Question

What is the particular solution of this differential equation 5y'''-y''-6y'=1+x^2

Answer 1

Reduce the order of the ODE by setting
`z=y'[tex], so that [tex]z'=y''` and
`z''=y'''`.


`5z''-z'-6z=1+x^2`

Consider the homogeneous ODE


`5z''-z'-6z=0`

which has characteristic equation


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

which has roots at
`r=\frac65` and
`r=-1`, so that the characteristic solution is


`z_c=C_1e^(6x/5)+C_2e^(-x)`

For the nonhomogeneous ODE,


`5z''-z'-6z=1+x^2`

we can expect a particular solution of the form


`z_p=ax^2+bx+c`

`{z_p}'=2ax+b`

`{z_p}''=2a`

Substituting these expressions into the ODE yields


`10a-(2ax+b)-6(ax^2+bx+c)=1+x^2`

`\iff-6ax^2+(-2a-b)x+(10a-b-6c)=x^2+1`

from which it follows that


`\begin{cases}-6a=1\\-2a-b=0\\10a-b-6c=1\end{cases}\implies a=-\frac16,b=\frac13,c=-\frac12`

and so the particular solution is


`z_p=-\frac16x^2+\frac13x-\frac12`

and the general solution for
`z` is


`z=z_c+z_p`

`z=C_1e^(6x/5)+C_2e^(-x)-\frac16x^2+\frac13x-\frac12`

Integrate both sides once to solve for
`y`:


`z=y'\implies\displaystyle\int z\,\mathrm dx=\int y'\,\mathrm dx=y`

`\implies y=\hat{C_1}e^(6x/5)+\hat{C_2}e^(-x)-\frac19x^3+\frac16x^2-\frac12x+\hat{C_3}`