Question

find a general solution for dy/dt=(ty)^2

Answer 1

dy = (ty)^2 dt = t^2*y^2 dt

Answer 2

\[\frac{ 1 }{ y^2 }dy = t^2 dt\]

integrate those

Answer 3

are you at the very beginning of DE class?

Answer 4

this helps! but yes I am in the first week

Answer 5

have you done anything else besides separable equations

Answer 6

usually i think, substitutions and exact equations are next

Answer 7

been a couple years since that class

Answer 8

no we haven't done much except basically the introduction. And she only touched on the separable equations that's why it's confusing to me

Answer 9

right, basically those are just equations where you can put each variable on its own side, then integrate to find the general function

Answer 10

It gets more fun later on

Answer 11

\[\int\limits \frac{ 1 }{ y^2 }dy =\int\limits t^2 dt\]

Answer 12

you get a function y(t)

Answer 13

\[\frac{ -1 }{ y } + C _{1} = \frac{ t^3 }{ 3 }+ C _{2}\]

Answer 14

You really dont need to remember ALL the integrating techniques, it helps, but basic U -Sub and Parts will get you through the course.

Answer 15

yeah it seems easy enough I just need to practice. what do you think about dy/dt = 2y+1

Answer 16

c*e^(2x) - 1/2

Answer 17

just used my calc, that is the answer

Answer 18

how do you do it so fast! -___- and how did you do it at all lol?

Answer 19

Oh i just used my calculator, for that one, since all you have done so far are separable equations, i assume it will be that

Answer 20

yeah that's the answer in the back of the book I just don't see how you got that

Answer 21

\[dy = (2y + 1) dt \]
\[\frac{ 1 }{ 2y+1 }dy = 1dt\]

Answer 22

integrate both sides

Answer 23

\[\frac{ \ln(2y+1) }{ 2 } = t + C\]

Answer 24

solve for y(t)

Answer 25

Both integrations produce a constant, C , you can combine the 2 into a single constant. Since it is arbitrary. Call it whatever you want.

Answer 26

I just put a C on the right side. C =C1-C2 where C2 was the constant from integrating the left side, and C1 the constant from the right side integration.

Answer 27

recall with exponents:
\[e^a *e^b = e^{a+b}\]

Answer 28

Add me as a fan, and tag me with @DanJS when you get later into the course. It would be fun to bust out the old notebook and homework from DE and give them a go again.

Answer 29

I think it's the solving for y(t) that I'm having a problem with. But I am new to the site so I'm just seeing how it's all working out. But I will DEFINITELY add you lol thank you for the help!

Answer 30

yeah, most of the work , like 80 or 90% will be algebra, expecially when you get to higher order DE, probably after the first test

Answer 31

Review partial fraction decompositions

Answer 32

ahhh yes, I've forgotten all about that. that Christmas break away from math did me no good

Answer 33

I took a year off school after finishing calc 3 and linear algebra, i know the feeling, jumping into DE after a year break. It isn't too bad though. There arent many of those crazy integrals to work out. Most are straight forward.

Answer 34

just a ton of algebra

Answer 35

I would Recommend MIT OCW Differential Equations. Arthur Mattuk
and
MIT OCW Calculus Revisited (Black and White from like 1950) there are 5 lectures on DE's that deal with variation of parameters, and undetermined coefficient, that will be like next months work. ...

ill find a link for you .. 1 sec

Answer 36

DE
http://ocw.mit.edu/courses/mathematics/18-03-differential-equations-spring-2010/video-lectures/

PART II Differential equations from this will help out ALOT, believe me
http://ocw.mit.edu/resources/res-18-008-calculus-revisited-complex-variables-differential-equations-and-linear-algebra-fall-2011/part-ii/

Answer 37

The first link is DE and Linear algebra together, some of the lectures are very helpful.

The black and white ones, PART II (5 or so lectures) are a gold mine. Those help out a BUNCH when you get to those topics

Those things got me an A in the course, and i understood it all very well

Answer 38

@DanJS help me pleaseee

Answer 39

yeah I'm about to go bookmark those now and look at them. I hope they help me as much as they helped you

Answer 40

I love the style of that old black and white teachers teaching. Easy to understand. Those will help out like next month when you get to second order DE's

Answer 41

cool i added you as a fan, tag me whenever you want @DanJS in a question.

Answer 42

Arranging your equation as
(dy/y^2)=tdt
now integrating both the sides we get
-1/y= (t^2)/2 + c
where c is a constant of integration
(t^2)/2+1/y+c=0
the final solution