Question

Given y'(x) =
`√(-4y(x) + 28)` and y(-2)=-2, find y(x)?

Answer 1

I saw this question the first time around missing those parentheses around the
`x`, which made the ODE much, much more difficult to solve. This is doable, especially because this ODE is separable.


`(\mathrm dy)/(\mathrm dx)=√(-4y+28)`

`(\mathrm dy)/(2√(7-y))=\mathrm dx`

`\displaystyle\frac12\int(\mathrm dy)/(√(7-y))=\int\mathrm dx`

`-√(7-y)=x+C`

`√(7-y)=-x+C`

`7-y=(C-x)^2`

`y=7-(C-x)^2`

Given that
`y(-2)=-2`, we have


`-2=7-(C+2)^2`

`\implies C=-5,C=1`

So there are two possible particular solutions:


`\begin{cases}y=-x^2-10x-18\\y=-x^2+2x+6\end{cases}`