Question

help

Answer 1

is my work right? and how do i find max

Answer 2

@Nnesha

Answer 3

looks good ! o^_^o

Answer 4

yay but how do i find the max?

Answer 5

what chapter is t?

Answer 6

1.2

Answer 7

basic idea and teminology

Answer 8

you there D:

Answer 9

@Vocaloid

Answer 10

Looks good.
I'm not sure why your paper says "integrate" though.
You took some derivatives :o

Answer 11

its suppose to say diff. lool

Answer 12

how you do this one

Answer 13

You're given a general solution (on the left)
and a differential equation (on the right).

Remember in the last one, you had to fill in y'' in order to verify it was a solution.
Same idea here, you need to differentiate the thing on the left, and plug it in for y'.

Answer 14

This one is a tad trickier though.
You have to remember back to first year calculus.
Implicit differentiation.

Answer 15

oml

Answer 16

i started it but i got lost with my work

Answer 17

solve the diff eq

Answer 18

@Zepdrix

Answer 19

\(y' = \dfrac{x (y^2 - 1)}{2 (x-2)(x-1)}\)

This is separable

\(\dfrac{1}{y^2-1}y' = \dfrac{x }{2 (x-2)(x-1)}\)

For LHS, use partial fractions:

\(\dfrac{1}{y^2-1}y'= \dfrac{y'}{2} \left( \dfrac{1}{y-1} - \dfrac{1}{y+1} \right)\)

And

\( \dfrac{1}{2} \int \dfrac{1}{y-1} - \dfrac{1}{y+1} ~ dy = \dfrac{1}{2} \ln \dfrac{y-1}{y+1} + c_1\)


For RHS, kinda the same:

\( \dfrac{x }{2 (x-2)(x-1)} = \dfrac{1}{2} \left(\dfrac{x - 1 + 1}{ (x-2)(x-1)} \right) \)

\(= \dfrac{1}{2} \left(\dfrac{1}{x-2} + \dfrac{ 1}{ (x-2)(x-1)} \right) \)

\(= \dfrac{1}{2} \left(\dfrac{1}{x-2} + \dfrac{1}{x-2} - \dfrac{1}{x-1} \right) \)

So RHS is:

\(\dfrac{1}{2} \int \dfrac{2}{x-2} - \dfrac{1}{x-1} ~ dx\)

\(\ = \ln (x-2) - \frac{1}{2} \ln (x-1) + c_2 \)

LHS = RHS and combining integration constants:

\(\dfrac{1}{2} \ln \dfrac{y-1}{y+1} = \ln \dfrac{x-2}{\sqrt{x-1}} + C \)

we can say \(\ln k = C\) so

\(\dfrac{1}{2} \ln \dfrac{y-1}{y+1} = \ln \dfrac{ k (x-2)}{\sqrt{x-1}} \)

And that is:

\( \dfrac{y-1}{y+1} = \dfrac{ k' (x-2)^2}{x-1} \)


That finishes off as:

\( y-1 = (y + 1) \dfrac{ k' (x-2)^2}{x-1} \)

\( y( 1 -\dfrac{ k' (x-2)^2}{x-1}) =1 + \dfrac{ k' (x-2)^2}{x-1} \)

\(y = \dfrac{x-1 + k' (x-2)^2 }{x- 1 - k' (x-2)^2}\)

If you are just learning how to do ODE's. this is a bad way to go about it. turgid, tedious algebra is no fun for anyone.

Answer 20

`is there any x value where y is undefined ?
for that exponential function x cna be any real number and exponential is defined for all real numbers
so the solution is valid over the (-infinity, infinity )

Answer 21

how would you do that partial fraction... i got lost there

Answer 22

work through this, AND MAKE A POINT OF doing the questions at the end:

https://www.mathsisfun.com/algebra/partial-fractions.html

the basic idea in this problem, taking the LHS s the example, we had:

\(\dfrac{1}{y^2 - 1}\)

we factor

\(=\dfrac{1}{(y - 1)(y+1)}\)

We imagine that that came from adding 2 prior fractions, so

\(\dfrac{1}{(y - 1)(y+1)} = \dfrac{A}{y - 1} + \dfrac{B}{y+1} \)

We multiply each side by \((y - 1)(y+1)\)

So we say that:

\(1 = A(y + 1) + B(y -1) \)

then we say:

- let y = -1 so 1 = -2B

- let y = 1 so 1 = 2A



from whence we conclude that:


\(\dfrac{1}{y^2 - 1} =\dfrac{1}{(y - 1)(y+1)}=\dfrac{1}{2} \left( \dfrac{1}{y - 1} - \dfrac{1}{y+1} \right)\)

which we can integrate easily.....that is why we decompose complicated stuff into simpler fractions in integral calculus

but do the mathisfun stuff. first.

PS:

when you think you get all of that, look at this:

https://math.stackexchange.com/questions/165118/how-does-partial-fraction-decomposition-avoid-division-by-zero

and then explain it to me because I don't get that bit at all. partial fracion decomp makes no sense really :( but it works

Answer 23

so we can factor out the 1/2?

Answer 24

when do we use partial fractions

Answer 25

to decompose fractions into simpler ones that we can integrate

Answer 26

So:

\(\int \dfrac{1}{y^2 - 1} ~ dy\) looks hard [unless you know the Hyperbolic functions]

But:

\( \dfrac{1}{2} \int \dfrac{1}{y - 1} - \dfrac{1}{y+1} ~ dy\) looks easy cos it's just natural logs

Answer 27

you can do this one with a substitution too, but the only one i can think of is hyperbolic

Answer 28

oh

Answer 29

if this is about learning to solve 1st order ODE's, why not forget about the algebra for a minute and try these ones, which I just made up:

\(1 \qquad y' = \dfrac{x}{y}\)

\(2 \qquad y' = \dfrac{y}{x}\)

\(3 \qquad y' = e^{y + x}\)

\(4 \qquad y' - 4 y = 29\)

Answer 30

@sillybilly123 can u clarify how u got the steps for the ln

Answer 31

im confused bc i got this

Answer 32

times each side by 2

Answer 33

can you show me that step please ..

Answer 34

sure

Answer 35

and guide me through the steps please. :)

Answer 36

starting from where?

Answer 37

where u said multiply by 2

Answer 38

@sillybilly123 D:

Answer 39

and there's @BlankSpace right up there !!!

Answer 40

wait what D:

Answer 41

someone called?

Answer 42

\[\frac{ 1 }{ 2 }\ln (y-1)+\frac{ 1 }{ 2 }\ln( y+1)= \frac{ 1 }{ 2 }\ln(x-2)-\ln(x-2)\]

Answer 43

im asking about number 9 which thats what i got for partial fractions but im still confuse on the algebra part.. so can you walk me through it lol

Answer 44

@sillybilly123