Chapter 1, Problem 1RP

Program Plan Intro

Program Description: Purpose ofproblem is to calculate the general solution of the differential equation $x^3+3y-x{y}^{\prime }=0$ .

Explanation of Solution

Given information:

The differential equation is $x^3+3y-x{y}^{\prime }=0$ .

Explanation:

The given differential equation can be written as,

  $\begin{array}{c}x\frac{dy}{dx}-3y=x^3\\ \frac{dy}{dx}-\frac{3}{x}y=\frac{x^3}{x}\\ \frac{dy}{dx}-\frac{3}{x}y=x^2\end{array}$

The above equation is a linear first order differential equation and its solution is calculated as,

  $\begin{array}{c}y{e}^{-3{\displaystyle \int \frac{1}{x}dx}}={\displaystyle \int x^2{e}^{-{\displaystyle \int \frac{3}{x}dx}}dx}\\ y{e}^{-3\ln x}={\displaystyle \int x^2{e}^{-3\ln x}dx}\\ y{e}^{\ln x^{-3}}={\displaystyle \int x^2{e}^{\ln x^{-3}}dx}\\ y{x}^{-3}={\displaystyle \int x^2{x}^{-3}dx}\end{array}$

Simplify the above equation.

  $\begin{array}{c}y{x}^{-3}={\displaystyle \int \frac{1}{x}dx}\\ y{x}^{-3}=C+\ln x\\ y=x^3\left(C+\ln x\right)\end{array}$

Conclusion:

Thus, the solution of the differential equation $x^3+3y-x{y}^{\prime }=0$ is $y=x^3\left(C+\ln x\right)$ .