# Solve the given differential equations or initial-value problems

1. Aug 9, 2010

### Marwanx

Hello guys,

I have these two questions that I spent s much time to solve them but couldn't.

solve the given differential equations or initial-value problems

* dy/dx = x/t

* dy/dt = 3 + 5y

It's about SEPARABLE FIRST-ORDER DIFFERENTIAL equations.

Thank you so much,

2. Aug 11, 2010

### ceus

first try to solve second eq. dy/dt -5y=3 is a linear differansial eq. , you will get t(y)
and put t(y) in the first eq. like: t(y).dy=x.dx, integrate this, here is the solution.

3. Aug 11, 2010

It looks like you have two completely separate question there?

Is there a typo or something in the first equation?

The second equation: it's separable! dy/(3 + 5y) = dt. If you can't solve that question, you haven't understood the material.

4. Aug 11, 2010

### HallsofIvy

Staff Emeritus
Your first equation doesn't make sense. There shouldn't be an "x" in the equation if nothing else is said about "x". You probably mean dy/dx= y/t.

These are called "separable" because you can separate y and t. In the first equation dy/dx= y/t becomes (1/y)dy/dt= t and now you can treat the derivative dy/dt as the ratio of differentials (dy)/(dt) so that (1/y)dy= t dt and integrate both sides, (1/y)dy with respect to y and t dt with respect to t.

Similarly, dy/dt= 3+ 5y becomes dy/(3+ 5y)= dt in "differential form" and you can again integrate both sides.