Question

How to you find the general solution of `dy/dx=xsqrt(5-x)`?

Answer 1

`y = 2/5(5-x)^(5/2) - 10/3(5-x)^(3/2) + C`, Provided `x<=5`

Explanation:

We have `dy/dx=xsqrt(5-x)` which is a First Order Separable Differential Equation. We can therefore e separable the variables, as follows;

`int dy = int xsqrt(5-x)dx`

And so `y = int xsqrt(5-x)dx + C`

For the RHS integral, Let `u=5-x => x=5-u`
And, `(du)/dx=-1 => int ... du = -int ... dx`

So `int xsqrt(5-x)dx = int (5-u)sqrt(u)(-1)du`
`:. int xsqrt(5-x)dx = int (5u^(1/2)-u^(3/2))(-1)du`
`:. int xsqrt(5-x)dx = int (u^(3/2)-5u^(1/2))du`
`:. int xsqrt(5-x)dx = u^(5/2)/(5/2) - 5u^(3/2)/(3/2)`
`:. int xsqrt(5-x)dx = 2/5u^(5/2) - 5*2/3u^(3/2)`
`:. int xsqrt(5-x)dx = 2/5u^(5/2) - 10/3u^(3/2)`

So the DE solution is;

`y = 2/5(5-x)^(5/2) - 10/3(5-x)^(3/2) + C`