Solve xy' + 2y = \frac{1}{x} y^\frac{2}{3}.


Solve {eq}xy' + 2y = \frac{1}{x} y^\frac{2}{3}. {/eq}

Solution of the Differential Equation:

The given differential equation is first order differential equation. A differential equation in the form {eq}y{}'+P(x)y=Q(x)y^n,\,(n\neq 0,1) {/eq} is called a Bernoulli differential equation. After some substitution the given differential equation convert in first order linear differential equation. We compare this differential equation with the differential equation {eq}\displaystyle \frac{\mathrm{d} y}{\mathrm{d} x}+P(x)y=Q(x) {/eq} and find the values of the function {eq}P(x) {/eq} and {eq}Q(x) {/eq} which is function of {eq}x. {/eq} To find the solution of the differential equation we find the integrating factor of the differential equation and then the solution of the differential equation will be written in the form of

{eq}\displaystyle y\times \textbf{I.F}=\int Q(t)\times \textbf{I.F}dt+c {/eq}

Where, {eq}c {/eq} is a constant of integration.

Answer and Explanation:

Consider the differential equation

{eq}\displaystyle xy' + 2y = \frac{1}{x} y^\frac{2}{3}\\ \displaystyle y' + \frac{2}{x}y = \frac{1}{x^{2}}...

See full answer below.

Become a member to unlock this answer! Create your account

View this answer

Learn more about this topic:

First-Order Linear Differential Equations

from GRE Math: Study Guide & Test Prep

Chapter 16 / Lesson 3

Related to this Question

Explore our homework questions and answers library