Solve the given initial-value problem (x + y)² dx + (2xy + x² - 5) dy = 0, y(1) = 1

Solution:

To answer this question, we will be using the topic of ordinary differential equations.

A differential equation that does not involve partial derivatives is called an Ordinary Differential equation.

First, we should check if the equation is an ordinary differential equation or not.

(x + y)² dx + (2xy + x² - 5) dy = 0

Take M = (x + y)² and N = (2xy + x² - 5)

We require the condition M\(_y\) = N\(_x\) for a differential equation to be an exact differential equation.

M\(_y\) = 2 (x + y)

N\(_x\) = 2y + 2x = 2 (x + y)

The given equation is an exact differential equation.

The solution should be of the form Ψ\(_{(x, y)}\) = C

dΨ = Ψ\(_x\) dx + Ψ\(_y\) dy = 0

Ψ\(_y\) = (x + y)² and Ψ\(_y\) = (2xy + x² − 5)

Now integrate both sides with respect to x

∫ Ψx dx = ∫ (x + y)² dx

Ψ\(_{(x, y)}\) = [(x + y)³] / 3 + f(y)

Differentiate both sides with respect to y

Ψ\(_y\) = 2xy + x² − 5 = (x + y)² + f '(y)

2xy + x² − 5 = x² + 2xy + y² + f '(y)

f '(y) = - 5 - y²

f (y) = - 5y - y³/3 + C

Ψ\(_{(x, y)}\) = [(x + y)³] / 3 - 5y - y³/3 = C -----------> (1)

[(x + y)³] / 3 - 5y - y³/3 = C

Given: y(1) = 1 

⇒ When x = 1, y = 1.

Substituting the values in the equation (1), we get 

C = -8/3

Ψ\(_{(x, y)}\) = [(x + y)³] / 3 - 5y - y³/3 = -8/3

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Solve the given initial-value problem (x + y)² dx + (2xy + x² - 5) dy = 0, y(1) = 1

Summary:

In the given initial value problem (x + y)² dx + (2xy + x² - 5) dy = 0 when y(1) = 1 is Ψ\(_{(x, y)}\) = [(x + y)³] / 3 - 5y - y³/3 = -8/3