Question

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

$\mathit{y}\mathbf{+}\mathbf{3}\mathit{y}\mathbf{=}\mathbf{-}\mathbf{48}{\mathit{x}}^{\mathbf{2}}{\mathit{e}}^{\mathbf{3}\mathbf{x}}$

Short answer

$y=c_1\cos \left(\sqrt{3}x\right)+c_2\sin \left(\sqrt{3}x\right)-4x^2{e}^{3x}+4x{e}^{3x}-\frac{4}{3}{e}^{3x}$

Step-by-step solution

General solution for homogeneous differential equation

We have the differential equation

$y+3y=-48x^2{e}^{3x}$

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 $y+3y=0$ to obtain the auxiliary equation.

$\begin{array}{c}{m}^2{e}^{mx}+3e^{mx}=0\\ e^{mx}\left(m^2+3\right)=0\end{array}$

Since $e^{mx}$ cannot be equal to 0 , then we have

$\begin{array}{c}{m}^2+3=0\\ m^2=-3\end{array}$

Then the roots are

${\lambda }_1=\sqrt{3}i , {\lambda }_2=-\sqrt{3}i$

which are conjugate complex.

Then the solution of the corresponding homogeneous equation $y+3y=0$ can be given by

$y_h=c_1\cos \left(\sqrt{3}x\right)+c_2\sin \left(\sqrt{3}x\right)$

General solution for non-homogeneous differential equation

Second, we have to find the particular solution of the non-homogeneous differential

equation$y+3y=-48x^2{e}^{3x}$as the following technique :

Assume that$y_p=A{x}^2{e}^{3x}+Bx{e}^{3x}+C{e}^{3x}$is a solution for the non-homogeneous differential equation where $g\left(x\right)=-48x^2{e}^{3x}$.

After that, differentiate the assumption with respect to x , then we obtain '

$\begin{array}{c}{y}_p\text{'}=2Ax{e}^{3x}+3A{x}^2{e}^{3x}+B{e}^{3x}+3Bx{e}^{3x}+3C{e}^{3x}\\ =3A{x}^2{e}^{3x}+\left(2A+3B\right)x{e}^{3x}+\left(B+3C\right)e^{3x}\end{array}$

Differentiate another time with respect to x , then we obtain

$\begin{array}{l}{y}_p\text{'}\text{'}=6Ax{e}^{3x}+9A{x}^2{e}^{3x}+\left(2A+3B\right)e^{3x}+3\left(2A+3B\right)x{e}^{3x}+3\left(B+3C\right)e^{3x}\\ y_p\text{'}\text{'}=9A{x}^2{e}^{3x}+\left(12A+9B\right)x{e}^{3x}+\left(2A+6B+9C\right)e^{3x}\end{array}$

Substitute$y_p=A{x}^2{e}^{3x}+Bx{e}^{3x}+C{e}^{3x}$ with $y_p\text{'}\text{'}$ and into the non-homogeneous differential equation, then we obtain

$\begin{array}{c}(9A{x}^2{e}^{3x}+\left(12A+9B\right)x{e}^{3x}+\left(2A+6B+9C\right)e^{3x})+3(A{x}^2{e}^{3x}+Bx{e}^{3x}+C{e}^{3x})=-48x^2{e}^{3x}\\ 12A{x}^2{e}^{3x}+\left(12A+12B\right)x{e}^{3x}+\left(2A+6B+12C\right)e^{3x})=-48x^2{e}^{3x}\end{array}$

Then by comparison between the right and left sides, we can have

$\begin{array}{c}A=-4\\ 12A+12B=0 \Rightarrow B=4\\ 2A+6B+12C=0 \Rightarrow C=-\frac{4}{3}\end{array}$

Then the particular solution becomes

$y_p=-4x^2{e}^{3x}+4x{e}^{3x}-\frac{4}{3}{e}^{3x}$

Then we obtain

$\begin{array}{l}y=y_h+y_p\\ =c_1\cos \left(\sqrt{3}x\right)+c_2\sin \left(\sqrt{3}x\right)-4x^2{e}^{3x}+4x{e}^{3x}-\frac{4}{3}{e}^{3x}\end{array}$

is the general solution of the non-homogeneous differential equation.

Therefore,the generalsolution of the given differential equation by undetermined coefficients is found to be.

$y=c_1\cos \left(\sqrt{3}x\right)+c_2\sin \left(\sqrt{3}x\right)-4x^2{e}^{3x}+4x{e}^{3x}-\frac{4}{3}{e}^{3x}$