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=-\frac12` 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}`