What is a solution to the differential equation #y'=3^-y(2x-4)# and y(5)=0?

Question

1486 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-07-25T08:54:36

#y =( ln (ln 3 (x^2-4x - 5) + 1) )/ ln 3#

#y'=3^-y(2x-4)#

this is separable

#3^y \ y'=2x-4#

#int \ 3^y \ y' \ dx=int \ 2x-4 \ dx#

we can explore the integral on the LHS a little

consider #u = C^v#
#ln u = v ln C#
#1/u (du)/(dv) = ln C#
#implies d/(dv)(C^v) = ln C \ C^v#
so #d/dx (3^y) = ln 3 \ 3^y \ y'#

and so the integral is
#int \d/dx( 1/(ln 3) 3^y ) \ dx=int \ 2x-4 \ dx#

#1/(ln 3) 3^y =x^2-4x + C #

# 3^y =ln 3 (x^2-4x) + C #

#y ln 3 = ln (ln 3 (x^2-4x) + C )#

#y =( ln (ln 3 (x^2-4x) + C ))/ ln 3#

applying the IV: #y(5) = 0#

#0 =( ln (ln 3 (25-20) + C ))/ ln 3 #

#implies ln 3 (5) + C = 1#

#y =( ln (ln 3 (x^2-4x - 5) + 1) )/ ln 3#