What is a solution to the differential equation #dy/dx=e^(y+2x)#?

2 Answers
Jul 16, 2016

# y=ln (2/(C - e^(2x))) #

Explanation:

it's separable!!

#dy/dx=e^(y+2x)#

#dy/dx=e^ye^(2x)#

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

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

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

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

#e^(-y) =C - 1/2 e^(2x) #

#ln(e^(-y)) =ln (C - 1/2 e^(2x)) #

#-y =ln (C - 1/2 e^(2x)) #

#y =-ln (C - 1/2 e^(2x)) #

#y =ln (1/(C - 1/2 e^(2x))) #

# y=ln (2/(C - e^(2x))) #

#e^(2x)+2e^(-y)+C=0#

Explanation:

The given differential equation is

#dy/dx=e^(y+2x)#

#dy/dx=e^(y)*e^(2x)#

Multiplying both sides by #dx#

#dy/dx*dx=e^(y)*e^(2x)*dx#

#dy/cancel(dx)*cancel(dx)=e^(y)*e^(2x)*dx#

#dy=e^(y)*e^(2x)*dx#

Dividing both sides by #e^y#

#dy/e^y=(e^(y)*e^(2x)*dx)/e^y#

#dy/e^y=(cancel(e^(y))*e^(2x)*dx)/cancel(e^y)#

#dy/e^y=e^(2x)*dx#

#e^(-y)*dy-e^(2x)*dx=0#

Integrating both sides of the equation

#int e^(-y)*dy-int e^(2x)*dx=int 0#

#-1*int e^(-y)*(-1)dy-1/2*int e^(2x)*2*dx=int 0#

#-1* e^(-y)-1/2* e^(2x)=C_1#

# e^(-y)+1/2* e^(2x)+C_1=0#

#e^(2x)+2e^(-y)+C=0#

God bless...I hope the explanation is useful.