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: (x³ + 3xy²)dx + (3x²y + y²)dy = 0

M = (x³ + 3xy²),       N = (3x²y + y²).

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 = (x³ + 3xy²),                    equation 2 -> ∂­­­u/∂­­y = (3x²y + y²).

step 3: Integrate equation 1 w.r.t x      ∂u =  (x³ + 3xy²) ∂­­x

equation A -> u = x⁴/4 + 3/2 x²y² + k(y).

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

N = ∂­­­u/∂­­y = 3x²y+ ∂­­­/∂­­y k(y)

now substitute N,

3x²y + y² = 3x²y+ ∂­­­/∂­­y k(y)

y² = ∂­­­/∂­­y k(y)

integrate to get the value of k(y),

k(y) = y³/3.

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

x⁴/4 + 3/2 x²y² + y³/3 = c.