Q.166. Differential Equation.

Question 166.
Solve the differential equation [4(x^3/y^2) + (3/y)]dx + [3(x/y^2) + 4y]dy = 0.

Answer 166.
[4(x^3/y^2) + (3/y)]dx + [3(x/y^2) + 4y]dy = 0

=> (4x^3 + 3y) dx + (3x + 4y^3) dy = 0
If M = 4x^3 + 3y and N = 3x + 4y^3,
∂M/∂y = ∂N/∂x = 3
=> This is a perfect differential equation.
The method for solving a perfect differential equation of the type Mdx + Ndy = 0 is

"First integrate the terms in Mdx as if y were constant, then integrate the terms in Ndy cosidering x as constant, and rejecting the terms already obtained, equate the sum of these integrals to a constant."

 Thus, ∫ Mdx (treating y as constant) = ∫ (4x^3 + 3y) dx = x^4 + 3xy
and ∫ Ndy (treating x as constant) = ∫ (3x + 4y^3) dy = 3xy + y^4
=> solution of the differential equation is
x^4 + 3xy + y^4 = C.
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Verification:
Differentiating the solution,
4x^3 dx + 3(xdy + ydx) + 4y^3 dy = 0
=> (4x^3 + 3y) dx + (3x + 4y^3) dy = 0
=> (4x^3 + 3y) / y^2 dx + (3x + 4y^3) / y^2 dy = 0 ... [if y is not zero]
=> [4(x^3/y^2) + (3/y)]dx + [3(x/y^2) + 4y]dy = 0.

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