Q.135. Differential Equation

Question 135.
Solve the differential equation y*e^(y) dx=(y^3 + 2*x*e^(y) ) dy.

Answer 135.
ye^y dx = (y^3 + 2xe^y) dy
=> y dx - 2x dy = y^3*e^-y dy
Dividing by y^3,
=> (1/y^2) dx - (2x/y^3) dy = e^-y dy
=> d(x/y^2) = e^-y dy
Integrating,
x/y^2 = - e^(-y) + c

=====================================…

I could solve the above problem after reading "A note on integrating factor" in a book of "Integral Calculus including Differential Equations" written by "B.C.Das and B.N.Mukherjee"

For the differential equation of the type
Mdx + Ndy = 0,
(i)    If (1/N) (∂M/∂y - ∂N/∂x) is a function of x only, say f(x),
       then e^∫f(x)dx is the integrating factor.
(ii)   If (1/M) (∂N/∂x - ∂M/∂y) is a function of y only, say f(y),
       then e^∫f(y)dy is the integrating factor.
(iii)  If M and N are both homogeneous functions
       in x, y of degree n,
       then 1/(Mx + Ny) is the integrating factor.
(iv)  If the equation is of the form, y f(xy) dx + x g(xy) dy = 0,
       then 1/(Mx -Ny) is the integrating factor [Mx -Ny ≠ 0].

I converted the equation into
ydx + (-y^3 e^-y - 2x)dy = 0.
M = y and N = - y^3 e^-y - 2x
=> ∂N/∂x = - 2 and ∂M/∂y = 1
=> (1/M)(∂N/∂x - ∂M/∂y) = - 3/y ... [condition (ii) above]
=> Integrating factor = e^∫(- 3/y) dy = e^(-3lny) = 1/y^3
I hope the rest of the steps are obvious.

LINK to YA!