Question

In Problems 29-36 solve the given initial-value problem.

$$\begin{gathered} \mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{2}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\left(0\right)\mathbf{=}\mathbf{5}\mathbf{,}\mathit{y}\mathbf{\text{'}}\left(0\right)\mathbf{=}\mathbf{10} \end{gathered}$$

Short answer

$y=5e^x+5x{e}^x$

Step-by-step solution

Step 1: Determine the initial value of differential equation

The second order differential equation:

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

Initial conditions:

$y\left(0\right)=5,y\text{'}\left(0\right)=10$

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

Substitute $y=e^{mx},y\text{'}=m{e}^{mx}$and $y\text{'}\text{'}=m^2{e}^{mx}$into $y\text{'}\text{'}-2y\text{'}+y=0$to obtain the auxiliary equation.

$$\begin{gathered} m^2{e}^{mx}-2m{e}^{mx}+e^{mx}=0 \\ {e}^{mt}\left(m^2-2m+1\right)=0 \end{gathered}$$

For any real$x,e^{mx}\ne 0$, we have

$$\begin{gathered} m^2-2m+1=0 \\ {\left(m-1\right)}^2=0 \end{gathered}$$

Roots are:

$m_{1,2}=1$

Find the additional linearly independent solution

Real and repeated roots $m_{1,2}=1$.

The we need to have an additional linearly independent solution for this repeated root, can obtain the general solution of our differential equation

$$\begin{gathered} y=n{e}^{\text{x}}\left(k+sx\right) \\ y=c_1{e}^{\text{x}}+c_{2}x{e}^{\text{x}}······\left(1\right) \end{gathered}$$

$c_1=n\times k,c_2=n\times s$arbitrary constants.

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$\left(x,y\right)=\left(0,5\right)$ in the general solution (1) as

$$\begin{gathered} 5=c_1{e}^{\text{0}}+c_2\left(0\right) \\ {c}_1=5 \end{gathered}$$

After that, we have to take the first derivative for the general solution (1) of the given differential equation as.

$y\text{'}=c_1{e}^{\text{x}}+c_2{e}^{\text{x}}+c_{2}x{e}^{\text{x}}$

Apply with the point$\left(x,y\text{'}\right)=\left(0,10\right)$

$$\begin{gathered} 10=c_1{e}^0+c_2{e}^0+c_2\left(0\right) \\ {c}_1+c_2=10 \end{gathered}$$

Then by solving the two equations

$c_1=5 \text{and} c_2=5$

Finally, substitute with the values of $c_1\text{and}{c}_2$into the general solution (1) of our given differential equation, then we have

$y=5e^x+5x{e}^x$

Hence, the general solution is$\mathit{y}\mathbf{=}\mathbf{5}{\mathit{e}}^{\mathbf{x}}\mathbf{+}\mathbf{5}\mathit{x}{\mathit{e}}^{\mathbf{x}}$