Q.132. Differential Equation

Question 132.
Find the value of A that makes the following differential equation exact.
4xy^2 + x + 2y + siny + (Ax^2y + 2x + xcosy) dy/dx = 0.
Using this value of A find an implicit solution when y(1) = 0.

Answer 132.
4xy^2 + x + 2y + siny + (Ax^2y + 2x + xcosy) dy/dx = 0
=> (4xy^2 + x + 2y + siny) dx + (Ax^2y + 2x + xcosy) dy = 0

For differencial equation of the type Mdx + Ndy = 0, where both M and N are functions of x and y,
if δM/δy = δN/δx, then the differential equation will be exact.

Now, δM/δy = δN/δx
=> 8xy + 2 + cosy = 2Axy + 2 + cosy
=> A = 4 for the equation to be exact.

The method of solving an exact differential equation 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." This will be the solution of the required equation.

For the given equation,
∫ Mdx
= ∫ (4xy^2 + x + 2y + siny) dx
= 2x^2 * y^2 + x^2/2 + 2xy + xsiny
∫ Ndy
= ∫(4x^2y + 2x + xcosy) dy
= 2x^2 * y^2 + 2xy + xsiny

Rejecting the terms 2x^2 * y^2, 2xy and xsiny already obtained, the general solution is
2x^2 * y^2 + x^2/2 + 2xy + xsiny = c.

y(1) = 0
=> 2*0 + 1/2 = c
=> Particular solution is
2x^2*y^2 + x^2/2 + 2xy + xsiny = 1/2
=> 4x^2y^2 + x^2 + 4xy + 2xsiny = 1.

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Verification:
Differentiating the general solution
2x^2 * y^2 + x^2/2 + 2xy + xsiny = c w.r.t. to x,
4xy^2 + 4x^2y dy/dx + x + 2y + 2x dy/dx + siny + xcosy dy/dx = 0
=> 4xy^2 + x + 2y + siny + (4x^2y + 2x + xcosy) dy/dx = 0.

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