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

1 Answer
Steve M
Nov 16, 2016

# 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 #

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