Solving Exact Differntial Equations by Integrating factor

If Mdx + Ndy = 0 is a homogeneous differential equation and Mdx + Ndy is not equal to 0 then this equation can be solve by multiplying the differential equation with integrating factor and the integrating factor for the same is, 1/(Mx+Ny) for equation of the form Mdx + Ndy = 0.

Example: Solve exact differential equation x²ydx – (x³+y³)dy = 0

Solution:

Comparing with Mdx + Ndy = 0

M = x²y, and N = -(x³+y³)

∂M/∂y = x² and ∂N/∂x = -3x²

So here ∂M/∂y is not equal to ∂N/∂x

Here, given equation is not exact differential equation

Considering Integrating factor

Integrating Factor = 1/Mx + Ny

=1/(x²y)x+(-x³-y³)y

=  1/(x³y-x³y-y⁴)=-1/y⁴

=-1/y⁴ is an integrating factor

Multiplying the equation with Integrating factor

=(-x²y/y⁴)dx+(x³+y³/y⁴)dy=0

=(-x²/y³)dx+(x³/y⁴+1/y)dy=0

Comparing with  M₁dx + N₁dy= 0

M₁= -x²/y²    and N₁ = x³/y⁴+1/y

Therefore ∂M₁/∂y =  ∂N₁/∂x

Finding ∫M₁dx +∫N₁dy =C      

∫(-x²/y³)dx+(x³/y⁴+1/y)dy=C

-x³/3y³+logy = C

It is the solution for given equation.

Equation that involves a differential co-efficient is called a Differential equation. Let’s take the equation P(x, y)dx + Q(x, y)dy = 0. This equation is called the exact differential equation if, ∂P/∂y = ∂Q/∂y.

In this article, we have covered in detail about exact differential equation.