# 2nd order differential equation: undetermined coefficients

y''-y=t-4e^(-t)

## Homework Equations

method of undetermined coefficients

## The Attempt at a Solution

solving for characteristic equation first

y''-y=0

r^2-1=0

c_1e^(-t)+c_2e^(t)

RHS

particular solution

t-4e^(-t)

y_p(t)= At+B+Ce^(-t)

y_pt'(t)=A-Ce^(-t)

y_p''(t)=Ce^(-t)

plug into LHS

Ce^(-t)-At-B-Ce^(-t)=t-4e^(-t)

-A=t
A=-1

B=0

C cancels on the left.... Im not sure how to any further?

Thanks

Mark44
Mentor

y''-y=t-4e^(-t)

## Homework Equations

method of undetermined coefficients

## The Attempt at a Solution

solving for characteristic equation first

y''-y=0

r^2-1=0

c_1e^(-t)+c_2e^(t)

RHS

particular solution

t-4e^(-t)

y_p(t)= At+B+Ce^(-t)
No, this won't work, since e-t is already a solution in the complementary equation. Do you know what to do when you run into this?
dmoney123 said:
y_pt'(t)=A-Ce^(-t)

y_p''(t)=Ce^(-t)

plug into LHS

Ce^(-t)-At-B-Ce^(-t)=t-4e^(-t)

-A=t
A=-1

B=0

C cancels on the left.... Im not sure how to any further?

Thanks

No, this won't work, since e-t is already a solution in the complementary equation. Do you know what to do when you run into this?

so the guess becomes At+B+Cte^(-t) right? When I plug that back in, it still cancels out

Mark44
Mentor

so the guess becomes At+B+Cte^(-t) right? When I plug that back in, it still cancels out
I'm not sure what you mean.
If yp = At + B + Cte-t, then yp'' - yp has to equal t - 4e-t. Take yp and its second derivative, and substitute them into your nonhomogeneous equation to get A, B, and C.

• dmoney123
Got it... I was making a mistake with my derivative... I feel stupid.

Thanks Mark44!