The equation is separable:
[tex]\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{x^2+x}{\sqrt x}\implies \mathrm dy=\left(x^{3/2}+x^{1/2}\right)\,\mathrm dx[/tex]
Integrate both sides to get
[tex]y=\dfrac25x^{5/2}+\dfrac23x^{3/2}+C[/tex]
Given that [tex]y(1)=0[/tex], we find
[tex]0=\dfrac25+\dfrac23+C\implies C=-\dfrac{16}{15}[/tex]
so the IVP has the solution
[tex]\boxed{y(x)=\dfrac25x^{5/2}+\dfrac23x^{3/2}-\dfrac{16}{15}}[/tex]
