Separable Differential Equation: (t^2 + 16)dx/dt = (x^2 + 36)

reconrusty

New member
Joined
Dec 13, 2017
Messages
3
I have a problem I just can't seem to get right.

(t^2 + 16)dx/dt = (x^2 + 36)

So I know you have to get all the x's to one side and the t's to the other so I got this:

1/(x^2 + 36) dx = 1/(t^2 + 16) dt

Differentiate each side and you get:

1/6arctan(x) = 1/4arctan(t/4) + C

Now the part where im confused is how to isolate for x = ...

I got arctan(x/6) = 6(1/4arctan(t/4) + C) and don't know how to proceed from here.

Any help? :)
 
I have a problem I just can't seem to get right.

(t^2 + 16)dx/dt = (x^2 + 36)

So I know you have to get all the x's to one side and the t's to the other so I got this:

1/(x^2 + 36) dx = 1/(t^2 + 16) dt

Differentiate each side and you get:

1/6arctan(x) = 1/4arctan(t/4) + C

Now the part where im confused is how to isolate for x = ...

I got arctan(x/6) = 6(1/4arctan(t/4) + C) and don't know how to proceed from here.

Any help? :)
Think about the definition of arctan(x) = Θ
 
Top