Question

Find the general solution. of the following equations. Express the solution explicitly as a function of the independent variable. x2y(x)=y2x^{2} y^{\prime}(x)=y^{2}

Solution

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It is given that

x2y(x)=y2x^2 y^\prime(x)=y^2

Let us rewrite this equation in a more suitable form:

x2y(x)=y2x2dydx=y2x2y2dydx=1dyy2=dxx2\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:

dyy2=dxx2y2dy=x2dxy2+12+1=x2+12+1+C1y=1x+C1y=1+Cxxy=x1+Cxy=x1Cxy=x1+Cx\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 general solution\textsf{general solution} of the given differential equation is

y=x1+Cx\boxed{y=\frac{x}{1+Cx}}

where CC is an arbitrary constant.

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