satellite73:
\[y=ae^x+be^{-x}\] is a solution of \[y''-y=0\]
satellite73:
first off, it is because it is, you can check. all this says is that the second derivative is equal to the function itself
satellite73:
now if \(y(o)=1\) what do you know? that \[1=ae^0+be^0=a+b\]
satellite73:
that is one equation then \[y'=ae^x-be^{-x}\]and since \[y'(0)=2\] you get \[2=a-b\]
satellite73:
solve \[a+b=1\\a-b=2\] and you are done
satellite73:
\[\csc(y)dx+\sec^2(x)dy=0\] right?
Vrefela:
yes
satellite73:
ok first step is algebra
satellite73:
you have to make it look like something(y)dy = someothething(x)dx then integrate, but the algebra is first
satellite73:
you want me to do the algebra, or you do it?
Vrefela:
please
satellite73:
ok i am kind of a bonehead so it takes me a few steps
Vrefela:
sounds good.
satellite73:
\[\csc(y)dx+\sec^2(x)dy=0\] \[\sec^2(x)dy=\csc(y)dx\] now divide to get \[\sin(y)dy=\cos^2(x)dx\] let me know when this is obvious
satellite73:
oh, and it is wrong
satellite73:
\[\sin(y)dy=-\cos^2(x)dx\]
satellite73:
i forgot the stupid minus sign
Vrefela:
looks good.
satellite73:
now integrate
satellite73:
\[\int\sin(y)dy=\int\cos^2(x)dx\]
satellite73:
again i forgot the minus sign lets put it on the left
Vrefela:
okie dokie.
satellite73:
so the left hand side is \[\cos(y)\]
Vrefela:
right
satellite73:
the right side is a very common integral, you use one of those double angle formulas that i always forget but you end up with \[\frac{1}{2}x+\frac{1}{4}\sin(2x)+c\]
satellite73:
if you want to refresh your memory on sine squared and cosine squared, look at a calc 2 book
Vrefela:
Lol i think i remember this part. But thats it?
satellite73:
actually we did a raft of them right?
Vrefela:
Most definitely.
satellite73:
yeah thats it you can write it in a bunch of different ways, but basically it is going to be \[\cos(y)=\frac{1}{2}x+\frac{1}{4}\sin(2x)+c\]
satellite73:
you cannot solve this for y, some of them you can
Vrefela:
So is that the final form?
Vrefela:
of the answer?
satellite73:
yes, unless you want to multiply both sides by 4 to clear the fraction
satellite73:
thats what they do to get the answer in the back of the book
Vrefela:
True but that's it?
satellite73:
yes dammit
satellite73:
want to try 3?
Vrefela:
Lol okie! Thank you!
satellite73:
or 13?
Vrefela:
Let's do it!
satellite73:
3 you actually solve for y do it yourself. make it look like dy=something in x dx
satellite73:
write your answer here, that takes only two algebra steps
Vrefela:
I kinda don't know where to start....
satellite73:
lol
satellite73:
subtract dx from both sides what do you get?
Vrefela:
(e^(y) + 1)^2*e^-y + (e^x +1)^3*3x = 0 dx
satellite73:
no cheating and looking up the answer
Vrefela:
i'm not
satellite73:
ooh you are doing 13, not 3 lets start with 3
Vrefela:
lollll ok
satellite73:
\[dx+e^{3x}dy=0\] go , subtract dx from both sides
Vrefela:
lol ok
Vrefela:
done
satellite73:
now the dx is on the right but there is still a function of x on the left divide by \(e^{3x}\)
satellite73:
what do you get?
Vrefela:
\[dx/(e^(3x))\]
satellite73:
nice latex
Vrefela:
lol sorry still new
satellite73:
what i wanted you to write was \[sy=e^{-3x}dx\]
satellite73:
\[dy=e^{-3x}dx\]
Vrefela:
show off
satellite73:
either way next is \[\int dy=\int e^{-3x}dx\] what do you get?
Vrefela:
y = (-1/3)e^-3x
satellite73:
plus c
satellite73:
which is really needed here
Vrefela:
okie
satellite73:
ok next is 5
satellite73:
let me know when you make \[x\frac{dy}{dx}=4y\] look like \[g(y)dy=f(x)dx\]
satellite73:
like ratios an proportions from arithmetic class in 4th grade
satellite73:
...
Vrefela:
x = 4y dy?
