# Thread: General solution to 2nd order

1. ## General solution to 2nd order

I am stuck on this question

A suitable form of the general solution to y''-2y'+y=e^t+t

My work so far is

(r-1)^2=0 so r=1 so C_1*e^t+C_2*t*e^t=y(t)

Y_1=At^2e^t
Y_1'=2Ate^t+At^2e^t
Y_1''=2Ae^t+2Ate^2+2At^2e^t

plugging in and simplifying i get At^2+2A=1, however i dont know where to go from here

also for t i did

Y_2=Bt+C
Y_2'=B
Y_2''=0

so plugging in i get -2B+Bt+C=t but i dont know where to go from here, please help.

2. Hmm. I tried your ansatz for the particular solution, and I don't think it works. You just don't quite have the number of degrees of freedom you need in order to make it work, I think. I think you're going to need something like

$y_{p}(t)=At^{2}e^{t}+Bte^{t}+Ce^{t}+Dt+E.$ Plug this form into your DE, and set coefficients of like terms equal to each other. That's the usual procedure, right?

3. Please disregard me previous post. You've got the right idea. I think the issue is that you're not taking the derivatives correctly. You need to fix those. Then you plug all that into your DE and equate coefficients of like terms on both sides.

4. have you been taught variation of parameters?

5. no i have not learned variation of parameters, looking over it and reworking i believe the answer is c_1te^t+c_2e^t+At^2e^t+Bt+C

6. Originally Posted by Juggalomike
I am stuck on this question

A suitable form of the general solution to y''-2y'+y=e^t+t

My work so far is

(r-1)^2=0 so r=1 so C_1*e^t+C_2*t*e^t=y(t)

Y_1=At^2e^t
Y_1'=2Ate^t+At^2e^t
Y_1''=2Ae^t+2Ate^2+2At^2e^t

plugging in and simplifying i get At^2+2A=1, however i dont know where to go from here

also for t i did

Y_2=Bt+C
Y_2'=B
Y_2''=0

so plugging in i get -2B+Bt+C=t but i dont know where to go from here, please help.

You're pretty much there.

Your expression $y(t)=C_{1}e^{t}+C_{2}te^{t}$ is called the Complementary Function (CF), and is correct.

Your choice of Particular Integral (PI) for the $e^{t}$ part of the RHS is correct, but you have made a mistake in calculating the second derivative. Correct that and you should be able to calculate the value of $A.$

Your choice of PI for the $t$ on the RHS and your equation for $B \mbox{ and } C$ are also correct. You just have to follow it through. What values do $B \mbox{ and } C$ need to take in order that the LHS is identical with the RHS ?

The general solution of the equation will then be the sum of the CF and the two PI's.

7. Originally Posted by Juggalomike
no i have not learned variation of parameters, looking over it and reworking i believe the answer is c_1te^t+c_2e^t+At^2e^t+Bt+C
ah because given the second order ODE $a(x)y''(x)+b(x)y'(x)+c(x)y(x)=f(x),$ the particular solution is $\displaystyle{{y}_{p}}(x)=\int_{{}}^{x}{\frac{f(t) }{a(t)W[{{y}_{1}},{{y}_{2}}](t)}\left| \begin{matrix}
{{y}_{1}}(t) & {{y}_{1}}(t) \\
{{y}_{2}}(x) & {{y}_{2}}(x) \\
\end{matrix} \right|\,dt},$
where $y_1,y_2$ are the solutions of the homogeneous equation and $W[{{y}_{1}},{{y}_{2}}](x)$ is the Wronskian.

if initial conditions are added, you just put it in the lower integration limit.