Question

Find the general solution to the differential equation.$x^2{y}^{\prime }=(x+1)y$

Short answer

The general solution is $y=C\cdot x\cdot e^{-1/x}$.

Step-by-step solution

Convert to Standard Form

Rearrange the differential equation to the standard form of a first-order linear differential equation. The given differential equation is $x^2{y}^{\prime }=(x+1)y$. Divide both sides by $x^2$:$$y^{\prime }-\frac{x+1}{x^2}y=0$$This is now in the standard form $y^{\prime }+P(x)y=Q(x)$ with $P(x)=-\frac{x+1}{x^2}$ and $Q(x)=0$.

Find the Integrating Factor

The integrating factor $\mu (x)$ is given by:$$\mu (x)=e^{\int P(x)dx}=e^{\int -\frac{x+1}{x^2}dx}$$Break it down:$$\int -\frac{x+1}{x^2}dx=\int -(\frac{1}{x}+\frac{1}{x^2})dx$$Compute each integral separately:$$\int -\frac{1}{x}dx=-\ln |x|$$$$\int -\frac{1}{x^2}dx=\frac{1}{x}$$Thus,$$\mu (x)=e^{-\ln |x|+\frac{1}{x}}=\frac{e^{1/x}}{x}$$

Multiply the Differential Equation by the Integrating Factor

Multiply the entire differential equation by the integrating factor $\frac{e^{1/x}}{x}$:$$\frac{e^{1/x}}{x}{y}^{\prime }-\frac{e^{1/x}}{x^2}(x+1)y=0$$

Simplify and Solve the Differential Equation

Recognize that the left-hand side is the derivative of $y\cdot \frac{e^{1/x}}{x}$, thus:$${(y\cdot \frac{e^{1/x}}{x})}^{\prime }=0$$Integrate both sides with respect to $x$:$$y\cdot \frac{e^{1/x}}{x}=C$$Therefore, solving for $y$ gives us:$$y=C\cdot x\cdot e^{-1/x}$$

Verify the General Solution

The solution $y=C\cdot x\cdot e^{-1/x}$ satisfies the differential equation because differentiating $y$ and plugging the value back into the original equation confirms it's true. Thus, the general solution is verified.

Key concepts

Integrating Factor

One often encounters first-order linear differential equations in the form of $y^{\prime }+P(x)y=Q(x)$. To solve these, we use a method called the integrating factor. The integrating factor, usually denoted by $\mu (x)$, simplifies the differential equation in a way that allows us to solve it more easily. An integrating factor is calculated using the formula: $$\mu (x)=e^{\int P(x)dx}$$In this exercise, the integrating factor was derived from the standard form equation's term $P(x)=-\frac{x+1}{x^2}$. Calculating the integral: $$\int -(\frac{1}{x}+\frac{1}{x^2})dx$$ helped us find the appropriate integrating factor. Integrating separately, we get:

• $\int -\frac{1}{x}dx=-\ln |x|$
• $\int -\frac{1}{x^2}dx=\frac{1}{x}$

This results in the integrating factor: $$\mu (x)=\frac{e^{1/x}}{x}$$ Using $\mu (x)$ simplifies solving the differential equation.

General Solution of Differential Equation

The general solution provides an expression for all possible solutions of a differential equation. To determine it, we often utilize the integrating factor to simplify the equation. In this exercise, after finding the integrating factor, we multiply the entire differential equation by this factor: $$\frac{e^{1/x}}{x}{y}^{\prime }-\frac{e^{1/x}}{x^2}(x+1)y=0$$Recognizing that the left side of the equation is now the derivative of a product, we rewrite it as: $${(y\cdot \frac{e^{1/x}}{x})}^{\prime }=0$$Integrating both sides with respect to $x$, we solve to find: $$y\cdot \frac{e^{1/x}}{x}=C$$Solving for $y$ gives the general solution: $$y=C\cdot x\cdot e^{-1/x}$$where $C$ is a constant determined by initial or boundary conditions. This solution represents the family of curves that satisfy the differential equation.

Standard Form of Differential Equation

To solve a first-order linear differential equation, it's essential to convert it into the standard form, which is expressed as: $$y^{\prime }+P(x)y=Q(x)$$In our given exercise, the differential equation began as: $$x^2{y}^{\prime }=(x+1)y$$By dividing both sides by $x^2$, the equation was rearranged to match the standard form: $$y^{\prime }-\frac{x+1}{x^2}y=0$$Here, $P(x)=-\frac{x+1}{x^2}$ and $Q(x)=0$. This rearrangement is a crucial step as it sets the framework for using the integrating factor method. Adjusting the equation into this form is foundational for solving it systematically and accurately.