Question

Problem with work. Can't locate error.

Answer 1

so what i have so far is

Answer 2

15sin(4t) = (6/2)(integral(v(t))) + v(t)

Answer 3

because v(t) = L di/dt.

Answer 4

i know there are two ways to do it, but i did it this way.
-60 sin(4t) = 3(v(t)) + dv/dt
-60e^(j4t) = 3Ae^(j4t) + 4jAe^(j4t)
-60/(3+4j) = A = -12e^(-j53.1))

Answer 5

v(t) = -12e^(-j53.1t) ==> the real value is just ==> -12cos(4t - 53.1)

Answer 6

hey!
Are you trying to solve the differential equation or do you want to show why that is the differential equation?

Answer 7

Um, solve it.

Answer 8

I'm pretty sure that is the differential equation. Well that's one of two.

Answer 9

cool
do you know how to find the integrating factor?

Answer 10

I can have that or 15cos(4t) = 6i + 2di/dt

Answer 11

yes.

Answer 12

\[\frac{di}{dt}+3i=\frac{15}{2} \cos(4t) \\ \text{ So integrating factor is } v=e^{\int\limits_0^t 3 dx}=e^{3t} \\ e^{3t} \frac{di}{dt}+3 e^{3t}i=\frac{15}{2} e^{3t} \cos(4t) \\ \text{ we multiplied by this integrating factor so we can write } \\ (e^{3t} i)'=\frac{15}{2} e^{3t} \cos(4t) \\ \text{ \to integrate we will need integration by parts }\]

Answer 13

wait, can we try to solve it with dv/dt in the equation. with di/dt, i get the right answer, but doing it with dv/dt, i don/t...

Answer 14

so, \[15\cos(4t) = 6/L \int\limits_{?}^{?} v(t) + v(t) \]

Answer 15

and then take the derivatiev of the whole equation.

Answer 16

v(t) = L di/dt.
ok so this is the relationship between v and i?

Answer 17

yes.

Answer 18

and our L was 2

Answer 19

is that right?

Answer 20

mhmm

Answer 21

so the overarching equation is 15cos(4t) = 6i + v(t). We can either plug in 'i' or v(t)

Answer 22

\[\frac{di}{dt}+3i=\frac{15}{2} \cos(4t) \\ \text{ differentiating gives } \frac{d^2i}{dt^2}+3\frac{di}{dt}=\frac{15(-4)}{2}\sin(4t) \\ \\ \frac{d^2 i}{dt^2}+3 \frac{di}{dt}=-30 \sin(4t) \\ \text{ and we have } v(t)=2 \frac{di}{dt} \\ \text{ so } v'(t)=2 \frac{d^2i}{dt^2} \\ \text{ so we can write our differential equation as } \\ \frac{1}{2} v'(t)+\frac{3}{2} v(t)=-30 \sin(4t)\]

Answer 23

is that what you were going for putting it in terms of v?

Answer 24

hmmm, what i had was
\[(1/L) * \int\limits_{?}^{?} v(t) = i(t). \]
and I plugged that into i(t).
So my equation was
\[15\cos(4t) = (6/2) \int\limits_{?}^{?} v(t) + v(t) \rightarrow \]
take the derivative of the whole equation.
\[-60\sin(4t) = 3v(t) + v'(t) \]

Answer 25

is that wrong?

Answer 26

nope that is actually the same thing I have
you can multiply both sides of my equation by 2 and we have same thing

Answer 27

this is good news
so now we can find the integrating factor since this is still a first order linear diff equation

Answer 28

right, and solving that equation boils down to -12cos(4t - 53.1) as the real solution.

Answer 29

we replace -60sin(4t) = -60e^(-j4t) with j being an imaginary number.

Answer 30

and replace V(t) with Ae^(j4t).

Answer 31

sorry, it should be -60sin(4t) = -60e^(j4t), no negative sign in the exponent*

Answer 32

and then solving for A gives me, -12e^(-j53.16t)

Answer 33

\[v'+3v=-60 \sin(4t) \\ (e^{3t} v)'=-60 e^{3t} \sin(4t) \\ e^{3t} v=-60\int\limits e^{3t} \sin(4t) dt \\ e^{3t} v=-\frac{60}{25} e^{3t} (3 \sin(4t)-4 \cos(4t))+ C \\ e^{3t} v=12e^{3t}(\frac{-3}{5} \sin(4t)+\frac{4}{5} \cos(4t))+C \\ v= 12(\frac{4}{5} \cos(4t)-\frac{3}{5} \sin(4t))+Ce^{-3t}\]

choose cos(a)=4/5 so that makes sin(a)=3/5
we see this actually makes sense since


What is a?
\[a=arcos(\frac{4}{5}) \approx 36.87 ^o\]
\[v=12(\cos(a)\cos(4t)-\sin(a)\sin(4t))+Ce^{-3t} \\ v=12\cos(a+4t)+Ce^{-3t} \\ v=12 \cos(36.87^o+4t)+Ce^{-3t}\]