learning diff-e-q...and stuck here...help.

[Semidevil]

so I am learning first order linear equation, and I am stuck in this problem

dy/dx - 2y = 3e^2x.

so we know that p = -2, Q = 3e^2x

the integrating factor is e^(integeral of -2) dx, which is e^-2x.

the next step is to multiply the integrating factor throughout the equation.

so we have (e^-2x * dy/dx) - (2y * e^-2x) = (3e^2x * e^-2x)

this is where I am stuck. what do I do next? The books says to recognize the left-hand side of the resulting equation as a derivative of a product... I dont understand...

anyone care to enlighten me on that part, and carry me through the rest to solve it?

 

[fyleow]

Add (2y * e^-2x) to both sides
Divide both sides by e^-2x

?

 

[Dissipate]

Its easy at this point. Dx(e^-2x*y) = 3
because the left side = Dx(e^-2x*y)

now you just integrate both sides (the Dx goes away on the left side) so... e^-2x*y=int(3)

You should be able to take it from there.

edit: small mistake. right side is just 3 because the integrating factor cancels with the right side.

 

[Semidevil]

Originally posted by: Dissipate
Its easy at this point. Dx(e^-2x*y) = 3e^5x

because the left side = Dx(e^-2x*y)

now you just integrate both sides (the Dx goes away on the left side) so... e^-2x*y=int(3e^5x)

You should be able to take it from there.

it is late at night, and I am getting a brain fart........

how did you conclude the left side to be Dx(e^-2x*y)??

and how did 3e^5x come about??

 

[cchen]

Dx(e^-2x*y) comes from the product rule
expand the derivative out and you'll see

 

[Dissipate]

Originally posted by: Semidevil

Originally posted by: Dissipate
Its easy at this point. Dx(e^-2x*y) = 3e^5x

because the left side = Dx(e^-2x*y)

now you just integrate both sides (the Dx goes away on the left side) so... e^-2x*y=int(3e^5x)

You should be able to take it from there.

it is late at night, and I am getting a brain fart........

how did you conclude the left side to be Dx(e^-2x*y)??

and how did 3e^5x come about??

Small mistake, read above. Right side should just be 3 (those negatives are hard to see on my screen). Sorry.

 

[Semidevil]

Originally posted by: cchen
Dx(e^-2x*y) comes from the product rule
expand the derivative out and you'll see

thanx, but I am still having trouble w/ the integrating factor.....If you give me the answer, I can see that it is the integrating factor because I can try it out and expand the produc rule. But when I got the expanded equation, I could not tell the integrating factor just by looking at it.....

Is there a technique in that? what if it is a really hairy equation?

 

[agnitrate]

I get my integrating factor to be e^(-2x). *shrug*

u*(y' - 2y ) = u*3*e^(2x)
( u * y)' = u*3*e^(2x)

u'y + uy' = uy' - 2yu

u'y = -2yu

u' = -2u

du/u = -2dx

integrate both sides

ln(u) = -2*x+C

u = C*e(-2x)

---> back to original equation

( u * y)' = u*3*e^(2x), integrate both sides

( e^(-2x) * y ) = integral of( e^(-2x) * 3 * e^(2x) )

e^(-2x) * y = 3x + C

y = (3x + C)/ (e^(-2x))

tada

-silver