Differential Equation problem

[cse63146]

Homework Statement

Compute the general solution of y'' + 6y' + 8y = 2t + e^t

Homework Equations

The Attempt at a Solution

after using determinants, I found the general solution to be y(t) = k₁e^-2 + k₂e^-4 + y_p

To find y_p, I would make it equal to y_p = ae^t find its first and second derivates, and substitute them back into y'' + 6y' + 8y , but what do I do about the 2t?

 

[Cyosis]

You can do some clever guess work. You can make sure that y'' is 0 by taking some linear function. This simplifies your problem to 6y'+8y=2t. Since we took a linear function y' will be a constant so this will reduce your problem to constant+8y=2t. A function of the form b+ct should suffice.

 

[cse63146]

would y_p = t be a good guess?

since y'_p = 1 and y''_p = 0

 

[Cyosis]

But $6+8t \neq 2t$, so no. Try b+ct as suggested in my previous post.

 

[cse63146]

how about $$\frac{1}{4}t - \frac{3}{16}$$

so 6(1/4) + 8($$\frac{1}{4}t - \frac{3}{16}$$) = 3/2 + 2t - 3/2 = 2t

so when considering e^t, I would let y_p =$$\frac{1}{4}t - \frac{3}{16} + e^t$$ ?

 

[Cyosis]

The linear term is correct now, however if we plug in e^t we get, e^t+6e^t+8e^t=15e^t!=e^t. So you have to divide by 15.

 

[cse63146]

oops

it should be $$\frac{1}{4}t - \frac{3}{16} + ae^t$$

and a would be 1/15

 

[Cyosis]

Correct.

 

[cse63146]

Thank you for all your help.

 

[Cyosis]

You're welcome.

 

[Mark44]

Homework Statement

Compute the general solution of y'' + 6y' + 8y = 2t + e^t

Homework Equations

The Attempt at a Solution

after using determinants, I found the general solution to be y(t) = k₁e^-2 + k₂e^-4 + y_p

To find y_p, I would make it equal to y_p = ae^t find its first and second derivates, and substitute them back into y'' + 6y' + 8y , but what do I do about the 2t?

Pretty sure you meant to write y(t) = k₁e^-2t + k₂e^-4t + y_p here.

 

[cse63146]

Oops (again).

Yeah, I did. Thanks.