Let y=y(x) solve x^3,dy+(xy-1),dx=0, x0, y(12)=3-e.

Question

Let $y=y(x)$ solve $x^3\,dy+(xy-1)\,dx=0$, $x>0$, $y\left(\dfrac12\right)=3-e$. Then $y(1)$ is:

Accepted Answer

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

I.F. $=e^{-1/x}$.
$$ye^{-1/x} = \int e^{-1/x}\frac{1}{x^3}\,dx.$$

With $t=-1/x$: this integral is $-\int te^t\,dt = -(te^t-e^t)$, giving
$$y = \frac1x+1+Ce^{1/x}.$$

Using $y(1/2)=3-e$: $3-e=2+1+Ce^2 \Rightarrow C=-\dfrac1e$.

At $x=1$: $y=1+1-\dfrac1e\cdot e = 1$.

Solutions · 1