# How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)*6

• Aug 11th 2012, 09:52 AM
JROCK
How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)*6
How is:

e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6

the same as:

(d/dx)*[e^(-3x)*y]=e^(-3x)*6
• Aug 11th 2012, 10:21 AM
HallsofIvy
Re: How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)
Are you saying that you do not know how to take the derivative of \$\displaystyle e^{-3x}y\$ with respect to x?
• Aug 11th 2012, 04:21 PM
JROCK
Re: How is: e^(-3x)*(dy/dx)-e^(-3x)*3y=e^(-3x)*6 equal to:(d/dx)*[e^(-3x)*y]=e^(-3x)
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.