Derivative Equations

Question

Answer 1

Given:

x²ydx+(y-x³)dy = 0

Check first if they are exact differential equation:

If M(x,y)dx + N(x,y)dy = 0 then

∂M/∂y = ∂N/∂x

∂M/∂y =x²

∂N/∂x = -3x²

Therefore it's not exact. We have to find the integrating factor:

Find one of these two:

(∂M/∂y - ∂N/∂x)/M = function of x alone

(∂N/∂x - ∂M/∂y)/N = function of y alone

(∂M/∂y - ∂N/∂x)/M = 4x²/(y-x³) => not function of x alone

(∂N/∂x - ∂M/∂y)/N = -4/y => function of y alone.

Then find the integrating factor using (∂N/∂x - ∂M/∂y)/N = -4/y.

e^{∫(∂N/∂x - ∂M/∂y)/N dy} = e^{∫-4/y dy}= e^{-4ln y}= y^-4 = 1/y⁴

Then multiply 1/y⁴ on both sides of the original equation:

(1/y⁴)(x²ydx+(y-x³)dy = 0)

(x²/y³)dx + (1/y³ - x³/y⁴)dy = 0

Now, they have to be exact. Let's check:

∂M/∂y =-3x²/y⁴

∂N/∂x = -3x²/y⁴. (Yes, they are exact.)

Now solve using the solution for exact DE:

ψ_x = M, ψ_y = N, ∂ψ/∂x = 0

∫ψ_x dx = ∫(x²/y³) dx + f(y). (This integration with respect to x)

ψ = x³/3y³ + f(y)

∂ψ/∂y = (1/3)x³(-3y^-4) = -x³/y⁴ +f'(y)

-x³/y⁴ +f'(y) = 1/y³ - x³/y⁴

f'(y) = 1/y³

f(y) = -1/(2y²)

Therefore our complete solution is:

x³/3y³-1/(2y²) = C