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I have been doing some self study on differential equations using Tom Apostol's Calculus Vol. 1. and I got stuck on a problem (problem 12, section 8.5, vol. 1).

Let K be a non zero constant. Suppose P and Q are continuous in an open interval I.

Let [tex] a\in I[/tex] and b a real number. Let [tex]v=g(x)[/tex] be the only solution to the initial value problem [tex]v' +kP(x)v=kQ(x)[/tex] in I, with [tex]g(a)=b[/tex]. If [tex] n\not= 1[/tex] and [tex]k=1-n[/tex], prove the function [tex] y=f(x)[/tex] not identically zero in I, is a solution to [tex] y'+P(x)y=Q(x)y^n[/tex] and [tex]{f(a)}^k=b[/tex] in I, if and only if the k-th power of [tex]f[/tex] is equal to [tex]g[/tex] in I.

The general solution of a linear first order differential equation is probably used.

I have tried several manipulations and substitutions, but I am kinda lost. I would appreciate any help.

## Homework Statement

Let K be a non zero constant. Suppose P and Q are continuous in an open interval I.

Let [tex] a\in I[/tex] and b a real number. Let [tex]v=g(x)[/tex] be the only solution to the initial value problem [tex]v' +kP(x)v=kQ(x)[/tex] in I, with [tex]g(a)=b[/tex]. If [tex] n\not= 1[/tex] and [tex]k=1-n[/tex], prove the function [tex] y=f(x)[/tex] not identically zero in I, is a solution to [tex] y'+P(x)y=Q(x)y^n[/tex] and [tex]{f(a)}^k=b[/tex] in I, if and only if the k-th power of [tex]f[/tex] is equal to [tex]g[/tex] in I.

## Homework Equations

The general solution of a linear first order differential equation is probably used.

## The Attempt at a Solution

I have tried several manipulations and substitutions, but I am kinda lost. I would appreciate any help.

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