Question

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

$$\begin{gathered} \mathit{y}\mathbf{\text{'}\text{'}}\mathbf{-}\mathbf{}\mathbf{10}\mathit{y}\mathbf{+}\mathbf{25}\mathit{y}\mathbf{=}\mathbf{0} \\ \mathit{y}\mathbf{\left(}\mathbf{0}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{,}\mathbf{}\mathit{y}\mathbf{}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{0} \end{gathered}$$

Short answer

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

Step-by-step solution

Determine the initial value of differential equation

Given,

$y\text{'}\text{'}-10y\text{'}+25y=0$

Initial conditions:

$y\left(0\right)=1,y\left(1\right)=0$

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

Substitute localid="1663841225323" $y=e^{mx},y^{\text{'}}=m{e}^{mx}$and $y^{\text{'}\text{'}}=m^2{e}^{mx}$into localid="1663841244694" $y\text{'}\text{'}-10y\text{'}+25y=0$to obtain the auxiliary equation.

localid="1663841211202" $$\begin{gathered} m^2{e}^{mx}-10m{e}^{mx}+25e^{mx}=0 \\ {e}^{mt}(m^2-10m+25)=0 \end{gathered}$$

For any real x, localid="1663841259220" $e^{mx}\ne 0,$we have

Then the roots are

localid="1663841289425" $m_{12}=5$

Find the additional linearly independent solution

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

$$\begin{gathered} y^{\text{'}\text{'}}-10y\text{'}+25y=0 \\ y=n{e}^x(k+sx) \\ y=c_1{e}^x+c_{2}x{e}^x----1 \end{gathered}$$

Real and repeated

Finally, since we have repeated root , $m_{1,2}=5$then we need to have an additional linearly independent solution for this repeated root, then we obtain the general solution of our differential equation

$$\begin{gathered} y=n{e}^{5x}(k+sx) \\ =c_1{e}^{5x}+c_{2}x{e}^{5x} \end{gathered}$$

Arbitary constant:${c}_1=n\times k,c_2=n\times s$

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 the point $(x,y)=(0,1)$in the general solution as

$$\begin{gathered} 1=c_1{e}^{°}+c_2\left(0\right) \\ {c}_1=1 \end{gathered}$$

After that, we have to apply with oint $(x,y)=(1,0)$in the general solution as

by solving the two equations $\begin{array}{l}0=c_1{e}^5+c_2{e}^5\\ c_2=-c_1\\ c_1=1\mathrm{an}d{c}_2=-1\end{array}$

Substitute with the values of $c_1$and $c_2$into the general solution of our given differential equation.

role="math" localid="1663840693253" $y=e^{5x}-x{e}^{5e}$

The solution of the differential equation is role="math" localid="1663840619770" $y=e^{5x}-x{e}^{5e}$

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