# Solve y'' + 3y' - 10y=x(e^{x} + 1)

Question
Differential equations
Solve $$y''\ +\ 3y'\ -\ 10y=x(e^{x}\ +\ 1)$$

2021-01-03
Given $$y''\ +\ 3y'\ -\ 10y=x(e^{x}\ +\ 1)$$
The auxiliary equation is given by $$m^{2}\ +\ 3m\ -\ 10=0$$
solving this we get $$m = -5,\ 2$$
Hence the complimentary function is $$y_{c} = C_{1}e^{-5x}\ +\ C_{2}e^{2x}$$
Now The P.I. of given differential equation is
$$y_{p}=\frac{1}{D^{2}\ +\ 3D\ -\ 10}x(x^{x}\ +\ 1)=\frac{1}{(D\ +\ 5)(D\ -\ 2)}xe^{x}\ +\ 1$$
$$e^{x}\frac{1}{(D\ +\ 1\ +\ 5)(D\ +\ 1\ -\ 2)}x\ +\ \frac{1}{(D\ +\ 5)(D\ -\ 2)}x$$
$$=e^{x}\frac{1}{(D\ +\ 6)(D\ -\ 1)}x\ +\ \frac{1}{(D\ +\ 5)(D\ -\ 2)}x$$
$$=e^{x}\frac{1}{-6\left(1\ +\ \frac{D}{6}\right)(1\ -\ D)}x\ +\ \frac{1}{-10\left(1\ +\ \frac{D}{5}\right)\left(1\ -\ \frac{D}{2}\right)}x$$
$$=\ -\frac{e^{x}}{6}\left(1\ +\ \frac{D}{6}\right)^{-1}(1\ -\ D)^{-1}x\ -\ \frac{1}{10}\left(1\ -\ \frac{D}{5}\right)^{-1}\left(1\ -\ \frac{D}{2}\right)^{-1}x$$
$$=\ -\frac{e^{x}}{6}\left(1\ +\ \frac{D}{6}\ +\ \cdots\right)(1\ +\ D\ +\ D^{2})x\ -\ \frac{1}{10}\left(1\ +\ \frac{D}{5}\ \cdots\right)\left(1\ +\ \frac{D}{2}\ +\ \cdots\right)x$$
$$=\ -\frac{e^{x}}{6}\left(x\ +\ 1\ +\ \frac{1}{6}\right)\ -\ \frac{1}{10}\left(x\ +\ \frac{1}{5}\ +\ \frac{1}{2}\right)$$ $$=\ -\frac{e^{x}}{6}\left(x\ +\ \frac{7}{6}\right)\ -\ \frac{1}{10}\left(x\ +\ \frac{7}{10}\right)$$
Hence the general solution is
$$y=y_{c}\ +\ y_{p}=C_{1}e^{-5}x\ +\ C_{2}e^{2}x\ +\ -e_{6}\left(x\ +\ \frac{7}{6}\right)\ -\ \frac{1}{10}\left(x\ +\ \frac{7}{10}\right)$$

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