Solve the differential equation
dy/dx = √[(x^3)(y)] with initial conditions x = 1, y = 2
so far i got y = (x^5)/25 + c
so do i plug in 2 = (1^5)/25 + c
2 - (1/25) = c
c = 49/25 = 1.96
is this correct?
dy/dx = √[(x^3)(y)]
dy/dx = x^(3/2) y^(1/2)
y^(-1/2) dy = x^(3/2) dx
y^(1/2) = 1/5 x^(5/2) + c
Now if you square both sides, you do not get x^5/25 + c, since (a+b) is not a^2+b^2
But proceeding to find c,
√2 = 1/5 + c
c = √2 - 1/5
Now square both sides and you get
y = (x^(5/2) + √2 - 1/5)^2