Solving a Second Order Differential Equation with Circuit Components

[formulajoe]

dv(t)/dt + 200v(t) = 10 cos(100t)
v(0)=0

find v(t)
i know v(t) is going to have an exponential and sinusoid component

i also know that 200*ke^-st=-ske^-st
so, s = -200. i don't know how to find k.
i also can't figure out how the sinusoid part is going to be. the equations from my circuits book don't look anything like what i need.

 

[dextercioby]

Take the particular solution of the nonhomogenous LODE as a combination
v_{p}(t)=A\sin 100t+B\cos 100t

Plug it in the equation and find A and B.Then find the constant of integration (the one which resulted when integrating the homogenous LODE) by asking that
v(0)=v_{hom}(0)+v_{p}(0)=0

Daniel.

 

[formulajoe]

how do i find the k for the exponential portion?

 

[Curious3141]

This is a first order differential equation that you can solve by multiplying by an integrating factor.

In an equation of the form :

v'(t) + f(t)v = g(t)

you can multiply both sides by e^{\int{f(t)}dt (this is called an integrating factor)

to get

\frac{d}{dt} ({v(t)e^{\int{f(t)}dt}) = g(t)e^{\int{f(t)}dt

which can then be solved by integrating both sides wrt t.

What is f(t) in this case ? What is the corresponding value of e^{\int{f(t)}dt ?

 

[dextercioby]

how do i find the k for the exponential portion?

Separate variables and integrate.
\frac{dv(t)}{dt}+200v(t)=0

Daniel.

 

[formulajoe]

oh man, I am a complete bonehead. i know how to do these.
first-order linear differential equation, for some reason it didnt click that that was what it was.

 

[dextercioby]

Look at the bright side...You finally got the "click"...

Daniel.

 

[formulajoe]

This is a first order differential equation that you can solve by multiplying by an integrating factor.

In an equation of the form :

v'(t) + f(t)v = g(t)

you can multiply both sides by e^{\int{f(t)}dt (this is called an integrating factor)

to get

\frac{d}{dt} ({v(t)e^{\int{f(t)}dt}) = g(t)e^{\int{f(t)}dt

which can then be solved by integrating both sides wrt t.

What is f(t) in this case ? What is the corresponding value of e^{\int{f(t)}dt ?

the final form of this equation has an integral, this can't be right.

heres what I am getting

integrating factor = e^-200t

so,e^200t int{(e^-200t)(10cos(100t))}dt
that means integration by parts. as far as i can tell, I am doing everything right.

 

[dextercioby]

Nope,his version is right,too.It's an alternative.Obviously will lead to the same result like mine.

Daniel.

 

[Curious3141]

Tell me wherever you're lost in the following :

In this case,

f(t) = 200

e^{\int{f(t)}dt = e^{200t}

You now have the differential equation :

\frac{d}{dt} (v(t)e^{200t}) = 10 \cos(100t)e^{200t}

Integration will give you :

v(t)e^{200t} = \int{10 \cos(100t)e^{200t}}dt + constant

The easiest way to evaluate the integral on the RHS is to make the substitution x = 100t then integrate by parts. You'll have to integrate by parts twice to solve this one. Don't forget to solve for the constant by setting v = 0 at t = 0

The final expression is :

v(t) = \frac{1}{50}(\sin(100t) + 2\cos(100t) - 2e^{-200t})

 

[formulajoe]

im not getting this stuff to work out like the book does. the answer in the back of the book has a exp part, a sin part, and a cos part. but I am getting left with integrals in my final equations.

and when i try to separate the variables and solve for k, the variables I am separating for are v and t. i end up with t= -ln200v.

 

[Curious3141]

im not getting this stuff to work out like the book does. the answer in the back of the book has a exp part, a sin part, and a cos part. but I am getting left with integrals in my final equations.

and when i try to separate the variables and solve for k, the variables I am separating for are v and t. i end up with t= -ln200v.

Is my answer the same as the book answer (it should be, because it satisfies the given d.e.) ?

 

[formulajoe]

okay, i follow all that. i was getting lost at the part where both sides are integrated.
i was setting it up wrong.
thanks

 

[Curious3141]

Good.

 

[formulajoe]

your answer works with the books.

 

[formulajoe]

Tell me wherever you're lost in the following :

In this case,

f(t) = 200

e^{\int{f(t)}dt = e^{200t}

You now have the differential equation :

\frac{d}{dt} (v(t)e^{200t}) = 10 \cos(100t)e^{200t}

Integration will give you :

v(t)e^{200t} = \int{10 \cos(100t)e^{200t}}dt + constant

The easiest way to evaluate the integral on the RHS is to make the substitution x = 100t then integrate by parts. You'll have to integrate by parts twice to solve this one. Don't forget to solve for the constant by setting v = 0 at t = 0

The final expression is :

v(t) = \frac{1}{50}(\sin(100t) + 2\cos(100t) - 2e^{-200t})

okay, I am still having a problem working through this on my own.

i have v(t) = 10\sin(100t) + 20\cos(100t) + 20\int{cos(100t)e^{200t}}dt
how do i get rid of the integral?

 

[Curious3141]

okay, I am still having a problem working through this on my own.

i have v(t) = 10\sin(100t) + 20\cos(100t) + 20\int{cos(100t)e^{200t}}dt
how do i get rid of the integral?

I'll go through the integration step by step. It's a little tedious, so I hope you're appreciative.

You want to work out \int{10 \cos(100t)e^{200t}}dt = 10\int{\cos(100t)e^{200t}}dt

Let I = \int{\cos(100t)e^{200t}}dt

Substitute 100t = x

I = \int{\cos(x)e^{2x}}dt = \frac{1}{100}\int{\cos(x)e^{2x}}dx

Now let F = \int{\cos(x)e^{2x}}dx

Integration by parts, \int udw + \int wdu = uw

First let u = e^{2x}, dw = \cos x

\int{e^{2x}\cos{x}}dx + 2\int{\sin x}{e^{2x}}dx = (e^{2x})(\sin x) ---eqn (1)

From the LHS of eqn (1), we need to find G = \int{(\sin x)e^{2x}}dx

Integrate again by parts. This time let u = e^{2x}, dw = \sin x

Then \int{(\sin x)e^{2x}}dx + 2\int{(-\cos x) (e^{2x})}dx = -e^{2x}\cos x

Therefore G = -e^{2x}\cos x + 2\int{(\cos x)(e^{2x})}dx ---eqn(2)

Put the expression for G from eqn (2) back into eqn (1) and solve for F, then use that to find I, put it all back in the original equation, and you have your solution.

As I said, tedious. I just hope I haven't slipped up somewhere in my LaTEX.

 

[dextercioby]

That's why i still recommend my version.

Daniel.

 

[formulajoe]

geez, that seems complicated.
thanks for going into such detail. that explains it for me.