Question

Please: Differential equation

(dy/dx)^2 + sin(x)cos^2(x)dy/dx - sin^4(x)=0

Answer is y=cos(x)+c or y=1/3 cos^3(x) - cos(x) + c but stuck.

Thanks

Answer 1

complete the square:
y'^2 + sinx cos^2x y' = sin^4x
y'^2 + sinx cos^2x y' + ((sinx cos^2x)/2)^2 = sin^4x + ((sinx cos^2x)/2)^2
(y' + 1/2 sinx cos^2x)^2 = sin^2x(1+sin^2x)^2/4
y' + 1/2 sinx cos^2x = ±sinx(1+sin^2x)/2
2y' = -sinx(1±(sin^2x))
If you play around with that, I think you can end up with what you want, maybe up to a constant somewhere.

Answer 2

Thanks I got cos(x)+c which matches my working. However, can you explain lines 3 and 4


i.e. sin^2x(1+sin^2x)^2/4 ---> ±sinx(1+sin^2x)/2

Answer 3

sin^2x(1+sin^2x)^2/4 = (sinx (1+sin^2x)/2)^2

Answer 4

So for line 3...

can I use (sin(x) (1+sin^2x)/2)^2 instead of sin^2x(1+sin^2x)^2/4 and then root the answer to get ±sin(x)(1+sin^2x)/2

Thanks again