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 $r=\dfrac65$ 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

$\begin{cases}-6a=1\\-2a-b=0\\10a-b-6c=1\end{cases}\implies a=-\dfrac16,b=\dfrac13,c=-\dfrac12$

and so the particular solution is

$z_p=-\dfrac16x^2+\dfrac13x-\dfrac12$

and the general solution for $z$ is

$z=z_c+z_p$
$z=C_1e^{6x/5}+C_2e^{-x}-\dfrac16x^2+\dfrac13x-\dfrac12$

Integrate both sides once to solve for $y$:

$z=y'\implies\displaystyle\int z\,\mathrm dx=\int y'\,\mathrm dx=y$
$\implies y=\hat{C_1}e^{6x/5}+\hat{C_2}e^{-x}-\dfrac19x^3+\dfrac16x^2-\dfrac12x+\hat{C_3}$