Question

if dv/dt=-2v-32 and v(0)=-50, solve the differential equation...please help

Answer 1

\[dv/dt=-2v-32\]\[-dv/(2v+32) = dt\]\[-2 \int\limits_{}^{}dv/(v+16) = \int\limits_{}dt\]\[-2\ln (v+16)=t +c\]\[v=ce^{-2t}-16\]

Answer 2

Since \[v(0)=-50\]\[-50=ce^{0}-16\]\[-34=c\]and\[v=-34e^{-2t}-16\]

Answer 3

@mercrutio should the -2 be
\[
- \frac 1 2
\]

Answer 4

yeah i was wondering where u got the -2 from?

Answer 5

Oh, my bad. I wrote the step quickly here, but had it right on my paper. The -2 should be a one half, so \[−(1/2)∫dv/(v+16)=∫dt\]

Answer 6

It comes from pulling it out of the -2v-32, which became a v+16.

Answer 7

so wouldn't ur second step look different? the v=?

Answer 8

No, I got the last line correct in the first step. Like this:\[−(1/2)∫dv/(v+16)=∫dt\]\[ln(v+16)=-2t\]\[v+16=e^{-2t}\]\[v=e^{-2t}-16\]

Answer 9

With the +c cropping up. Oops.
\[ln(v+16)=-2t+c\]\[v+16=e^{-2t+c}\]\[v=ce^{-2t}-16\]
There we go.

Answer 10

what happened to the 1/2

Answer 11

You multiply it up, and that gives you 2t on the right.

Answer 12

uhm would you mind helping me with another one?

Answer 13

No problem.

Answer 14

thank you so much. dy.dx=xy/(x^2+4)
so this is what i ended up having
\[\int\limits_{?}^{?} 1/y dy=\int\limits_{?}^{?} x/(x^2+4) dx\] uhm whats next?

Answer 15

Integrate both sides

Answer 16

So \[\int\limits dy/y = \int\limits x dx/(x^2 +4)\] Then compute the integrals. Let u=x^2+4\[ln(y)=(1/2)\int du/u = (1/2)ln(u)=ln(x^2+4)/2\] And so, \[y^2=x^2+4\]

Answer 17

can you check one i did?

Answer 18

Sure.

Answer 19

dz/dz=2+sin z
dz=(2+sin z) dz
\[\int\limits_{}^{} dq= \int\limits_{}^{} 2 + \sin z dz\]

q= z + cos z

Answer 20

oh i meant dq/dz

Answer 21

One mistake.\[\int \sin x = -\cos x\] so for your problem q= z-cos z.

Also, you might want to write \[\int(2+\sin z)dz\] instead, for clarity.

Answer 22

okay and then the rest of the problem says and q=5, when x=0, findthe solution

Answer 23

Oh, they want you to keep track of the +C again. So your solution is actually \[q=z-\cos z + C\]Then, since q(0)=5,\[5=0-\cos 0 + C\]\[5+\pi = C\]and your solution is\[q=z - \cos z + 5 + \pi\]

Answer 24

thank u so much, okay i have one last one. I tried to solve it so please correct me where i am wrong.

Answer 25

Okay, can do.

Answer 26

xy'-3y^2=0
dy/dx=3y^2/x
\[\int\limits_{}^{} 1/y^2 dy= \int\limits_{}^{} 3/x dx\]

-1/y + c = (i kind of forgot the antiderivative for this, isn't it ln(3)?

Answer 27

oh its 3lnx

Answer 28

It's 3ln(x). \[\int3dx/x=3\int dx/x\]

Answer 29

So far so good, but you still need to solve for y.

Answer 30

okay so
-y^-1 + c= 3ln x + c

Answer 31

You can combine the 'c's.

Answer 32

uhm what else do ido?

Answer 33

And then invert and multiply by negative one. It comes out like this:\[y=-1/(3ln(x))+c\]

Answer 34

Erm, excuse me,\[y=-1/(3ln(x)+c)\]

Answer 35

well that's it, thank you so much:)!!

Answer 36

You're very welcome.