# Cant seem to solve RL differential equation

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#### Megamox

##### New Member
Hi All,

As some differential equation solving practise, I've been trying to derive the solution shown in a textbook on the left. Im trying to use ACos(wt)+BSin(wt) to solve for a particular solution, then later use Aexp^(Kt) later to solve for the homogeneous solution for the transient - is this the correct method? It's been a while since I've done these. However I'm already stuck and I'm not sure if I'm doing something wrong. I can already solve for the steady state response using phasors, but I'd like to try solving the differential this time and get in some much needed practise.

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Thanks!
Megamox

#### Megamox

##### New Member
Just an update, still cant solve it trigonometrically, if anyone can that'd be great. Decided to go back to trusty phasors to get the steady state response and then solve for the homogeneous transient response. Sorry it's a bit messy but it gives the correct answer at the bottom right corner.

Megamox

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#### Ratchit

##### Well-Known Member
Megamox,

Do what the attachment suggests. Find the integrating factor and solve the equation.

Ratch

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#### Megamox

##### New Member
Hi Ratchit, thanks for the help. I've managed to solve it with trigonometric manipulation, was quite exhausting.
Megamox

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#### Ratchit

##### Well-Known Member
Megamox,

Glad to help, but it looks like you are working yourself into a lather needlessly. The differential equation is a first order first degree equation that can easily be solved after multiplying the equation by the integrating factor. Any DE book can show you how. I believe you will find the integrating factor method much easier to to use for this type of equaton.

Ratch

#### Megamox

##### New Member
Hi Ratchit, thanks for the advice! I've solved it using the integrating factor method below. The integration by parts was the most time consuming as I'm still a little rusty. Phasors has been so far the easiest, I think.

Megamox

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#### Ratchit

##### Well-Known Member
Megamox,

...The integration by parts was the most time consuming as I'm still a little rusty....

A lot of folks who need to integrate complex functions use a table of integrals, which are readily available in calc books and on the web.

Ratch

#### Megamox

##### New Member
Yes absolutely, I found the entry in an integral table below which looks like it would have been just the ticket!
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I think this has been good practise for preparation to moving on to Laplace transforms. It will be interesting to compare the methods. I do understand the laplace method involves incorporating the initial conditions into the transform and so the result should be a total response, with transient included. There does appear quite a few results in tables to remember too. Perhaps it'll be as easy to apply as phasors are, once I get well versed with them.

Thanks,
Megamox

#### Ratchit

##### Well-Known Member
Megamox,

...I do understand the laplace method involves incorporating the initial conditions into the transform and so the result should be a total response, with transient included...

The Laplace method is the best way to go, but the DE must be linear.

Ratch

#### Megamox

##### New Member
Just for completeness, I've added the solution using the Laplace transform below. The partial fractions were the toughest part, almost as tough as the double integration by parts using the integrating factor method earlier.

Megamox

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#### Ratchit

##### Well-Known Member
Megamox,

You should be able to find the inverse Laplace directly from a table of inverse Laplacian transforms of moderate size without the need to do partial fractions.

Ratch

#### MrAl

##### Well-Known Member
Most Helpful Member
Just for completeness, I've added the solution using the Laplace transform below. The partial fractions were the toughest part, almost as tough as the double integration by parts using the integrating factor method earlier.

Megamox

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Hi there,

It looks like you did a pretty good job there

To show the beauty and simplicity of the Laplace Transform methods a little more, here is another quick approach when initial conditions are all zero...

We have a source Vm*cos(wt), and we have an impedance of R in series with L driven by that source.
The Laplace Transform of the impedance Z is:
Z=R+s*L

and the Laplace Transform of the source with Vm=1 is (Vm=1 for illustration simplicity):
Vs=s/(s^2+w^2)

so the current Is is just the source divided by the impedance:
Is=Vs/Z=(s/(s^2+w^2))/(R+s*L)

which simplified:
Is=s/((s^2+w^2)*(R+s*L))

so within a few seconds we've arrived at the basic solution.

And partial fraction expansion:
Is=(s*R+w^2*L)/((s^2+w^2)*(R^2+w^2*L^2))-(L*R)/((R+s*L)*(R^2+w^2*L^2))

and you can factor that any way you wish.

Partial fraction expansion isnt that hard once you do a few, but sometimes you'd have to have the numerical values of the components to proceed because the raw analytical form cant always be factored.

Ratchit offers some good advice here too...look for a table of Laplace Transforms and Operations as that will help sometimes...just dont depend on that to get all the answers all of the time. For example the sine and cos source transforms can be found in most LT tables, however you might not find those if they also contain a phase shift like cos(wt+ph).

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