What is a particular solution to the differential equation #dy/dx=2xy# with #y(0)=3#?

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Answer 1 · 2016-09-08T09:10:12

# y = 3e^(x^2)#

#y' = 2xy#

separate it out

#1/y y' = 2x#

integrate

#int 1/y y' \ dx = int 2x \ dx#

#implies int 1/y \ dy = int 2x \ dx#

#ln y = x^2 +C qquad triangle#

# y = e^( x^2 +C) = Ce^(x^2)# [C's just any constant]

apply the IV

# 3 =Ce^(0^2) implies C = 3#

So

# y = 3e^(x^2)#

strictly speaking, #triangle# should be #ln |y| = x^2 +C# with # |y| = Ce^(x^2)#