- #1

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so p(x) =1 q(x)=1

in order to get a homogeneous solution

the solution takes this equation

v'+v=1

and throws away the parameter "a"

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- Thread starter transgalactic
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- #1

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so p(x) =1 q(x)=1

in order to get a homogeneous solution

the solution takes this equation

v'+v=1

and throws away the parameter "a"

- #2

Cyosis

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- #3

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av'+v=1

q(x)=1

p(x)=1

[tex]

t=\int p(x)dx=x

[/tex]

[tex]z=e^t[/tex]

[tex]y=\frac{1}{z}[\int q(x)zdx+c]=\frac{1}{e^x}[xzd+c] [/tex]

i get a different equation

??

- #4

Cyosis

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That means you've to solve [itex]aq'(x)+q(x)=0[/itex] first. Try to do that.

To find the particular solution you set the homogeneous equation equal to 1. It would be nice to get rid of the v' term, what kind of 'guess' function would do the trick?

- #5

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and i dont have the expression that i saw in the solution

- #6

Cyosis

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It appears you want to do it the hard way by solving [itex]v'(x)+\frac{1}{a}v=\frac{1}{a}[/itex] with [itex]P(x)=\frac{1}{a}, \; Q(x)=\frac{1}{a}[/itex]?

- #7

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a*dq/q=1/dx

(i cant do integral on both sides)

dq and dx must be on the numenator

i cant get them bot there

??

- #8

dx

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a(dv/dx) + v = 1

a(dv/dx) = 1 - v

(a dv)/(1 - v) = dx

- #9

Cyosis

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If you have [itex]a q'(x)+q(x)=0[/itex] then [itex]a q'(x)=-q(x)[/itex] and [itex]a q'(x)/q(x)=-1[/itex]. Try to integrate this expression by using a substitution u=q(x).

- #10

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a(dv/dx) + v = 1

a(dv/dx) = 1 - v

(a dv)/(1 - v) = dx

you are correct

i get:

a*ln|1-v|=x

but its dont have the step where i throw away the parameter "a"

like they did in the proccess

??

- #11

Cyosis

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Edit: Hint plug the answer from your book into the differential equation and check if both sides match. Whenever you're working with differential equations it's very easy to check your answer and you should teach yourself to do so.

- #12

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here is the full solution that i was presented

i cant see what is the continuity.

from the KCL eqaution i will get

v+cv'=1

i cant understand the disappearance of the parameter in the next line

it makes no sense

and this is a formal solution so it has to be correct

??

- #13

Cyosis

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Also in the first equation there are two v_cs and one v. Is that supposed to be a v_c as well.

Oh before I forget plug THEIR answer into YOUR equation and see if the equation holds.

- #14

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your 9th post doesnt explain the continuety part

and they dont use the "q" like you did

and they dont use the "q" like you did

- #15

Cyosis

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So I use the variable q and they name it v. What is the big deal? And for the fourth time now have you plugged their answer into your equation? If you do you notice that their answer is not a solution to your differential equation for all a-s. This means there are two options, either the differential equation you listed is the wrong one or their answer is wrong. I also asked you

which you just ignored. I am guessing it isn't.

Also in the first equation there are two v_cs and one v. Is that supposed to be a v_c as well.

which you just ignored. I am guessing it isn't.

Last edited:

- #16

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i cant understand where this solution starts

and where it ends

can you make steps?

and where it ends

can you make steps?

- #17

Cyosis

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- #18

Mark44

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Correct - the parameter should not disappear as it seems to have done in the next line. This is either a mistake that your instructor made or that you made when you were taking notes. On the other hand, where does c come from? I don't see anything in your drawing that shows this. Isn't this the capacitance, which you show as 4 F.?

here is the full solution that i was presented

i cant see what is the continuity.

from the KCL eqaution i will get

v+cv'=1

i cant understand the disappearance of the parameter in the next line

it makes no sense

and this is a formal solution so it has to be correct

??

So it appears that the problem is:

cv' + v = 1

From your notes, it appears that your initial condition is v(0) = 4. Shouldn't this be 2 volts? That's the voltage from the battery or power supply. The part about continuity is that the voltage across the capacitor has to be the same just before t = 0 as just after t = 0, which is the same as it must be right at t = 0.

Now, put the solution you have somewhere that you can't see it.

You need to do two things:

- Solve the homogeneous equation: cv' + v = 0, or maybe it is 4v' + v = 0. Cyosis and dx have given you at least a couple of ways to do this. Your solution to the homogeneous equation, v
_{h}will be Ae^{-t/c}or maybe Ae^{-t/4}, if c is as I think it is. From your initial condition you can solve for A. - Find a particular (not private) solution to the nonhomogeneous equation. I would try v
_{p}= K, and substitute this into the equation cv' + v = 1 (or maybe 4v' + v = 1), to see what K needs to be.

You can check that it actually is a solution by substituting it back into your differential equation cv' + v = 1 (or maybe it's 4v' + v = 1). It should also be true that v(0) = 4 (or whatever it's supposed to be, as I mentioned above.

If your solution satisfies the DE and the initial condition, then that's the solution. If the one that you threw away is different in any appreciable way,

- #19

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4(dv/dx) = 1 - v

(4 dv)/(1 - v) = dx

-4ln(1-v)=x

e^x=(1-v)^(-4)

v(0) => 1=(1-v)^(-4)

1=1/((1-v)^(4))

(1-v)^(4)=1

v=1

i dont get this kind of expression

[tex]

Ae^{-t/4}

[/tex]

??

- #20

Cyosis

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transgalactica said:4(dv/dx) + v = 1

4(dv/dx) = 1 - v

(4 dv)/(1 - v) = dx

-4ln(1-v)=x

e^x=(1-v)^(-4)

Almost correct, you forgot the integration constant.

transgalactic said:v(0) => 1=(1-v)^(-4)

1=1/((1-v)^(4))

(1-v)^(4)=1

v=1

Wrong.

How can v possibly remain in the equation when you're calculating v(0)?

First think clearly what you want to solve. You want to find an expression for v(x). Therefore solve for v(x) first. After you have done that correctly you can use v(0)=4 (or whatever since this still isn't clear) to find the integration constant.

- #21

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what do i do after this step

-4ln(1-v)=x+c

??

-4ln(1-v)=x+c

??

- #22

Cyosis

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Easiest way is to divide by -4 on both sides.

- #23

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ln(1-v)=\frac{x+c}{-4}

[/tex]

[tex]

1-v=e^{\frac{x+c}{-4}}

[/tex]

==

[tex]

v=-e^{\frac{x+c}{-4}}+1

[/tex]

what now

??

- #24

Cyosis

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- #25

Cyosis

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Edit: Seeing as you're constantly editing your original question in its current shape you now have to use the boundary condition to find the value of the integration constant.

- #26

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i fixed it

[tex]

ln(1-v)=\frac{x+c}{-4}

[/tex]

[tex]

1-v=e^{\frac{x+c}{-4}}

[/tex]

==

[tex]

v=-e^{\frac{x+c}{-4}}+1

[/tex]

what now

??

- #27

Cyosis

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Use the boundary condition to find the value of the integration constant.

- #28

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even if i will find the C

i will not have the homogeneous solution like in the solution

Be^ ....

??

i will not have the homogeneous solution like in the solution

Be^ ....

??

- #29

Cyosis

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To make it worse we still haven't been told what the voltage of the source is whether or not v and v_c are different variables etc etc.

Calculate c and see that it will give the correct solution.

And read Mark's post again he tells you clearly what all the ambiguities are with the problem you've provided us.

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