[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{3x^2+4x}y\iff y\,\mathrm dy=(3x^2+4x)\,\mathrm dx[/tex]
[tex]\implies\displaystyle\int y\,\mathrm dy=\int(3x^2+4x)\,\mathrm dx[/tex]
[tex]\implies \dfrac12y^2=x^3+2x^2+C[/tex]
When [tex]x=1[/tex] you have [tex]y=\sqrt{10}[/tex], so
[tex]\dfrac12(\sqrt{10})^2=1^3+2(1)^2+C\implies 5=1+2+C\implies C=2[/tex]
and so the particular solution to the ODE is
[tex]\dfrac12y^2=x^3+2x^2+2[/tex]
Then when [tex]x=0[/tex], you get
[tex]\dfrac12y^2=0^3+2(0)^2+2=2[/tex]
[tex]\implies y^2=4[/tex]
[tex]\implies y=2[/tex]
where we omitted the negative root because it's given that [tex]y>0[/tex].
