Question

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

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

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

$$\begin{gathered} m^{\text{2}}{e}^{\text{mx}}-10m{e}^{\text{mx}}+25e^{\text{mx}}=0 \\ {e}^{\text{mt}}\left(m^{\text{2}}-10m+25\right)=0 \end{gathered}$$

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

Then the roots are

$m_{1,2}=5$

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 .

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

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

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

$\begin{array}{rcl}y& =& n{e}^{5x}\left(k+sx\right)\\ & =& c_1{e}^{5x}+c_{2}x{e}^{5x}\end{array}$

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

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

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

by solving the two equations

$$\begin{gathered} 0=c_1{e}^5+c_2{e}^5 \\ {c}_2=-c_1 \\ {c}_1=1\text{and}{c}_2=-1 \end{gathered}$$

Substitute with the values of $c_1\text{and}{c}_2$into the general solution of our given differential equation.

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

The solution of the differential equation is$y=e^{5x}-x{e}^{5e}$

Hence, the general solution is data-custom-editor="chemistry" $y=e^{5x}-x{e}^{5e}$.