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Q.2 Solve the Initial-value problem


(dy)/(dx) = \frac{ {x}^(2) + x }{ √(x) } . \: y(1) = 0




1 Answer

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The equation is separable:


(\mathrm dy)/(\mathrm dx)=(x^2+x)/(\sqrt x)\implies \mathrm dy=\left(x^(3/2)+x^(1/2)\right)\,\mathrm dx

Integrate both sides to get


y=\frac25x^(5/2)+\frac23x^(3/2)+C

Given that
y(1)=0, we find


0=\frac25+\frac23+C\implies C=-(16)/(15)

so the IVP has the solution


\boxed{y(x)=\frac25x^(5/2)+\frac23x^(3/2)-(16)/(15)}

User Geremy
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