Solve the given differential equations or initial-value problems

[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,

 

[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.

 

[adriank]

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.

 

[HallsofIvy]

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,

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.