Question

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

Answer 1

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

Explanation:

`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

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

Explanation:

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)`