anonymous
  • anonymous
**Calc 2 Help Needed** Solve the initial value problem below: (x^2 + 81)dy/dx = 3 y(9)=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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amistre64
  • amistre64
calc2 has the easy separables
anonymous
  • anonymous
I know this needs to be separated, but how?
amistre64
  • amistre64
\[(x^2+81)\frac{dy}{dx}=3\] \[\frac{dx}{(x^2+81)}\left((x^2+81)\frac{dy}{dx}=3\right)\] \[dy=\frac{3}{(x^2+81)}dx\]

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amistre64
  • amistre64
then the left is a tan-1 type integration
anonymous
  • anonymous
ok so dy integrated is y
amistre64
  • amistre64
\[x=9tan(t)\] \[dx=9sec^2(t)dt\] \[dy=\frac{3}{((9tan(t))^2+81)}dx\] \[dy=\frac{3*9sec^2(t)}{(81tan^2(t)+81)}dt\] \[dy=\frac{27sec^2(t)}{81(tan^2(t)+1)}dt\] \[dy=\frac{1}{39}\frac{sec^2(t)}{sec^2(t)}dt\] then again, it might be pretty easy unless i messed it someplace
amistre64
  • amistre64
1/3 , not 1/39
anonymous
  • anonymous
hmmm. I'm still lost. haha
anonymous
  • anonymous
where did you get the x = 9tanx?
amistre64
  • amistre64
\[y=\frac13t\] \[x = 9tan(t)\] \[\frac x9 = tan(t)\] \[tan^{-1}(\frac x9) = t\] \[y=\frac13 tan^{-1}(\frac x9)+C\] maybe lol
amistre64
  • amistre64
were did I get it? from a trig substitution technique
amistre64
  • amistre64
i know if I had 81tan^2(t) + 81 I can manimpulate it into a 81sec^2(t) making the integration that much simpler
amistre64
  • amistre64
i got an A in typing class, honest :)
anonymous
  • anonymous
haha.. well I'm horrible at calculus..
amistre64
  • amistre64
it just takes getting used to that you can actually adjust things in math. We grow up thinking this thin is static and unforgiving
amistre64
  • amistre64
when I first learned this stuff im like; How do they expect us to know to do that? :)
anonymous
  • anonymous
that's how I feel now. The class was relatively easy until 2 weeks ago.
anonymous
  • anonymous
ok so doesn't there need to be a constant K on this problem? which I would plug back in with y(9) = 0?
amistre64
  • amistre64
i use C as an "c"onstant
amistre64
  • amistre64
we can call it M or W or \(\Large \phi \) if we want to
anonymous
  • anonymous
is the final answer just an integer?
amistre64
  • amistre64
by integer i take it yo mean some real number value; let see :) \[y=\frac13 tan^{-1}(\frac x9)+C\] \[0=\frac13 tan^{-1}(\frac 99)+C\] \[0=\frac13 tan^{-1}(1)+C\] \[0=\frac13 \frac{\pi}{4}+C\] \[0=\frac{\pi}{12}+C\] \[-\frac{\pi}{12}=C\] therefore \[y=\frac13 tan^{-1}(\frac x9)-\frac{\pi}{12}\]
amistre64
  • amistre64
the final answer is an equation that will solve the initial condition
anonymous
  • anonymous
makes sense! somewhat! haha. thanks for your help, you're a dang genius.
amistre64
  • amistre64
youre welcome :)
anonymous
  • anonymous
I have another one similiar if you want to help??
amistre64
  • amistre64
maybe in a minute or so if noone else takes it; i gotta rest form this one ...
anonymous
  • anonymous
ok thanks!

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