What is the particular solution of the differential equation # 2yy' = e^(x-y^2) # with #y=-2# when #x=4#?
1 Answer
Oct 2, 2017
# y = -sqrt(x) #
Explanation:
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
# 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 = +- sqrt(x) #
And with the initial conditions, only:
# y = -sqrt(x) #
forms a valid solution.
