# Make and solve the given equation d^{3} frac{y}{dx^{3}} - d^{2} frac{y}{dx^{2}}=2x + 3

Question
Differential equations
Make and solve the given equation $$d^{3}\ \frac{y}{dx^{3}}\ -\ d^{2}\ \frac{y}{dx^{2}}=2x\ +\ 3$$

2020-12-16
Given $$d^{3}\ \frac{y}{dx^{3}}\ -\ d^{2}\ \frac{y}{dx^{2}} = 2x\ +\ 3$$
The above equation can be written as $$(D^{3}\ -\ D^{2})y = 2x\ +\ 3$$
Now the auxiliary equation is $$m^{3}\ -\ m^{2} = 0$$
which gives $$m = 0,\ 0,\ 1$$
Hence the complimentary function is $$C.F. = C_{1}\ +\ C_{2}x\ +\ c_{3}e^{x}$$
Now we need to find the Particular Integral
$$y_{p}=\frac{1}{D^{2}\ +\ 3D\ -\ 10}x(x^{x}\ +\ 1)=\frac{1}{(D\ +\ 5)(D\ -\ 2)}xe^{x}\ +\ 1$$
$$=e^{x}\frac{1}{(D\ +\ 1\ +\ 5)(D\ +\ 1\ -\ 2)}x\ +\ \frac{1}{(D\ +\ 5)(D\ -\ 2)}x$$
$$=e^{x}\frac{1}{(D\ +\ 6)(D\ -\ 1)}x\ +\ \frac{1}{(D\ +\ 5)(D\ -\ 2)}x$$
$$=e^{x}\frac{1}{-6\left(1\ +\ \frac{D}{6}\right)(1\ -\ D)}x\ +\ \frac{1}{-10\left(1\ +\ \frac{D}{5}\right)\left(1\ -\ \frac{D}{2}\right)}x$$
$$=\ -\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}xSK \(=\ -\ \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\ \pm\ e_{6}\left(x\ +\ \frac{7}{6}\right)\ -\ \frac{1}{10}\left(x\ +\ \frac{7}{10}\right)$$

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