# Differential equatn involving cos^2 and sin^2 in denominator

Gold Member
Homework Statement
(9cos^2(x) + 4sin^2 (x)).dy/dx = 6

The attempt at a solution
Rearranging:

(1/6).dy = 1/(9cos^2(x) + 4sin^2 (x))

(1/6).dy = 1/((3cos(x))^2 + (2sin(x))^2)

L.H.S becomes y/6 after integrating.

The R.H.S. (denominator):
(3cos(x))^2 + (2sin(x))^2
I take out cos^2(x)
This leaves me with:
(9 + 4tan^2(x))
Then i use the standard integral formula for arctan
And i get: (1/3)arctan (2tanx/3)

Then i'm stuck...

## Answers and Replies

What are you trying to find?

dextercioby
Homework Helper
It's almost perfect. The LHS is y/6 + C. The RHS is 1/6 tan^(-1)((2 tan(x))/3).

What's y(x) then ?

Gold Member
What are you trying to find?
Solving the ODE to get y in terms of x.

dextercioby, for the R.H.S, after i worked it out, i get 1/2 tan^(-1)((2 tan(x))/3)
I used substitution t=tanx to integrate R.H.S.

Based on the formula: integral of 1/(a^2 + x^2) is 1/a tan^(-1)(x/a) + C

So, i don't understand how you got 1/6 instead of 1/2

I got the 1/2 by: 1/3 x 3/2 (obtained by dividing by the differentiation of 2t/3)

Last edited:
Gold Member
Any help?

dextercioby
$$... = \frac{1}{9}\int \frac{dx}{\left(\frac{2}{3}\sin x\right)^2 + \cos^2 x} = ...= \frac{3}{18} \int \frac{dp}{p^2 +1} = ...$$