(x+y)^2dx+(2xy+x^2-1)dy=0 Y(1)=1

6 min read Jun 17, 2024

Solving the Differential Equation (x+y)^2 dx + (2xy + x^2 - 1) dy = 0 with Initial Condition y(1) = 1

This article will walk through the steps of solving the given differential equation and applying the initial condition to find a particular solution.

1. Identifying the Type of Differential Equation

The given equation is a first-order, non-linear differential equation. This is because it involves the first derivative of y (dy/dx) and has terms with powers of x and y.

2. Checking for Exactness

A differential equation of the form M(x,y)dx + N(x,y)dy = 0 is exact if ∂M/∂y = ∂N/∂x.

In our case:

M(x,y) = (x+y)^2
N(x,y) = 2xy + x^2 - 1

Calculating the partial derivatives:

∂M/∂y = 2(x+y)
∂N/∂x = 2y + 2x

Since ∂M/∂y ≠ ∂N/∂x, the given differential equation is not exact.

3. Finding an Integrating Factor

To solve a non-exact differential equation, we can try to find an integrating factor, a function that, when multiplied by the original equation, makes it exact.

One way to find an integrating factor is by using the following formula:

If (∂N/∂x - ∂M/∂y)/M is a function of y alone, then the integrating factor is μ(y) = exp(∫(∂N/∂x - ∂M/∂y)/M dy).
If (∂M/∂y - ∂N/∂x)/N is a function of x alone, then the integrating factor is μ(x) = exp(∫(∂M/∂y - ∂N/∂x)/N dx).

Let's try the first formula:

(∂N/∂x - ∂M/∂y)/M = (2y + 2x - 2(x+y))/(x+y)^2 = 0/(x+y)^2 = 0

This is a function of y alone (it's a constant).

Therefore, the integrating factor is:

μ(y) = exp(∫0 dy) = exp(C) = C

We can choose C = 1 for simplicity.

4. Multiplying by the Integrating Factor

Multiplying the original differential equation by μ(y) = 1, we get:

(x+y)^2 dx + (2xy + x^2 - 1) dy = 0

This equation is now exact because the integrating factor made ∂M/∂y = ∂N/∂x.

5. Solving the Exact Differential Equation

To solve the exact differential equation, we need to find a function F(x,y) such that:

∂F/∂x = M(x,y) = (x+y)^2
∂F/∂y = N(x,y) = 2xy + x^2 - 1

Integrating the first equation with respect to x, we get:

F(x,y) = ∫(x+y)^2 dx = (1/3)(x+y)^3 + g(y)

Here, g(y) is an arbitrary function of y that will be determined later.

Now, differentiate this expression for F(x,y) with respect to y:

∂F/∂y = (x+y)^2 + g'(y)

We know this should be equal to N(x,y) = 2xy + x^2 - 1. Comparing the expressions, we get:

g'(y) = 2xy + x^2 - 1 - (x+y)^2 = -1

Integrating this equation with respect to y, we get:

g(y) = -y + C

Where C is another constant of integration.

Therefore, the general solution of the differential equation is:

F(x,y) = (1/3)(x+y)^3 - y + C = 0

6. Applying the Initial Condition

We are given the initial condition y(1) = 1. Substituting this into the general solution, we get:

(1/3)(1+1)^3 - 1 + C = 0

Solving for C, we find C = -7/3.

7. The Particular Solution

The particular solution that satisfies the given initial condition is:

(1/3)(x+y)^3 - y - 7/3 = 0

This is the implicit solution of the differential equation with the given initial condition.

Note: It might be possible to solve this equation for y explicitly, but it is not necessary in most cases. The implicit solution is perfectly valid and represents the relationship between x and y.