Question

dy / (3x ^ 2) = (1 + y ^ 2) ^ 1.5 * d * x

Answer 1

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

Integrate both sides:

$\intop \frac{dy}{(1+y^2)^{\frac{3}{2}}}=\intop3x^2.dx$

Put $y=tan \theta \\$

$\therefore dy=sec^2 \theta.d\theta$

$\intop \frac{sec^2\theta }{(1+tan^2 \theta )^{\frac{3}{2}}} d\theta =x^3+c$

$\Rightarrow \intop \frac{sec^2\theta}{sec^3\theta} d\theta =x^3+c\\ \Rightarrow \intop cos\theta d\theta=x^3+c\\ \Rightarrow sin\theta=x^3+c\\ \Rightarrow \frac{y}{\sqrt{y^2+1}}=x^3+c$