}{2}+k_2{e}^{\frac{\mathrm{\pi }}{2}}\sin \frac{\mathrm{\pi }}{2} \\ {k}_2{e}^{\frac{\mathrm{\pi }}{2}}\left(1\right)=0 \\ {k}_2=0 \end{gathered}$$

Apply the point $\left(x,y\right)=\left(\mathrm{\pi },-1\right)$ in the equation (3), then :

$$\begin{gathered} -1=k_1{e}^{\mathrm{\pi }}\mathrm{cos\pi }+k_2{e}^{\mathrm{\pi }}\sin \mathrm{\pi } \\ {\mathrm{k}}_1{\mathrm{e}}^{\mathrm{\pi }}\left(-1\right)=-1 \\ {\mathrm{k}}_1{\mathrm{e}}^{\mathrm{\pi }}=1 \\ {\mathrm{k}}_1={\mathrm{e}}^{-\mathrm{\pi }} \end{gathered}$$

Substitution

Lastly, we substitute the value of$k_1$and$k_2$in equation (3), then we obtain,

$$\begin{gathered} y=e^{-\mathrm{\pi }}{e}^x\cos x+0 \\ y=e^{\left(x-\mathrm{\pi }\right)}\cos x \end{gathered}$$

Hence,the solution for the given differential equation at the given conditions is,

$y=e^{\left(x-\mathrm{\pi }\right)}\cos x$