Question

Solving a First-Order Linear Differential Equation In Exercises \(5-14,\) solve the first-order linear differential equation. $$ y^{\prime}+2 x y=10 x $$

Short answer

General solution is \(y(x)= \frac{1}{\exp(x^{2})}\int 10x\exp(x^{2}) dx\) which require further techniques to calculate the remaining integral.

Step-by-step solution

Identifying the form of the differential equation

The given differential equation is \(y^{\prime}+2 x y = 10 x\). This is a first-order linear differential equation. It is in the standard linear form \(y'+p(x)y = g(x)\) where here, \(p(x) = 2x\) and \(g(x) = 10x\).

Calculating the integrating factor

The integrating factor is calculated by exponentiating the integral of \(p(x)\). So, the integrating factor \(I(x) = \exp\left(\int p(x) dx\right) = \exp\left(\int 2x dx\right) = \exp(x^{2})\).

Multiplying the differential equation by the integrating factor

Multiply every term in the differential equation by the integrating factor, we get \(\exp(x^{2})y' + 2x\exp(x^{2})y = 10x\exp(x^{2})\).

Recognize the left-hand side as a derivative

The left-hand side of this equation can be rewritten as \((\exp(x^{2})y)'\). This can be verified by applying the product rule of differentiation.

Integrating both sides

We integrate both sides of the equation with respect to \(x\): \(\int (\exp(x^{2})y)' dx = \int 10x\exp(x^{2}) dx\). By the fundamental theorem of calculus, the left side simplifies to \(\exp(x^{2})y\). The integral on the right is non-trivial and cannot be expressed in terms of elementary functions.

Solving for y

We can solve for y by dividing every term by the integrating factor, leading to \(y(x)= \frac{1}{\exp(x^{2})}\int 10x\exp(x^{2}) dx\). The remaining challenge is to evaluate the remaining integral, which is an exercise left to the reader.

Key concepts

Integrating Factor

To solve a first-order linear differential equation like \( y' + 2xy = 10x \), we use an important tool called the integrating factor. The integrating factor helps to transform the differential equation into a simpler form that we can solve more easily.

Here's how it works:

• First, identify \( p(x) \) from the linear form \( y' + p(x)y = g(x) \). In this equation, \( p(x) = 2x \).
• Next, we calculate the integrating factor, denoted by \( I(x) \), using the formula \( I(x) = \exp(\int p(x) \, dx) \).
• For our example, \( I(x) = \exp(\int 2x \, dx) = \exp(x^2) \).

The integrating factor simplifies the process of solving these equations by allowing us to multiply through and rewrite the equation in a more manageable form. Once the integrating factor is applied, the equation becomes easier to integrate, helping us to find the solution for \( y \).

Linear Differential Equations

A linear differential equation is a type of equation that describes the relationship between a function and its derivatives. These equations are essential in modeling various real-world phenomena such as population growth or heat transfer.

Our main focus here is on first-order linear differential equations, which can be written in the general form \( y' + p(x)y = g(x) \). Here, \( y' \) represents the derivative of \( y \). Such equations can be solved using systematic methods that often involve finding an integrating factor.

• The given equation, \( y' + 2xy = 10x \), fits the linear form with \( p(x) = 2x \) and \( g(x) = 10x \).
• Recognizing this form is the first step to correctly apply further techniques like integrating factors to find the general solution.

These types of differential equations are prevalent in mathematical modeling and offer a structured approach to finding solutions that are otherwise challenging to derive.

Integration Techniques

Integration techniques become crucial after transforming a differential equation using the integrating factor. Once we have the equation in the form \( (\exp(x^2)y)' = 10x\exp(x^2) \), we need to integrate both sides to find the solution.

Here's how integration is applied:

• Integrate the left-hand side: The expression \( (\exp(x^2)y)' \) simplifies directly to \( \exp(x^2)y \) thanks to the fundamental theorem of calculus.
• Integrate the right-hand side: This involves finding the integral \( \int 10x\exp(x^2) \, dx \), which is more complex and doesn’t result in elementary functions.

Integration is about finding the anti-derivative, and with practice, you learn various methods to deal with complex integrals. Some integrals, like the one on the right here, might require numerical methods or special functions for explicit evaluation.

Product Rule of Differentiation

The product rule of differentiation is an essential concept when dealing with differential expressions. It states how to differentiate products of two functions. This rule is pivotal in recognizing the left-hand side of the equation as a derivative.

Consider the expression \( (\exp(x^2)y)' \). Based on the solution, we apply the product rule:

• For two functions \( u(x) = \exp(x^2) \) and \( v(x) = y \), the differential is \( u'v + uv' \).
• Here, \( u' = \frac{d}{dx}[\exp(x^2)] \) and \( v' = y' \).

By reconstructing \( (\exp(x^2)y)' \) using the product rule, we verify the transformation and ensure the left-hand side correctly represents the differential form. Understanding this principle helps in rearranging terms appropriately for integration and solution derivation.