Exact Differential Equation
SOLVE FOR THE DIFFERENTIAL EQUATION:
Step 1: Identify M(x,y) and N(x,y)
M(x,y)
N(x,y)
Step 2: Get the derivative My(x,y) and Nx(x,y) to
identify wheter the equation is exact or not
• Note:
– derivative of M(x,y) with respect to y....... all value of x will be considered
a constant.
– derivative of N(x,y) with respect to x....... all value of y will be considered
a constant.
• M(x,y) = 2xy2 + 3x2
• N(x,y) = 2x2y+4y3
----------> M(x,y)dy = 2x(2)(y)2-1 + 0
= 4xy
----------> Nx(x,y)dx = 2y(2)x2-1 + 0
= 4xy
Step 3: get the implicit integral of f(x,y)dx or f(x,y)dy to get f(x,y)
• Note for the function fx(x,y) = M(x,y)dx and f(x,y) = Ny(x,y)dy
f(x,y)dx = (2xy2 + 3x2)dx
f(x,y)dy = (2x2y+4y3)dy
f(x,y) = 2x2y2 + 3x3 + h(y)
2
3
= x2y2 + x3 + h(y)
step 4: get the partial derivative of f(x,y) with respect to y
f(x,y) = x2y2 + x3 + h(y)
f(x,y)dy = x2 (2)y2-1 + 0 + h’(y)
= 2x2y + h’(y)
Step 5: since we got f(x,y)dy equate it to the other
value from step 3 and get the value of h’(y)
= 2x2y + h’(y)
2x2y + 4y3 = 2x2y + h’(y)
h’(y) = 4y3
Step 6: integrate h’(y) and Substitute back to f(x,y)
• h’(y) = 4y3
h(y) = 4y 3+1 + C
3+1
h(y) = y4+C
f(x,y) = x2y2 + x3 + h(y)
therefore f(x,y) = x2y2 + x3 + y4+C
Step 7: equate to Zero
x2y2 + x3 + y4+C = 0
Solution
SOLVE FOR THE EXACT D.E.
Assignment