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.
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.