# Solve the Given Initial-Value Problem. (x + y)^{2} dx + (2xy + x^{2} − 2) dy = 0, y(1) = 1

We will be using the concept of ordinary differential equations to answer this.

### Answer: The soultion of Initial-Value Problem (x + y)^{2} dx + (2xy + x^{2} − 2) dy = 0 is [(x + y)^{3}] / 3 - 2y - y^{3}/3 = C

Let us solve this step by step.

**Explanation:**

A differential equation that does not involves partial derivatives is known as an Ordinary differential equation.

We need to check whether the given equation is an ordinary differential equation or not.

Given that, (x + y)^{2} dx + (2xy + x^{2} − 2) dy = 0

M = (x + y)^{2}

N = (2xy + x^{2} − 2)

For a differential equation to be an exact differential equation, we need condition M_{y} = N_{x}

M_{y} = 2(x + y)

N_{x} = 2y + 2x = 2(x +y)

So, the given equation is an exact differential equation.

We are looking for the solution of the form Ψ(x, y) = C

dΨ = Ψ_{x} dx + Ψ_{y} dy = 0

⇒Ψ_{x} = (x + y)^{2} and Ψ_{y} = (2xy + x^{2} − 2)

Integrate both sides with respect to x,

∫ Ψ_{x} dx = ∫ (x + y)^{2} dx

⇒Ψ(x, y) = [(x + y)^{3}] / 3 + f(y)

Differentiate with respect to y,

Ψ_{y} = 2xy + x^{2} − 2 = (x + y)^{2} + f '(y)

2xy + x^{2} − 2 = x^{2} + 2xy + y^{2} + f '(y)

f '(y) = - 2 - y^{2}

f (y) = - 2y - y^{3}/3 + C

Ψ(x, y) = [(x + y)^{3}] / 3 - 2y - y^{3}/3 = C

[(x + y)^{3}] / 3 - 2y - y^{3}/3 = C