In general, the integrating factor method can solve differential equations of the form
#("d"y)/("d"x) + "P"(x)*y = "Q"(x)#.
Observe that,
#"d"/("d"x)(e^(\int "P"(x) "d"x)y) = e^(\int "P"(x) "d"x)*("d"y)/("d"x)+P(x)e^(\int "P"(x) "d"x)y#
#"d"/("d"x)(e^(\int "P"(x) "d"x)y) = e^(\int "P"(x) "d"x)(("d"y)/("d"x) + "P"(x)*y)#
Then,
#"d"/("d"x)(e^(\int "P"(x) "d"x)y) = e^(\int "P"(x) "d"x)"Q"(x)#,
#e^(\int "P"(x) "d"x)y = \int e^(\int "P"(x) "d"x)"Q"(x) "d"x#.
This is the general solution to the differential equation.
We call #e^(\int "P"(x) "d"x)# the integrating factor.
In your specific case, #"P"(x) = 1#, #"Q"(x) =e^(3x)#. The integrating factor is therefore
#e^(\int 1 "d"x)=e^(x)#
Then, substituting,
#e^x*y = \int e^(x)*e^(3x) "d"x#
#e^x*y = \int e^(4x) "d"x#
#e^x*y = e^(4x)/4 + C#
#y = (e^(4x)/4+C)/e^x#
#y=1/4e^(3x) + Ce^(-x)#
So, yes, you are correct. A method to check your answer in the future would be to differentiate #y# and substitute #("d"y)/("d"x)# and #y# into your differential equation.
