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## Homework Statement

If y = y(t) is the solution of the initial value problem

y' + (2 t + 1) y = 2 cos(t)

y(0) = 2

What is y''(0)?

## Homework Equations

## The Attempt at a Solution

Since this is a first order linear, I started out by finding the integrating factor so I can find what y is, and then just take the second derivative of it, and then put it to 0.

[itex] IF = e^{ \int {2t + 1} } => IF = e^{t^2 + t} [/itex]

Then, I end up with this equation, if we set IF = k.

[itex] (ky)' = k (2 cos(t)) [/itex]

From here, I don't think I can integrate that equation, so now I'm pretty much stuck. Is there a method I'm missing? Or could I find the value of the second derivative using the value of y at 0?

Any help is appreciated, thank you.