Question

Find the general solution to each differential equation. $$\frac{d y}{d x}+x y=2 x$$

Short answer

The general solution of the differential equation \( \frac{d y}{d x} + x y = 2 x \) is \( y = 1 + Ce^{-\frac{1}{2} x^2} \) where \( C \) is an arbitrary constant.

Step-by-step solution

Recognize the Type of Differential Equation

Identify the differential equation given as a first-order linear differential equation, which can be written in the standard form \( \frac{d y}{d x} + P(x) y = Q(x) \). Here, \( P(x) = x \) and \( Q(x) = 2x \) respectively.

Calculate the Integrating Factor

Calculate the integrating factor, which is \( e^{\int P(x) \, dx} \). In this case, it is \( e^{\int x \, dx} = e^{\frac{1}{2} x^2} \) since the integral of \( x \) is \( \frac{1}{2} x^2 \)

Multiply Through by the Integrating Factor

Multiply the entire differential equation by the integrating factor to obtain \( e^{\frac{1}{2} x^2}(\frac{d y}{d x} + x y) = e^{\frac{1}{2} x^2}(2x) \) which simplifies to \( \frac{d}{d x}(e^{\frac{1}{2} x^2} y) = 2x e^{\frac{1}{2} x^2} \) due to the product rule.

Integrate Both Sides

Integrate both sides with respect to \( x \): \(\int \frac{d}{d x}(e^{\frac{1}{2} x^2} y) \, dx = \int 2x e^{\frac{1}{2} x^2} \, dx \). The left side simplifies to \( e^{\frac{1}{2} x^2} y \) and the right side requires a u-substitution to integrate.

Solve the Integral on the Right

Use the substitution \( u = \frac{1}{2}x^2 \) with \( du = x dx \). Hence, the integral becomes \( \int e^u du \) which is \( e^u \), and upon back substitution, we get \( e^{\frac{1}{2} x^2} \).

Apply the Constant of Integration

After integrating and applying the constant of integration, we have \( e^{\frac{1}{2} x^2} y = e^{\frac{1}{2} x^2} + C \) where \( C \) is the constant of integration.

Solve for \( y \) (General Solution)

Isolate \( y \) to find the general solution, \( y = 1 + Ce^{-\frac{1}{2} x^2} \) where \( C \) is the constant of integration representing the family of solutions.

Key concepts

Integrating Factor

In the world of differential equations, solving them can sometimes feel like deciphering a secret code. An 'integrating factor' is a clever tool that simplifies this process. Imagine you're trying to untangle a knotted string. The integrating factor is like a magic dust that, when sprinkled over the knot, loosens it, making it easier to undo. For a first-order linear differential equation like \( \frac{d y}{d x} + P(x) y = Q(x) \), the integrating factor is calculated as \( e^{\int P(x) \, dx} \).

Think of \( P(x) \) as a part of the tangled knot and the integrating factor as something that transforms the equation into a form that's easier to solve. It's derived from the original equation and once determined, it's like the whole equation gets multiplied by this factor, changing its form but not its essence, which leads to a smoother path towards the solution. It's a bit like using a key to open a locked door, providing direct access to the heart of the problem.

Differential Equation Solution

Finding the solution to a differential equation is akin to completing a maze. There are twists, turns, and sometimes dead ends, but eventually, with the right strategy, you can reach the end. The solution to a 'differential equation' represents a function or a set of functions that satisfy the given equation. In our exercise, the aim was to solve the first-order linear differential equation \( \frac{d y}{d x}+ xy = 2x \). By obtaining an integrating factor and strategically multiplying it through the equation, we unraveled a clearer path.

The process transformed the equation into a perfect differential, allowing for straightforward integration. Upon integrating and applying the constant of integration, we arrived at a general solution, which represents an infinite number of possible functions, each corresponding to a different initial condition or value of the constant \( C \). This result is profound as it maps out the entire landscape of solutions for the original problem posed by the equation.

U-Substitution Integration

When you encounter an integral that resembles a knotty expression and it's hard to see a clear way forward, 'u-substitution integration' becomes a valuable technique. It's like solving a puzzle by rearranging the pieces to form a picture you can recognize. In math terms, you pick a part of the integral, call it \( u \), and then rewrite the integral in terms of \( u \). This simplification can often turn a daunting integral into a more friendly one.

In our step-by-step solution, we employed u-substitution for the integral on the right side, setting \( u = \frac{1}{2}x^2 \) which corresponded with \( du = x dx \). The complex-looking integral transformed into the much simpler \( \int e^u du \), which is a basic integral that equals \( e^u \), and this leads us to our solution upon reversing the substitution. U-substitution is like finding the hidden path in a dense forest; it's always there, but without the right perspective, you might never find it.