Question

In Problems 1–26 solve the given differential equation by undetermined coefficients.

$\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{-}\mathbf{10}\mathit{y}\mathbf{\text{'}}\mathbf{+}\mathbf{25}\mathit{y}\mathbf{=}\mathbf{30}\mathit{x}\mathbf{+}\mathbf{3}$

Short answer

$y=c_1 e^{5x}+c_2 x{e}^{5x}+\frac{6}{5}x+\frac{3}{5}$

Step-by-step solution

Solving equation

We have the non-homogeneous differential equation

$y\text{'}\text{'}-10y\text{'}+25y=30x+3$

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

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

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

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

$\begin{array}{l}{m}^2-10m+25=0\\ {\left(m-5\right)}^2=0\end{array}$

Then the roots are,

$m_{1,2}=5$

which are real and repeated.

Since we have repeated roots $m_{1,2}=5$, then we need to have two linearly independent solutions, then we obtain the homogeneous solution as

$\begin{array}{c}{y}_h=n e^{5x} \left(k+l x\right)\\ =c_1 e^{5x}+c_2 x{e}^{5x}\end{array}$

where we have $c_1=nk\mathrm{and}{c}_2=nl$

Solution of the non-homogeneous differential equation

Second, we have to find the particular solution of the non-homogeneous differential equation$y\text{'}\text{'}-10y\text{'}+25y=30x+3$as the following technique:

Assume that$y_p=Ax+B$is a solution for the non-homogeneous differential equation.

After that, differentiate the assumption with respect to x, and substitute $y{\text{'}}_p=A,y\text{'}{\text{'}}_p=0$into$y\text{'}\text{'}-10y\text{'}+25y=30x+3$to obtain the auxiliary equation.

$\begin{array}{c}0-10A+25A x+25B=30x+3\\ 25A x+\left(25B-10A\right)=30x+3\\ A=\frac{6}{5} B=\frac{3}{5}\end{array}$

Then the particular solution becomes$y_p=\frac{6}{5}x+\frac{3}{5}$

Then we obtain

$\begin{array}{c}y=y_h+y_p\\ =c_1 e^{5x}+c_2 x{e}^{5x}+\frac{6}{5}x+\frac{3}{5}\end{array}$

Hence the final solution is

$y=c_1 e^{5x}+c_2 x{e}^{5x}+\frac{6}{5}x+\frac{3}{5}$