Question

Find an explicit solution of the given initial - value problem . x ^ 2 * (dy)/(dx) = y - xy , y(- 1) = - 2

Answer 1

The ODE,

$x^2\dfrac{\mathrm dy}{\mathrm dx}=y-xy$

is separable, since we can write

$\dfrac{\mathrm dy}y=\dfrac{1-x}{x^2}\,\mathrm dx$

Integrate both sides to get

$\ln|y|=-\dfrac1x-\ln|x|+C$

Solve for y :

$y=e^{-\frac1x-\ln|x|+C}$

$y=C\dfrac{e^{-\frac1x}}x$

Use the initial condition to solve for C :

$-2=C\dfrac{e^{-\frac1{-1}}}{-1}=-Ce\implies C=\dfrac2e$

So the particular solution to the IVP is

$y(x)=\dfrac{2e^{-\frac1x}}{ex}=\dfrac{2e^{-\frac1x-1}}x$

Answer 2

The explicit solution is: $\ln\left|y\right|=-\frac{1}{x}-\ln\left|x\right|+\ln\left(2\right)-1$

The given differential equation is: $x^{2}\cdot\frac{dy}{dx}=y-xy$

We will use separation of variable to solve it.

$x^{2}\cdot\frac{dy}{dx}=y-xy\\ x^{2}\cdot\frac{dy}{dx}=y\left(1-x\right)\\ \frac{1}{y}dy=\frac{\left(1-x\right)}{x^{2}}dx\\ \int_{ }^{ }\frac{1}{y}dy=\int_{ }^{ }\frac{\left(1-x\right)}{x^{2}}dx\\ \int_{ }^{ }\frac{1}{y}dy=\int_{ }^{ }\left(\frac{1}{x^{2}}-\frac{1}{x}\right)dx\\ \ln\left|y\right|=-\frac{1}{x}-\ln\left|x\right|+C$

Given that y(- 1) = - 2.

$\ln\left|-2\right|=-\frac{1}{-1}-\ln\left|-1\right|+C\\ \ln\left(2\right)=1-0+C\\ \ln\left(2\right)-1=C$

So the explicit solution is: $\ln\left|y\right|=-\frac{1}{x}-\ln\left|x\right|+\ln\left(2\right)-1$

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