The solution to the Differential Equation

[Northbysouth]

Homework Statement

The Initial value problem:

y' - (3/t)y = 0

y(1) = -10

Has the solution:

I have attached an image of the question

Homework Equations

The Attempt at a Solution

Firstly I found the integrating factor:

u(t) = e^{∫-3/t dt}

u(t) = -3/t

I plugged this back into the original equation:

(-3/t)y' + (9/t²y = 0

∫[-(3/t)y]' dt = ∫ 0 dt

(-3/t)y = C

y = -Ct/3

But I don't understand where I go from here. The possible answers all include t³ and I don't understand where that has come from. What step did I miss?

 

[Saitama]

Homework Statement

The Initial value problem:

y' - (3/t)y = 0

y(1) = -10

Has the solution:

I have attached an image of the question

Homework Equations

The Attempt at a Solution

Firstly I found the integrating factor:

u(t) = e^{∫-3/t dt}

u(t) = -3/t

What is the integration of 1/t?

 

[Northbysouth]

Yes, I see my mistake now.

u(t) = t^-3