## Calculus 10th Edition

Published by Brooks Cole

# Chapter 6 - Differential Equations - 6.1 Exercises: 4

#### Answer

$y^2 -2\ln{y} =x^2$ is a solution to the differential equation

#### Work Step by Step

Start by taking the derivative of the Solution $y^2 -2\ln{y} =x^2$ $2y\frac{dy}{dx} - 2\frac{1}{y} \frac{dy}{dx}= 2x$ Then begin simplifying $2(y-\frac{1}{y}) \frac{dy}{dx} =2x$ , divide both sides by two and the $(y- y^{-1})$ $\frac{dy}{dx}= \frac{x}{y-\frac{1}{y}}$, then take a $\frac{1}{y}$ factor out of the denominator and put it in the numerator $\frac{dy}{dx}= \frac{xy}{y^2 -1}$ This satisfies the differential equation, so it is a solution

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