Question

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

Answer 1

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
\(y=y_{c}\ +\ y_{p}=C_{1}e^{-5}x\ +\ C_{2}e^{2}x\ \pm\ e_{6}\left(x\ +\ \frac{7}{6}\right)\ -\ \frac{1}{10}\left(x\ +\ \frac{7}{10}\right)\)