Let x=x(y) be the solution of the differential equation [y^2,dx+(x-1y)dy=0, x(1)=1.] Then …

Question

Let $x=x(y)$ be the solution of the differential equation
$$y^2\,dx+\left(x-\frac1y\right)dy=0,\qquad x(1)=1.$$
Then $x\!\left(\dfrac12\right)$ is:

Accepted Answer

$$\frac{dx}{dy}+\frac{x}{y^2} = \frac{1}{y^3}.$$

I.F. $=e^{-1/y}$.
$$xe^{-1/y} = \int e^{-1/y}\frac{1}{y^3}\,dy.$$

Substituting $-1/y=t$ and integrating by parts gives
$$xe^{-1/y} = e^{-1/y}\left(\frac1y+1\right)+C.$$

Using $x(1)=1$: $e^{-1}=2e^{-1}+C \Rightarrow C=-e^{-1}$.

$$x = \frac1y+1-e^{-1+1/y}.$$

At $y=1/2$: $x=2+1-e^{-1+2}=3-e$.

Solutions · 1