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