how do you solve first order differential equations?

Question

i have no idea how to solve them! if anyone also has any idea how to solve this one that would be a great help too!

Find value(s) of m such that u=t^m is a solution to 2tu'=u.

Answer 1

When they give you an answer to try, it's much easier.

You're given:

u = t^m

Using polynomial differentiation:

u' = m × t^(m - 1)

Then start with the equation:

2tu' = u

Plug in your values for u' and u:

2 × t × (m × t^(m - 1)) = t^m

Then, t × t^(m - 1) = t^1 × t^(m - 1) = t^(1 + m - 1) = t^m:

2 × m × t^m = t^m

Subtract t^m from both sides:

2 × m × t^m - t^m = 0

Factor out t^m:

t^m(2m - 1) = 0

So either t^m = 0 (which is impossible for any value of m), or (2m - 1) = 0:

2m - 1 = 0

Which means:

m = 1/2