Q.2 Solve the Initial-value problem (dy)/(dx) = \frac{ {x}^(2) + x }{ √(x) } . \: y(1) = 0 ​

Question

Q.2 Solve the Initial-value problem


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


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Answer 1

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)}`