Question

Differential Equations problem.

Solve the initial Value Problem: (see the given, below)

I have the solution, and I know that we have to use integration by parts, but I'd like a better (clearer) solution, please. Just send me the whole thing. I want to see the way you solve it, in one run. Thanks!

Answer 1

\[\frac{ dy }{ dt } = y + 2t\]
\[y(0) = -2\]

Answer 2

As a start, multiply \(e^{-t}\) through out

Answer 3

\(\dfrac{ dy }{ dt } = y + 2t\)

\(\dfrac{ dy }{ dt } -y = 2t\)

\(\color{red}{e^{-t}}\dfrac{ dy }{ dt } - \color{red}{e^{-t}}y = \color{red}{e^{-t}}2t\)

Now, do you notice anythign special about the left hand side ?

Answer 4

Can this form be defined as \(\dfrac{dy}{dt} - P\cdot y = Q\) ?

Answer 5

Oh I kind of see it now.

Answer 6

Yes, it's indeed a linear equation..

Answer 7

Note the integrating factor is \[e^{-t}\] now what happens when you take the derivative of this \[(e^{-t}y)'\]