Linear differential equation y' + 5y = t with y(0) = 0

[vpathak]

Hi, I have a Calculus test coming up and am very confused on how to start approching these types of problems (initial value problems) If possible can someone work these 2 different types of problems out and show some work so I know how to proceed with the rest of them? I don't think they are that hard, for some reason I'm just confused. Thanks


Problem 1: y' + 5y = t, y(0) = 0

Problem 2: y' = y/sqrt(t), y(1) = -1

Thanks again for all those who help!

 

[Muzza]

Problem 1: We first find a particular solution to y' + 5y = t. Suppose y_p = at + b, then the equation becomes

a + (5at + 5b) = t

<=>

{ 5a = 1
{ a + 5b = 0

<=>

{ a = 1/5
{ b = -1/25

Thus y_p = t/5 - 1/25 is a particular solution.

We must then solve the homogenous equation:

y' + 5y = 0.

This is a rather well-studied equation, y_h = Ce^(-5t) is the general solution (it could be solved with an integrating factor also). Thus, the general solution to y' + 5y = t can be written:

y = y_p + y_h = t/5 - 1/25 + Ce^(-5t). You can then use your initial condition to calculate C.

Problem 2: y' = y/sqrt(t) <=> dy/dt = y/sqrt(t) <=> 1/y * dy = 1/sqrt(t) * dt. What happens if you integrate both sides?

 

[arildno]

Just adding to Muzza here:
Problem 1)
If you know about the "integrating factor"-technique, you may use this technique instead to derive the solution.