## Related questions with answers

Find the general solution. of the following equations. Express the solution explicitly as a function of the independent variable. $x^{2} y^{\prime}(x)=y^{2}$

Solution

VerifiedIt is given that

$x^2 y^\prime(x)=y^2$

Let us rewrite this equation in a more suitable form:

$\begin{align*} x^2 y^\prime(x)&=y^2\\ x^2 \frac{\, dy}{\, dx}&=y^2\\ \frac{x^2}{y^2} \frac{dy}{dx}&=1\\ \frac{dy}{y^2}&=\frac{dx}{x^2} \end{align*}$

Integrating both sides of the equation we obtain:

$\begin{align*} \int \frac{dy}{y^2}&=\int \frac{dx}{x^2}\\ \int y^{-2}\, dy&=\int x^{-2}\, dx\\ \frac{y^{-2+1}}{-2+1}&=\frac{x^{-2+1}}{-2+1}+C\\ -\frac{1}{y}&=-\frac{1}{x}+C\\ -\frac{1}{y}&=\frac{-1+Cx}{x}\\ -y&=\frac{x}{-1+Cx}\\ y&=\frac{x}{1-Cx}\Rightarrow y=\frac{x}{1+Cx} \end{align*}$

We conclude that the $\textsf{general solution}$ of the given differential equation is

$\boxed{y=\frac{x}{1+Cx}}$

where $C$ is an arbitrary constant.

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