# solve xy'+y=x^4y^2 let z=y^-2 and convert to linear i.e \frac{dz}{dx} +p^2(x) = Q^2(x)

## Question:

solve {eq}xy'+y=x^4y^2 {/eq}

let {eq}z=y^-2{/eq} and convert to linear i.e {eq}\frac{dz}{dx} +p^2(x) = Q^2(x){/eq}

## Linear first order differential equation:

A first-order linear differential equation is one that can be written in the form

{eq}\frac{{dy}}{{dx}} + p(x)y = q(x), {/eq}

where {eq}p(x) {/eq} and {eq}q(x) {/eq} are continuous functions of {eq}x {/eq}.

First we find integrating factor for the first order linear differential equation

I.F. = {eq}{e^{\int p (x)dx}} {/eq}.

Solution of first order linear differential equation is given by

{eq}y.{e^{\int p (x)dx}} = \int {q(x).{e^{\int p (x)dx}}dx + c} {/eq}.

where {eq}c {/eq} is constant of integration.

We use the following formula of integration:

{eq}\int {{x^n}} dx = \frac{{{x^{n+1}}}}{{n + 1}} + c, {/eq}

where {eq}c {/eq} is constant of integration.

## Answer and Explanation:

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View this answerGiven D.E. {eq}\displaystyle xy'+y=x^4y^2, {/eq}

we can re-write above differential equation as,

{eq}\displaystyle \eqalign{ & \frac{{dy}}{{dx}} +...

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Chapter 16 / Lesson 3In this lesson you'll learn how to solve a first-order linear differential equation. We first define what such an equation is, and then we give the algorithm for solving one of that form. Specific examples follow the more general description of the method.