# Solving Exact Differential Equations

Here I introduce you to a method for solving the first-order differential equation

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

for the special case in which this equation represents the exact differential of a function z = f(x, y).

**step 1: Test for Exactness **

Let M and N have continuous partial derivatives. The differential equation is exact if and only if

∂M/∂y = ∂N/∂x.

**step 2: **

equation 1 -> ∂u/∂x = M equation 2 -> ∂u/∂y = N

**step 3: Integrate equation 1 with respect to x**

equation A -> ∂u = M ∂x + k(y)

**Note:** If you integrate equation 1 then take constant as k(y), as above. Otherwise if you integrate equation 2, which is with respect to y, then you’ll have to take h(x) as a constant.

**step 4:** **Partially differentiate with respect to y**

∂u/∂y = ∂/∂y M ∂x + ∂/∂y k(y)

use the equation 1 and 2 from the step 2 to substitute,

∂u = M ∂x, M ∂x/∂y = N.

So, left and right side cancels,

∴ ∂/∂y k(y) = 0, k(y) = 0.

**step 5: Put the value of k(y) or h(x) in equation A & finally substitute u=c.**

Example Question: (x^{3} + 3xy^{2})dx + (3x^{2}y + y^{2})dy = 0

M = (x^{3} + 3xy^{2}), N = (3x^{2}y + y^{2}).

Differential w.r.t y -> My = 6xy,

Differential w.r.t x -> Nx = 6xy.

step 1: My = Nx (Exact).

step 2: equation 1 -> ∂u/∂x = (x^{3} + 3xy^{2}), equation 2 -> ∂u/∂y = (3x^{2}y + y^{2}).

step 3: Integrate equation 1 w.r.t x ∂u = (x^{3} + 3xy^{2}) ∂x

equation A -> u = x^{4}/4 + 3/2 x^{2}y^{2} + k(y).

step 4: Partially differentiate equation A w.r.t y

N = ∂u/∂y = 3x^{2}y+ ∂/∂y k(y)

now substitute N,

3x^{2}y + y^{2} = 3x^{2}y+ ∂/∂y k(y)

y^{2} = ∂/∂y k(y)

integrate to get the value of k(y),

k(y) = y^{3}/3.

step 5: substitute k(y) and u=c,

x^{4}/4 + 3/2 x^{2}y^{2} + y^{3}/3 = c.