Question

In Exercises \(16-24\) find the general solution. $$ x y^{\prime}+\left(1+2 x^{2}\right) y=x^{3} e^{-x^{2}} $$

Short answer

Question: Find the general solution for the given first-order linear differential equation: \(x(y^{\prime})+(1+2x^{2})y=x^3e^{-x^2}\). Answer: The general solution for the given differential equation is: \(y(x) = \frac{1}{4}x^3e^{-x^2} + Cxe^{-x^2}\).

Step-by-step solution

Rewrite the differential equation in standard form

Divide both sides of the equation by \(x\): $$ y^{\prime}+\frac{1+2x^{2}}{x}y=\frac{x^2e^{-x^2}}{x} $$ Now, we can rewrite the given differential equation in the standard form: $$ y^{\prime}+p(x)y=r(x) $$ Here, \(p(x) = \frac{1+2x^{2}}{x}\) and \(r(x) = xe^{-x^2}\).

Find the integrating factor

Now, we need to find the integrating factor \(\mu(x)\). It is given by $$ \mu(x) = e^{\int p(x)dx}. $$ So, let's find the integral of \(p(x)\): $$ \int p(x)dx = \int \frac{1+2x^{2}}{x}dx = \int \frac{1}{x}dx + \int 2x dx. $$ $$ \int p(x)dx = \ln |x| + x^2 + C. $$ Now, we can find the integrating factor \(\mu(x)\): $$ \mu(x) = e^{\ln |x| + x^2} = xe^{x^2}. $$

Multiply the given equation by the integrating factor

Multiply the standard form of the given equation by the integrating factor \(xe^{x^2}\): $$ xe^{x^2}(y^{\prime}+\frac{1+2x^{2}}{x}y)=x^2e^{-x^2}xe^{x^2} $$ The left-hand side of this equation is the exact derivative of \(\mu(x)y\) with respect to \(x\). Therefore, we can rewrite the equation as: $$ \frac{d}{dx}(xe^{x^2} y)=x^3 $$

Integrate both sides of the equation

Now, we will integrate both sides of the equation with respect to \(x\) to find the general solution: $$ \int \frac{d}{dx}(xe^{x^2} y)dx = \int x^3 dx $$ $$ xe^{x^2}y = \frac{1}{4}x^4 + C $$

Solve for the general solution y(x)

Finally, we will solve for the general solution y(x) by dividing by the integrating factor \(\mu(x)\): $$ y(x) = \frac{1}{xe^{x^2}}(\frac{1}{4}x^4 + C) $$ So, the general solution of the given differential equation is: $$ y(x) = \frac{1}{4}x^3e^{-x^2} + Cxe^{-x^2} $$

Key concepts

General Solution

The general solution of a differential equation is a formula that includes all possible solutions for the equation. In the context of the problem, our task is to determine the form of solutions that fulfill the given differential equation. The solution often incorporates a constant, *C*, representing any number, which addresses the arbitrary nature of certain initial conditions.

For our given equation \( x y^{\prime}+(1+2x^2) y = x^3 e^{-x^2} \), the solution process forms such an expression, including \( C \), that satisfies the equation fully. This means the general solution \( y(x) = \frac{1}{4}x^3e^{-x^2} + Cxe^{-x^2} \) takes into account all potential specific solutions of the given problem. With this solution, by changing the value of \( C \), you can find a specific solution that may meet certain initial conditions.

Integrating Factor

The integrating factor is a crucial tool used to transform a non-exact differential equation into an exact one. An exact equation is easier to solve, as it can be represented as a total derivative. In this exercise, we found the integrating factor \( \mu(x) \) by taking the exponential of the integral of \( p(x) \), which was derived from dividing the differential equation into standard form.

This integrating factor appeared as \( \mu(x) = x e^{x^2} \). By utilizing this factor, we could multiply through the entire differential equation, making it possible to express the left-hand side as a derivative of a product. It then aligned the equation in such a form that enabled straightforward integration, paving the way to derive the general solution.

Standard Form

The standard form of a first-order linear differential equation simplifies the equation and makes it approachable for solving using techniques like integrating factors. To achieve this, we initially express the equation in the form \( y^{\prime} + p(x)y = r(x) \).

In our example, the original equation was \( x y^{\prime}+(1+2x^2) y = x^3 e^{-x^2} \). By dividing through by \( x \), we transformed it into the standard form: \( y^{\prime} + \frac{1+2x^{2}}{x}y = xe^{-x^2} \). This separation allowed us to identify \( p(x) = \frac{1+2x^2}{x} \) and \( r(x) = xe^{-x^2} \). This standardization is essential for leveraging the integrating factor method, which turns the non-exact equation into an exact one.

Mathematical Integration

Mathematical integration is the operation that essentially reverses differentiation, adding up parts to find whole quantities, especially in differential equations. Solving the differential equation involves various integrals. For instance, after applying the integrating factor, the equation \( \frac{d}{dx}(xe^{x^2} y) = x^3 \) required integrating both sides to find \( y \).

We solve \( \int \frac{d}{dx}(xe^{x^2} y)dx = \int x^3 dx \), using the fact that the left side is readily integrable as a complete derivative. This operation yields: \( xe^{x^2}y = \frac{1}{4}x^4 + C \).

Integration simplifies and consolidates the differential equation, allowing us to isolate and solve for \( y \). Hence, it is a vital part of deriving the general solution from transformed standard forms using exact equations.