Question

Find the general solution to the differential equation \((x-2) y^{\prime}+y=3 x^{2}+2 x\). Assume \(x>2\)

Short answer

The general solution is \(y = \frac{x^3 + x^2 + C}{x-2}\).

Step-by-step solution

Identify the type of differential equation

The given differential equation \[(x-2) y^{\prime}+y=3x^{2}+2x\]is a first-order linear differential equation in the standard form: \[a(x) y' + b(x) y = c(x)\]where \(a(x) = x-2\), \(b(x) = 1\), and \(c(x) = 3x^2 + 2x\).

Rewrite in standard form

Divide the entire equation by \(x-2\) to obtain the standard form:\[y' + \frac{1}{x-2}y = \frac{3x^2+2x}{x-2}\].

Find the integrating factor

Calculate the integrating factor \(\mu(x)\) using the formula: \[\mu(x) = e^{\int \frac{1}{x-2} \, dx} = e^{\ln|x-2|} = |x-2|\].Since \(x>2\), \(|x-2| = x-2\), so the integrating factor is \(x-2\).

Multiply equation by the integrating factor

Multiply the entire differential equation by the integrating factor \((x-2)\):\[(x-2) \left(y' + \frac{1}{x-2}y\right) = (x-2) \cdot \frac{3x^2+2x}{x-2}\] which simplifies to:\[(x-2)y' + y = 3x^2 + 2x\].

Integrate both sides

Recognize the left side as the derivative of \((x-2)y\):\[ \frac{d}{dx}((x-2)y) = 3x^2 + 2x\] Integrate both sides with respect to \(x\):\[(x-2)y = \int (3x^2 + 2x)\, dx\].

Solve the integration

Perform the integration:\[\int (3x^2 + 2x)\, dx = x^3 + x^2 + C\]where \(C\) is the constant of integration.

Solve for \(y\)

Express \(y\) in terms of \(x\):\[y = \frac{x^3 + x^2 + C}{x-2}\].This is the general solution to the differential equation.

Key concepts

Standard form of differential equations

First-order linear differential equations are a foundational element in calculus. They appear in the form of \[ a(x) y' + b(x) y = c(x) \]where:

• \(a(x)\) is a function of \(x\). In our exercise, this is \(x-2\).
• \(b(x)\) is another function of \(x\), here it is simply \(1\).
• \(c(x)\) is a function representing the "right-hand side" of the equation, such as \(3x^2 + 2x\).

The goal is to rewrite any given differential equation into this standard form. To do so, first recognize the roles of each function involved and second, perform simple algebraic manipulations like division or multiplication. Thiswill help create a clear path for applying methods to find solutions.

Integrating factor method

The integrating factor method simplifies solving first-order linear differential equations. Once an equation is in standard form, finding the integrating factor \(\mu(x)\) is the next step. This factor helps transform the differential equation into an easily integrable form.
Calculate the integrating factor using the formula:\[ \mu(x) = e^{\int \frac{b(x)}{a(x)} \, dx} \] In this exercise, \(b(x) = 1\) and \(a(x) = x-2\) giving us:\[ \mu(x) = e^{\int \frac{1}{x-2} \, dx} = e^{\ln|x-2|} = |x-2| \] Since it is given \(x>2\), we simplify it to \(x-2\).
The next step is multiplying the entire original equation by this factor, turning it into a straightforward equation where the left side becomes the derivative of another function, making the equation integrable with respect to \(x\).

General solution of differential equations

The general solution of a differential equation includes all possible particular solutions with an arbitrary constant of integration. For our example, after simplifying using the integrating factor, we reach an equation of the form:\[ \frac{d}{dx}((x-2)y) = 3x^2 + 2x \] Solving this requires integrating both sides with respect to \(x\).
Once integrated, it becomes:\[ (x-2)y = \int (3x^2 + 2x)\, dx = x^3 + x^2 + C \]Finally, isolate \(y\) to express the solution:\[ y = \frac{x^3 + x^2 + C}{x-2} \]This gives us the general solution, where:

• \(C\) is an arbitrary constant that accounts for the family of solutions satisfying the differential equation.

The provided solution is general because it encompasses all values of \(C\), demonstrating the flexibility of solutions.