Help with using integrating factor to solve dy/dt = 2x + 4t (ans: -2t -1 + ce^(2t))

Hi,

First post here, so bear with me if I'm breaking forum etiquette

I'm working out exercises in my Croft & Davis engineering mathematics book. I'm supposed to solve the equation :

dy/dt = 2x + 4t

By using an integrating factor. According to the book, the answer should be :

-2t -1 + ce^2t

While my answer is just -2t + ce^2t. What am I doing wrong? And where does the -1 come from?

Thank you!

MisterMD said:

I'm supposed to solve the equation :

dy/dt = 2x + 4t

By using an integrating factor. According to the book, the answer should be :

-2t -1 + ce^2t

While my answer is just -2t + ce^2t. What am I doing wrong? And where does the -1 come from?

Click to expand...

Unfortunately, it is not possible to troubleshoot work that we cannot see. So please reply with a clear listing of all of your steps. Thank you!

MisterMD said:

Hi,

First post here, so bear with me if I'm breaking forum etiquette

I'm working out exercises in my Croft & Davis engineering mathematics book. I'm supposed to solve the equation :

dy/dt = 2x + 4t

Click to expand...

Is this the correct ODE? You are showing 3 variables!

Ah, sorry for not replying. I actually figured it out though! Again, I hope to make it clear here. Bear with me. Also thanks to Wolfram Alpha.

It's solving the equation by using an integrating factor. The first thing I did was then make it into the standard form dy/dx + P(x) = Q(x). So I turned dx/dt = 2x + 4t into dx/dt - 2x = 4t.

P is then -2, Q is 4t.

Then I worked out the integrating factor (called U) : e^{(integral(P) dx)} which worked out to e^-2t.

Multiplying everything with this factor makes it into this : (dx/dt * e^-2t) - (2xe^-2t) = (4te^-2t). Using the product rule, that can be simplified to :

d(xe^-2t)/dt = 4te^-2t

Integrate both to produce : xe^-2t = integral(4te^-2t)dt, which is rather hairy. I did it with integration by parts.

4 (integral(te^-2t)dt. Removing the factor of 4. Integration by parts is : Integral(UV') = UV - Integral(U'V)

U = t, U' = 1
V = (-e^-2t)/2, V' = e^-2t

So it becomes : (-te^-2t)/2 - Integral((-e^-2t)/2)

Becoming : (-te^-2t)/2 - (e^-2t)/4 + C


Rmoving fractions (multiplying all by 4) : -2te^-2t - e^-2t + C

And then the final step which is dividing all by U : x = -2t -1 + Ce^2t which is exactly the solution in the book. Still! Thanks for replying