(dy/dx)^2 + sinx cosx (dy/dx) - sin^4x = 0
You sure about that? Because --
http://www.wolframalpha.com/input/?i=(dy%2Fdx)%5E2+%2B+sinx+cosx+(dy%2Fdx)+-+sin%5E4x+%3D+0
On the other hand, if you meant
y" + sinx cosx y' = sin^4x
Then you can substitute u=y' and get
u' + sinx cosx u = sin^4x
Then use the integrating factor e^(∫sinx cosx dx) = e^(1/2 sin^2x) to solve for u, and then let y = ∫u du
A differential equation question
Find the general solution to the equation
(dy/dx)^2 + sinxcos^2x(dy/dx) - sin^4x = 0
Thanks
2 answers
I took a 2nd look at this, and it occurred to me how to solve what you had typed.
y'^2 + sinx cosx y' = sin^4x
y'^2 + sinx cosx y' + sin^2x cos^2x/4 = sin^4x + sin^2x cos^2x/4
4y'^2 + 4sinx cosx y' + sin^2x cos^2x = 4sin^4x + sin^2x cos^2x
(2y' + sinx cosx)^2 = sin^2x(4sin^2x-cos^2x)
2y' + sinx cosx = ±sinx √(1-3cos^2x)
4y' = -sin2x ±u√(1-3u^2) where u=cosx
y' = 1/2 cos2x ± 1/9 (1-3cos^2x)^(3/2) + c
Double-check. sometimes wolframalpha.com comes up with complicated and convoluted solutions.
y'^2 + sinx cosx y' = sin^4x
y'^2 + sinx cosx y' + sin^2x cos^2x/4 = sin^4x + sin^2x cos^2x/4
4y'^2 + 4sinx cosx y' + sin^2x cos^2x = 4sin^4x + sin^2x cos^2x
(2y' + sinx cosx)^2 = sin^2x(4sin^2x-cos^2x)
2y' + sinx cosx = ±sinx √(1-3cos^2x)
4y' = -sin2x ±u√(1-3u^2) where u=cosx
y' = 1/2 cos2x ± 1/9 (1-3cos^2x)^(3/2) + c
Double-check. sometimes wolframalpha.com comes up with complicated and convoluted solutions.
1 answer