(x+xy^2)dx+e^x^2ydy=0

Jasonbradley

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Jun 17, 2024

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Solving the Differential Equation (x + xy^2)dx + e^(x^2)ydy = 0

This article explores the solution to the given differential equation:

(x + xy^2)dx + e^(x^2)ydy = 0

This equation is a first-order, non-linear differential equation. Let's solve it using the method of exact differential equations.

Identifying Exact Equations

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

∂M/∂y = ∂N/∂x

Let's check if our equation is exact:

• M(x,y) = x + xy^2
• N(x,y) = e^(x^2)y

Calculating the partial derivatives:

• ∂M/∂y = 2xy
• ∂N/∂x = 2xe^(x^2)y

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

Making the Equation Exact

To make the equation exact, we need to find an integrating factor. An integrating factor is a function μ(x,y) such that:

μ(x,y)M(x,y)dx + μ(x,y)N(x,y)dy = 0

is an exact differential equation.

In this case, we can use an integrating factor that depends only on x:

μ(x) = e^(∫(∂N/∂x - ∂M/∂y)/N dx)

Let's find μ(x):

μ(x) = e^(∫(2xe^(x^2)y - 2xy) / (e^(x^2)y) dx) μ(x) = e^(∫(2x - 2x/e^(x^2)) dx) μ(x) = e^(x^2)

Now, we multiply the original equation by μ(x):

e^(x^2)(x + xy^2)dx + e^(x^2) * e^(x^2)ydy = 0

This simplifies to:

(xe^(x^2) + x^2y^2e^(x^2))dx + e^(2x^2)ydy = 0

This equation is now exact since:

• ∂(xe^(x^2) + x^2y^2e^(x^2))/∂y = 2xy^2e^(x^2)
• ∂(e^(2x^2)y)/∂x = 2xy^2e^(x^2)

Solving the Exact Equation

Now that we have an exact equation, we can find its solution. We know that the solution can be represented as a function F(x,y) = C, where C is a constant, and:

• ∂F/∂x = M(x,y)
• ∂F/∂y = N(x,y)

Let's integrate ∂F/∂x = xe^(x^2) + x^2y^2e^(x^2) with respect to x:

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

Here, g(y) is an arbitrary function of y. Now, let's find g(y) by differentiating F(x,y) with respect to y and equating it to N(x,y):

∂F/∂y = x^2ye^(x^2) + g'(y) = e^(2x^2)y

From this, we can see that g'(y) = 0. Integrating, we get g(y) = C1, where C1 is another constant.

Therefore, the general solution to the differential equation is:

F(x,y) = (1/2)e^(x^2) + (1/2)x^2y^2e^(x^2) + C1 = C

Combining the constants, we get:

(1/2)e^(x^2) + (1/2)x^2y^2e^(x^2) = C

This is the implicit general solution to the differential equation (x + xy^2)dx + e^(x^2)ydy = 0.