How is:
e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6
the same as:
(d/dx)*[e^(-3x)*y]=e^(-3x)*6
No, I'm learning the method of solving linear 1st-order differential equations, and I have the example:
dy/dx + 3y = 2xe^(-3x)
I found e^(3x) to be the integrating factor, so then I multiplied it through the original equation giving me:
e^(3x)*(dy/dx) + e^(3x)*3y = 2x
this next step is what is confusing me, (and I realize that it is probably something very simple) I'm supposed to use the product rule where: u=e^(3x), dv=(dy/dx), du=e^(3x), & v=3y
Then the resulting equation is:
[e^(3x)*y]=2x, so my question is:
Can you show me the individual steps for where:
e^(3x)*(dy/dx) + e^(3x)*3y = 2x
becomes:
[e^(3x)*y]=2x
Thank you.