Solve #dy/dx=xye^(3x)# ?

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Answer 1 · 2017-03-20T19:44:13

#y=1/9e^(1/9e^(3x)(3x-1)#

#dy/dx=xye^(3x)#

#int1/ydy=intxe^(3x)dx#

#ln|y|=1/9e^(3x)(3x-1)+lnk#

#ln1=1/9e^0(3(0)-1)+lnk#

#k=1/9#

#ln|y|-ln(1/9)=1/9e^(3x)(3x-1)#

#ln|9y|=1/9e^(3x)(3x-1)#

#9y=e^(1/9e^(3x)(3x-1)#

#y=1/9e^(1/9e^(3x)(3x-1)#

Answer 2 · 2017-03-20T19:52:03

#y = e^((x/3-1/9)e^(3 x)+1/9)#

This is a separable differential equation. It can be arranged as

#f_1(x)dx=f_2(y)dy#

with #f_1(x)=x e^(3x)# and #f_2(y)=1/y#

Integrating we have

#F_1(x)+C=F_2(y)#

with

#F_1(x)= (x/3-1/9)e^(3 x)# and

#F_2(y)=log(y)#

so

#log(y)=(x/3-1/9)e^(3 x)+C# or

#y = C_1e^((x/3-1/9)e^(3 x))#

solving for initial conditions

#1=C_1 e^((0/3-1/9)e^(3 xx 0))# giving

#C_1=e^(1/9)# and finally

#y=e^(1/9)e^((x/3-1/9)e^(3 x)) = e^((x/3-1/9)e^(3 x)+1/9)#