Question

Find the general solution of the following differential equations (complementary function + particular solution). Find the particular solution by inspection or by $\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{18}\mathbf{)}\mathbf{,}\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{23}\mathbf{)}\mathbf{,}$or. $\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{24}\mathbf{)}\mathbf{,}$Alsofind a computer solution and reconcile differences if necessary, noticing especially whether the particular solution is in simplest form [see$\mathbf{(}\mathbf{6}\mathbf{.}\mathbf{26}\mathbf{)}\mathbf{,}$and the discussionafter$\mathbf{(}\mathbf{6}\mathbf{.15}\mathbf{)}$].

${\mathit{y}}^{\mathbf{\text{'}}\mathbf{\text{'}}}\mathbf{-}\mathbf{6}{\mathit{y}}^{\mathbf{\text{'}}}\mathbf{+}\mathbf{9}\mathit{y}\mathbf{=}\mathbf{12}\mathit{x}{\mathit{e}}^{\mathbf{3}\mathbf{x}}$

Short answer

The general solution given by differential equation is$y\left(x\right)=(C_{1}x+C_2)e^{3x}+2x^3{e}^{3x}$

Step-by-step solution

Given data. 

Given equation is$y^{\text{'}\text{'}}-6y^{\text{'}}+9y=12x{e}^{3x}$

General solution of differential equation.

A general solution to the nth order differential equation is one that incorporates a significant number of arbitrary constants. If one uses the variable approach to solve a first-order differential equation, one must insert an arbitrary constant as soon as integration is completed.

Find the general solution of given differential equation.y''−6y'+9y=12xe3x  

The given differential equation is

$\begin{array}{l}{y}^{\text{'}\text{'}}-6y^{\text{'}}+9y=12x{e}^{3x}\\ (D^2-6D+9)y=12x{e}^{3x}\end{array}$

The auxiliary equation can be written as

$\begin{array}{c}{m}^2-6m+9=0\\ m^2-3m-3m+9=0\\ m(m-3)-3(m-3)=0\end{array}$

The roots of the equation are

$\Rightarrow m=3,3$

$C.F=(C_{1}x+C_2)e^{3x}$

Now find the P.I.

$\begin{array}{l}y=x^2{e}^x(Ax+B)\\ y^{\text{'}}=x{e}^{3x}\left(3Ax\right(1+x)+B(2+3x)\\ 3Ax+B=6x\\ A=2\text{andB}=0\end{array}$

The obtained results are

$\begin{array}{l}\text{P.}I=2x^3{e}^{3x}\\ \text{C.}S=(C_{1}x+C_2)e^{3x}+2x^3{e}^{3x}\end{array}$

The solution of the differential equation is.$y\left(x\right)=(C_{1}x+C_2)e^{3x}+2x^3{e}^{3x}$