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How to solve this first order ODE? Very confusing, I have a picture (imgur)?
http://i.imgur.com/kos6znr.png
I don't think it's separable. Is it an exact equation? I'm trying that right now.
1 Answer
- intc_escapeeLv 77 years agoFavorite Answer
(-2 e^(2x) cos(y) + e^(2y) sin(x)) dx + (e^(2x) sin(y) - 2 e^(2y) cos(x)) dy = 0
seek a solution of the form ψ(x,y) = c
let ∂ψ/∂x = -2 e^(2x) cos(y) + e^(2y) sin(x)
and ∂ψ/∂y = e^(2x) sin(y) - 2 e^(2y) cos(x)
so that ∂ψ/∂x + ∂ψ/∂y dy/dx = 0 ............. μ is the integrating factor
∂/∂y (∂ψ/��x) = ∂/∂x (∂ψ/∂y) = 2 (e^(2y) sin(x) + e^(2x) sin(y)) .... proving the equation is exact
ψ = ∫ ∂ψ/∂y dy = -e^(2y) cos(x) - e^(2x) cos(y) + h(x)
∂ψ/∂x = e^(2y) sin(x) - 2e^(2x) cos(y) + h'(x) = -2 e^(2x) cos(y) + e^(2y) sin(x) .. ⇒ h=c
ψ = -e^(2y) cos(x) - e^(2x) cos(y)
Answer: e^(2y) cos(x) + e^(2x) cos(y) = C
