Question

given y>0 and dy/dx=(3x^2+4x)/yif the point (1,√10) is on the graph relating x and y, then what is y when x=0

Answer 1

$\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2+4x}y\iff y\,\mathrm dy=(3x^2+4x)\,\mathrm dx$
$\implies\displaystyle\int y\,\mathrm dy=\int(3x^2+4x)\,\mathrm dx$
$\implies \dfrac12y^2=x^3+2x^2+C$

When $x=1$ you have $y=\sqrt{10}$, so

$\dfrac12(\sqrt{10})^2=1^3+2(1)^2+C\implies 5=1+2+C\implies C=2$

and so the particular solution to the ODE is

$\dfrac12y^2=x^3+2x^2+2$

Then when $x=0$, you get

$\dfrac12y^2=0^3+2(0)^2+2=2$
$\implies y^2=4$
$\implies y=2$

where we omitted the negative root because it's given that $y>0$.