Separation of Variables Question 2

If $\dfrac{dy}{dx}=\dfrac{3x^2+5}{y} $ and $y=4 $ when $x= 1$, then when $x=2 $, $y=$

$4$

$40$

$\pm\sqrt{40}$

$\pm\sqrt{66}$

Approach

$$ \frac{dy}{dx}=\dfrac{3x^2+5}{y} $$ $$ y dy=3x^2+5 dx $$ $$ \int y dy= \int 3x^2+5 dx $$ $$ \frac{1}{2}y^2=x^3+5x+C $$ $$ y^2=2x^3+10x+C $$

Using our initial condition:

$$ (4)^2=2(1)^3+10(1)+C $$ $$ 16=2+10+C $$ $$ C=4 $$

Rewriting and substituting $x=2$:

$$ y^2=2(2)^3+10(2)+4 $$ $$ y^2 = 16+20 +4 $$ $$ y=\boxed{\pm\sqrt{40}}$$