What is the particular solution of the differential equation # dy/dx = xy^(1/2) # with #y(0)=0#?

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Answer 1 · 2018-02-04T00:32:05

# y = x^4/16 #

We have:

# dy/dx = xy^(1/2) # with #y(0)=0#

This is a First Order Separable ODE, so we can can write:

# y^(-1/2) \ dy/dx = x #

Now, we separate the variables to get

# int \ y^(-1/2) \ dy = int \ x \ dx #

Which consists of standard integrals, so we can integrate:

# y^(1/2)/(1/2) = x^2/2 + C #

Applying the initial condition, we have:

# 0 = 0 + C => C = 0 #

Thus, we have:

# y^(1/2)/(1/2) = x^2/2 #
# :. y^(1/2) = x^2 /4 #
# :. y = x^4/16 #