Vrefela:
no no
satellite73:
no no is right
Vrefela:
4y^-1 dy = x
satellite73:
no
Vrefela:
damn it
Vrefela:
im toaast
satellite73:
a) multiply both sides by dx
Vrefela:
I thought so but i seen it as dy/dx
Vrefela:
like one identity
satellite73:
yeah i know get used to it
satellite73:
you need \[g(y)dy=f(x)dx\] remember
satellite73:
dx should be on line line, not in a denominator
Vrefela:
ok so x dy = 4y dx
satellite73:
great
satellite73:
now you need the x's with the dx and the y's with the dy
Vrefela:
so 4y^-1 dy = x^-1 dx
satellite73:
yay
satellite73:
i wrote \[\frac{dy}{y}=\frac{4dx}{x}\] oh wait, your 4 is on the wrong side doh
Vrefela:
Oh oops
satellite73:
ok now integrate \[\int\frac{dy}{y}=\int\frac{4dx}{x}\]
Vrefela:
ln(y) = 4 ln(x)
Vrefela:
?
satellite73:
yes but what did you forget? it is very important here
Vrefela:
+ c
satellite73:
ok good so \[\ln(y)=4\ln(x)+c\] solve for \(y\)
satellite73:
if you get stuck here, i will show you
Vrefela:
e ^ (4 ln(x) + c )
satellite73:
yes, but you can do better
Vrefela:
....?
satellite73:
remember two things
satellite73:
first, what is another way to write \[4\ln(x)\]?
Vrefela:
.... i don't remember?
satellite73:
jeez really? laws of logariddims?
Vrefela:
I'm old
satellite73:
does \[\log(x^n)=n\log(x)\] ring a bell?
Vrefela:
Oh crap
satellite73:
soo \[e^{\ln(x^4)+c}\] is a start
Vrefela:
And thats the final form right?
satellite73:
next note that \[e^{\ln(x^4)+c}=e^{\ln(x^4)}\times e^c\]
Vrefela:
So basically simplify the form?
satellite73:
you are not done yet
satellite73:
note that the \(+c\) from the original integral is now a product, after you got rid of the log
satellite73:
now, for 100 points, what is \[\huge e^{\ln(\color{red}{x^4})}\]
Vrefela:
x^4
satellite73:
yay
satellite73:
final answer ?
Vrefela:
y = x^4 * e^c
satellite73:
yes, but don't be a mooncalf
satellite73:
\[e^c\] is just a constant some other constant you can call it C
Vrefela:
Are you serious?! all that work for x^4 + c
satellite73:
no no careful
satellite73:
\[cx^4\]
satellite73:
it was times right? not plus
satellite73:
yea all that work check the answer
satellite73:
replace \(y\) by \(cx^4\) in \[x\frac{dy}{dx}=4y\] and see that it is true
satellite73:
make sure you understand those steps, the solving for y part it comes up quite a bit
Vrefela:
Wait
satellite73:
ok i wait
Vrefela:
Why would that be true though? if we plugged in the answer we got for y?
satellite73:
\[x\frac{dy}{dx}=4y\] put \(y=cx^4\) get \[x\times 4cx^3=cx^4\] oh doh
satellite73:
i made a typo \[x\times 4cx^3=4cx^4\]
satellite73:
clear or no?
Vrefela:
OH!
Vrefela:
Goodness I'm so slow.
satellite73:
whew once you see it it is obvious right?
Vrefela:
Thank you!
Vrefela:
Yes lol absolutely
satellite73:
too much or one more?
Vrefela:
Alright I have to get to bed but thank you soooo much for tonight.
Vrefela:
Can I annoy you tomorrow?
satellite73:
ok me too you better do some of these many of these
satellite73:
the steps are clear
Vrefela:
Lol yea I definitely have to brush up. It's my last math class. I can do this lol
satellite73:
make it look like \[g(y)dy=f(x)dx\] then integrate don't screw up with the algebra
Vrefela:
I won't
satellite73:
sure tomorrow maybe earlier than 10:30 i am old
Vrefela:
Please and thank you.
satellite73:
\[\color\magenta\heartsuit\]
Vrefela:
Thank you so much Professor, I seriously appreciate this
Vrefela:
<3 You're the man
satellite73:
that was the sign off of my alter ego ttyl
Vrefela:
Lol goodnight
satellite73:
night
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