What is the particular solution of the differential equation # 2yy' = e^(x-y^2) # with #y=-2# when #x=4#?

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Answer 1 · 2017-10-02T13:43:50

# y = -sqrt(x) #

We have:

# 2yy' = e^(x-y^2) #

This is a non-linear First Order ODE which we can write as:

# 2y dy/dx = e^(x)e^(-y^2) #
#:. (2y)/e^(-y^2) dy/dx = e^(x) #
#:. 2ye^(y^2) dy/dx = e^(x) #

Which in this form is separable, so we can "seperate the variables" , to get:

# int \ 2ye^(y^2) \ dy = int \ e^(x) \ dx#

Which conveniently is directly integrable:

# e^(y^2) = e^(x) + C #

Using the initial condition #y=-2# when #x=4#:

# e^(4) = e^(4) + C => C = 0 #

Thus, the Particular Solution is:

# e^(y^2) = e^(x) #
# y^2= x #

Note that if we form an explicit solution for #y# we get:

# y = +- sqrt(x) #

And with the initial conditions, only:

# y = -sqrt(x) #

forms a valid solution.