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

## Question:

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.

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}}...

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