Solve the inital value problem:  `dy/dx` = `x*sqrt(x - 1)`    and y =2 when x =1. Confirm it conforms to the differential equation slope field.

justaguide | Certified Educator

The differential equation to be solved is `dy/dx = x*sqrt(x - 1)`

`dy/dx = x*sqrt(x - 1)`

=> `dy = x*sqrt(x - 1) dx`

take the integral of the two sides:

`int dy = int x*sqrt(x - 1) dx`

=> `y = int x*sqrt(x - 1) dx`

Let x - 1 = z, dx = dz

`int x*sqrt(x - 1) dx`

=> `int (z + 1)*sqrt z dz`

=> `int z^(3/2) + z^(1/2) dz`

=> `(z^(5/2))/(5/2) + (z^(3/2))/(3/2) + C`

=> `(2*z^(5/2))/5 + (2*z^(3/2))/3 + C`

Substitute z = x - 1

=> `(2*(x-1)^(5/2))/5 + (2*(x - 1)^(3/2))/3 + C`

At x = 1, y = 2

`2 = (2*(1-1)^(5/2))/5 + (2*(1 - 1)^(3/2))/3 + C`

=> C = 2

The solution of the differential equation is `y = (2*(x-1)^(5/2))/5 + (2*(x - 1)^(3/2))/3 + 2`

Plotting `y = (2*(x-1)^(5/2))/5 + (2*(x - 1)^(3/2))/3 + 2` , it is seen that at the point (1, 2) the slope of the graph is 0.

The solution of the differential equation is `y = (2*(x-1)^(5/2))/5 + (2*(x - 1)^(3/2))/3 + 2` .