Question

find a general solution to the given equation.

$y\text{'}\text{'}\text{'}-2y\text{'}\text{'}-5y\text{'}+6y=e^x+x^2$

Short answer

$y\left(x\right)=c_1{e}^{-2x}+c_2{e}^x+c_3{e}^{3x}-\frac{1}{6}x{e}^x+\frac{37}{108}+\frac{5x}{18}+\frac{x^2}{6}$

Step-by-step solution

Find the corresponding auxiliaryequation

Theauxiliary equationof corresponding homogeneous equation

$r^3-2r^2-5r+6=(r-1)(r-3)(r+2)=0$

The solutions of the auxiliary equation are

$r=-2,r=1,r=3$

Therefore a general solution to the homogeneous equation is

$y_h\left(x\right)=c_1{e}^{-2x}+c_2{e}^x+c_3{e}^{3x}$

Find particular solution

Let the particular solution be

$y_p\left(x\right)=ax{e}^x+b+cx+d{x}^2$

Then

$$\begin{gathered} y_p^{\text{'}}\left(x\right)=a{e}^x+ax{e}^x+c+2dx \\ {y}_p^{\text{'}\text{'}}\left(x\right)=2a{e}^x+ax{e}^x+c+2d \\ {y}_p^{\text{'}\text{'}\text{'}}\left(x\right)=3a{e}^x+ax{e}^x \end{gathered}$$

Then

$$\begin{gathered} y_p^{\text{'}\text{'}\text{'}}\left(x\right)-2y_p^{\text{'}\text{'}}\left(x\right)-5y_p^{\text{'}}\left(t\right)+6y_p\left(x\right) \\ =3a{e}^x+ax{e}^x-4a{e}^x-2ax{e}^x-4d-5a{e}^x-5ax{e}^x-5c-10dx+6ax{e}^x+6b+6cx+6d{x}^2 \\ =-6a{e}^x+(6b-5c-4d)+(6c-10d)x+6d{x}^2 \end{gathered}$$

If$-6a{e}^x+(6b-5c-4d)+(6c-10d)x+6d{x}^2=e^x+x^2$

Then$-6a=1,6b-5c-4d=0,6c-10d=0$

Then$a=-\frac{1}{6},b=\frac{37}{108},c=\frac{5}{18},d=\frac{1}{6}$

Hence$y_p\left(x\right)=-\frac{1}{6}x{e}^x+\frac{37}{108}+\frac{5x}{18}+\frac{x^2}{6}$

y(x)=yh+yp

Then$y\left(x\right)=c_1{e}^{-2t}+c_2{e}^t+c_3{e}^{3t}-\frac{1}{6}x{e}^x+\frac{37}{108}+\frac{5x}{18}+\frac{x^2}{6}$

Is the general solution of$y\text{'}\text{'}\text{'}-2y\text{'}\text{'}-5y\text{'}+6y=e^x+x^2$