Question

In Problems 37-40 solve the given initial-value problem.

$$\begin{gathered} {\mathit{y}}^{\mathbf{\text{'}\text{'}}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{2}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{\left(}\mathbf{\pi }\mathbf{\right)}\mathbf{=}\mathbf{1} \end{gathered}$$

Short answer

NO solution

Step-by-step solution

Determine the initial value of differential equation

The second order differential equation:

$y^{\text{'}\text{'}}-2y^{\text{'}}+2y=0$

initial conditions

$y\left(0\right)=1,y\left(\mathrm{\pi }\right)=1$

Consider, $y=e^{mx}$as the solution of the differential equation,

substitute, role="math" localid="1663907745427" $y=e^{mx},y^{\text{'}}=m{e}^{mx},y^{\text{'}\text{'}}=m^2{e}^{mx}$, $y^{\text{'}\text{'}}-2y^{\text{'}}+2y=0$to obtain the auxiliary equation.

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

For any real x, role="math" localid="1663908078867" $e^{mx}\ne 0$, we have

$m^2-2m+2=0$

Roots are

Conjugate and complex

Find the  general solution of the homogeneous equation

Homogeneous Linear Second Order Equation is a linear combination of $y_{1}and{y}_2$are real and distinct.

General solution of this homogeneous equation:

$y\text{'}\text{'}-2y^{\text{'}}+2y=0$

Given,

$$\begin{gathered} y=c_1{e}^{m1x}+c_2{e}^{m_{2}x} \\ y=c_1{e}^{(1+i)}+c_2{e}^{(1-i)} \\ y=c_1{e}^x(\cos x+\sin x)+c_2{e}^x(\cos x-\sin x) \\ y=k_1{e}^x\cos x+k_2{e}^x\sin x----1 \end{gathered}$$

Arbitrary constants:

localid="1663911551029" $k_1=c_1+c_2,k_2=c_1-c_2$

Fourth, now we have to apply the given conditions in the general solution of our given differential equations as the following technique: We have to apply with the point in the general solution as

Apply with the point $(x,y)=(0,1)$the general solution (3) as

$$\begin{gathered} 1=k_1{e}^{\theta }\cos \left(0\right)+k_2{e}^{\theta }\sin \left(0\right)1 \\ =k_1+0 \\ {k}_1=1 \end{gathered}$$

Apply with the point $(x,y)=(\mathrm{\pi },1)$the general solution (3) as

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

Two values of $k_1$for this condition, we have no solution second order differential equation.

No solution for D.E at these conditions.

Hence, Nosolution for D.E at these conditions.