Using the Method of Undetermined Coefficients to find a particular solution

[shamieh]

Note that the general solution to $y'' - y = 0$ is $y_h = C_1e^t + C_2e^{-t}$

In the following, use the Method of Undetermined Coefficients to find a particular solution.

a)$y'' - y = t^2$

So here is what I have so far

$y_p = At^2 + Bt + C$
$(y_p)'' = 2A$

Ive got $A = -1, B = 0 , C = 0$

so would the $y_p = -t^2$ be correct?

 

[MarkFL]

When you substitute $y_p$ into the given ODE, you obtain:

\(\displaystyle (2A)-\left(At^2+Bt+C\right)=t^2\)

or:

\(\displaystyle -At^2-Bt+2A-C=1t^2+0t+0\)

Equating coefficients, we obtain the system:

\(\displaystyle -A=1\)

\(\displaystyle -B=0\)

\(\displaystyle 2A-C=0\)

And so we find:

\(\displaystyle (A,B,C)=\left(-1,0,-2\right)\)

Thus, we have:

\(\displaystyle y_p(t)=-\left(t^2+2\right)\)

Does this make sense?

 

[chisigma]

Note that the general solution to $y'' - y = 0$ is $y_h = C_1e^t + C_2e^{-t}$

In the following, use the Method of Undetermined Coefficients to find a particular solution.

a)$y'' - y = t^2$

So here is what I have so far

$y_p = At^2 + Bt + C$
$(y_p)'' = 2A$

so my mind has went blank...how do I solve this? $2A - At^2 + Bt + C = t^2$ Any advice would be great

I think it would be $A = 1/2, B= 0, C =0$

... may be that the solution is more simple You think...

$A = -1$

$B=0$

$2\ A + C = 0$

A and B are known... what is C?...

KInd regards

$\chi$ $\sigma$

 

[shamieh]

YES! So $y_p = -t^2 - 2$