Question
If $x^{3}+y^{3}-\operatorname{sex} y=0$, then $d^{2} y / d x^{2}$ at $(3 a / 2,3 e / 2)=\ldots \ldots$
Step 1
We have $x^3 + y^3 - \operatorname{sex} y = 0$. Differentiating both sides with respect to x, we get: $3x^2 + 3y^2 \frac{dy}{dx} - \operatorname{sex} \frac{dy}{dx} = 0$ Show more…
Show all steps
Your feedback will help us improve your experience
Adriano Chikande and 90 other Calculus 3 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
$\frac{d y}{d x}+\frac{3 y}{x}=3 x^{2} ; \quad y=\frac{x^{3}}{2}+\frac{c}{x^{3}}$
$$ x \frac{\mathrm{d} y}{\mathrm{~d} x}-y=x^{3}+3 x^{2}-2 x $$
First-order differential equations
Further problems F.25
$$ x \frac{d y}{d x}-y=x^{3}+3 x^{2}-2 x $$
Further problems
Transcript
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD