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

1 Answer
Steve M
Nov 16, 2016

y = 2/5(5-x)^(5/2) - 10/3(5-x)^(3/2) + C y=25(5−x)52−103(5−x)32+C, Provided x<=5 x≤5

Explanation:

We have dy/dx=xsqrt(5-x)dydx=x√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 ∫dy=∫x√5−xdx

And so y = int xsqrt(5-x)dx + C y=∫x√5−xdx+C

For the RHS integral, Let u=5-x => x=5-u 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

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