Question

Given: 2y"+ 3y' + y = 13x2 Determine yp Determine ye ns ) Determine the general solution satisfying the following conditions: y' (2) + y" (2) = 2(13 -e1) & 0.5y"(2) + y(2) = 13 +2e-2 Ans.

Answer 1

I'm not sure what to make of "yp" and "ye ns", or the given initial conditions, so I'll let you handle that and just focus on the general solution.

$2y''+3y'+y=13x^2$

The corresponding homogeneous ODE has characteristic equation

$2r^2+3r+1=(2r+1)(r+1)=0$

with roots at $r=-\dfrac12$ and $r=-1$, so the characteristic solution to the ODE is

$y_c=C_1e^{-x/2}+C_2e^{-x}$

For the non-homogeneous ODE, assume a solution of the form

$y_p=ax^2+bx+c$

$\implies{y_p}'=2ax+b$

$\implies{y_p}''=2a$

Substituting $y_p$ and its derivatives into the ODE gives

$2(2a)+3(2ax+b)+(ax^2+bx+c)=13x^2$

$ax^2+(6a+b)x+(4a+3b+c)=13x^2$

$\implies\begin{cases}a=13\\6a+b=0\\4a+3b+c=0\end{cases}\implies a=13,b=-78,c=182$

so that the particular solution is

$y_p=13x^2-78x+182$

and the general solution is

$\boxed{y(x)=C_1e^{-x/2}+C_2e^{-x}+13x^2-78x+182}$