Divide the given equation
(√y)y' + √(y³) = 1,…..(1) by √y, giving
y' + y = y^(—1/2). . . . (2), which is in Bernoulli's form.
Put u = y^(1 +1/2) = y^(3/2).
Then (3/2)y^(1/2).y' = u'.
Multiply (2) by √y and we get
y^(1/2)y' + y^(3/2) = 1.
(2/3)u' + u = 1 or u' + (3/2)y = 3/2, which is a first order linear equation with an integrating factor exp[Int{(2/3)dx}] = exp(2x/3) and a general solution is
u.e^(2x/3) = (3/2) Int[e^(2x/3)dx] +c or
y^(3/2).e^(2x/3) = (9/4)e^(2x/3) +c.
To obtain the particular solution satisfying y(0) = 4, we put x=0, y=4, obtaining
8 = (9/4)+c ==> c=(23/4). Hence the required particular solution is given by the relation:
4y^(3/2) = 9e^(-2x/3) + 23
