Question

Solve the following differential equations by power series and also by an elementary method. Verify that the series solution is the power series expansion of your other solution. $$x{y}^{\prime }-y=x^2$$

Short answer

The solution is $y=x^2+Cx$ where the series solution and the elementary method solution match when arbitrary constants are chosen appropriately.

Step-by-step solution

- Rewrite the Differential Equation

Start by rewriting the given differential equation in the standard form: $$x{y}^{\prime }-y=x^2$$ Divide every term by x to simplify: $$y^{\prime }-\frac{y}{x}=x$$

- Solve using Elementary Methods

This is a first-order linear differential equation. Use the integrating factor method: Compute the integrating factor (\mu): $$\mu =e^{\int -\frac{1}{x}dx}=e^{-\ln |x|}=\frac{1}{x}$$ Multiplying both sides of the differential equation by the integrating factor: $$\frac{1}{x}{y}^{\prime }-\frac{y}{x^2}=1$$ Notice now that the left-hand side is the derivative of $\frac{y}{x}$: $${\left(\frac{y}{x}\right)}^{\prime }=1$$ Integrate both sides with respect to x: $$\frac{y}{x}=\int 1dx=x+C$$ Solve for y: $$y=x^2+Cx$$

- Solve using Power Series

Assume a power series solution of the form $y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$. Differentiate term-by-term: $$y^{\prime }=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$$ Substitute these into the original differential equation: $$x\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=x^2$$ Simplify and match coefficients to solve for the $a_n$. $a_0$ and $a_1$ are arbitrary constants and $a_2=1$: The solution is: $$y=a_0+a_{1}x+x^2$$

- Verify the Series Solution Matches the Elementary Solution

Compare the series solution $y=a_0+a_{1}x+x^2$ with the elementary solution $y=Cx+x^2$. Notice they are the same since $a_0$ and $a_1$ can be chosen to fit the form of the elementary method solution.

Key concepts

Power Series Solution

Solving differential equations using a power series involves expressing the unknown function as a sum of powers of the independent variable, usually denoted as a Maclaurin or Taylor series. In our case, we assume a solution of the form: $$y=\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n$$ Differentiating term by term, we obtain: $$y^{\prime }=\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}$$ Substituting these into the original equation $x{y}^{\prime }-y=x^2$, we get: $$x\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^{n-1}-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=x^2$$ This simplifies to: $$\sum _{n=1}^{\mathrm{\infty }}n{a}_n{x}^n-\sum _{n=0}^{\mathrm{\infty }}{a}_n{x}^n=x^2$$ By matching coefficients, we can solve for the series coefficients $a_n$. The result is: $$y=a_0+a_{1}x+x^2$$ This demonstrates that a power series solution allows us to represent our differential equation solution as an infinite sum of terms, making it very powerful for understanding the function's behavior.

Integrating Factor Method

The integrating factor method is a technique used to solve first-order linear differential equations of the form: $$y^{\prime }+p(x)y=q(x)$$ The integrating factor $\mu (x)$ is defined as: $$\mu (x)=e^{\int p(x)dx}$$ Multiplying both sides of the differential equation by this factor transforms it into an exact equation, simplifying integration. For the differential equation $y^{\prime }-\frac{y}{x}=x$, we find: $$\mu (x)=e^{\int -\frac{1}{x}dx}=e^{-\ln |x|}=\frac{1}{x}$$ Multiplying through by $\frac{1}{x}$, we get: $$\frac{1}{x}{y}^{\prime }-\frac{y}{x^2}=1$$ Recognizing the left-hand side as a derivative, we rewrite it as: $${\left(\frac{y}{x}\right)}^{\prime }=1$$ Integrating both sides with respect to $x$, we find: $$\frac{y}{x}=x+C$$ So, solving for $y$, we obtain: $$y=x^2+Cx$$

First-Order Linear Differential Equation

A first-order linear differential equation has the general form: $$y^{\prime }+p(x)y=q(x)$$ It is called 'first-order' because it involves the first derivative $y^{\prime }$; and 'linear' because $y$ and its derivatives appear to the first power and are not multiplied together. Such equations often model various physical phenomena, from population dynamics to electrical circuits. In our specific example, $y^{\prime }-\frac{y}{x}=x$, it's evident how solving this differential equation using methods like the integrating factor makes it tractable. Recognizing and transforming such equations into an easily solvable form is a central skill in differential equations. Combining these solutions through power series or straightforward integration offers multiple perspectives to grasp the underlying function.