Question

Determine

d²y

dx²

, if 2x³ − 3y² + 7xy = 0

Answer 1

$2x3 − 3y2 + 7xy = 0$

$\implies2\times 3x^2-3\times 2y\frac{dy}{dx}+7(\frac{dy}{dx} +y)=0$

$6x^2-6y\frac{dy}{dx}+7x\frac{dy}{dx}+7y=0\implies\frac{dy}{dx}=\frac{6y-7x}{6x^2+7y}$

$12x-6(yy\prime \prime+y\prime^2)+7(xy\prime\prime+y\prime)+7y\prime=0$

$\implies12x-6yy\prime \prime-6y\prime^2+7xy\prime\prime+7y\prime+7y\prime=0$

$\implies y\prime \prime(7x-6y)=6y\prime^2-12x-14y\prime$

$\implies y\prime \prime=\frac{6y\prime ^2-12x-14y\prime}{7x-6y}$

$\implies y\prime \prime =\frac{6(\frac{6y-7x}{6x^2+7y})-12x-14(\frac{6y-7x}{6x^2+7y})}{7x-6y}$