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

User Tchen
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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}
User Undreren
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