If e^y+xy=e, the ordered pair (dydx,d^2ydx^2) at x=0 is equal to:

Question

If $e^y+xy=e$, the ordered pair $\left(\dfrac{dy}{dx},\dfrac{d^2y}{dx^2}\right)$ at $x=0$ is equal to:

Accepted Answer

At $x=0$: $e^y=e\Rightarrow y=1$.

Differentiating: $e^yy'+xy'+y=0$. At $(0,1)$: $ey'+1=0\Rightarrow y'=-\dfrac1e$.

Differentiating again: $e^y(y')^2+e^yy''+xy''+2y'=0$. At $(0,1)$ with $y'=-\dfrac1e$: $e\cdot\dfrac{1}{e^2}+ey''-\dfrac2e=0\Rightarrow ey''=\dfrac2e-\dfrac1e=\dfrac1e\Rightarrow y''=\dfrac{1}{e^2}$.

Solutions · 1