anonymous
  • anonymous
don't know how to solve this differential equation
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
1 Attachment
freckles
  • freckles
\[\frac{dy}{y}= 6x \sqrt{3x^2-1} dx\]
freckles
  • freckles
integrate both sides

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anonymous
  • anonymous
okay real quick wouldn't it be dy/dx on the left side?
UsukiDoll
  • UsukiDoll
separation of variables.... in the form of h(y) dy = f(x) dx
freckles
  • freckles
i multiplied both sides by dx
UsukiDoll
  • UsukiDoll
you need all y's to the left and all x's to the right... then integrate both sides..
freckles
  • freckles
i also divided both sides by y and multiplied both sides by 6x
anonymous
  • anonymous
okay i see one sec let me integrate
anonymous
  • anonymous
okay so im getting \[\frac{ 2(3x^2 -1)^3/2 }{ 3}+C\]
anonymous
  • anonymous
not sure how to do the left side
freckles
  • freckles
\[\frac{2}{3}(3x^2-1)^\frac{3}{2}+C \text{ is \right for the \right hand side}\]
freckles
  • freckles
the left hand hand is just ln|y|+k but you only need one constant of integration in your equation
freckles
  • freckles
\[\ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K\]
anonymous
  • anonymous
wait how did you get that from dy/y?
freckles
  • freckles
\[\frac{1}{6x} \frac{dy}{dx}=y \sqrt{3x^2-1} \\ \text{ divide both sides by } y \\ \frac{1}{6x} \frac{1}{y} \frac{dy}{dx} = \sqrt{3x^2-1} \\ \text{ multiply both sides by } 6x \\ \frac{1}{y} \frac{dy}{dx}= 6x \sqrt{3x^2-1} \\ \text{ multiply both sides by } dx \\ \frac{1}{y} dy= 6x \sqrt{3x^2-1} dx \\ \text{ or this can also be written as } \frac{dy}{y}=6x \sqrt{3x^2-1} dx\]
UsukiDoll
  • UsukiDoll
1/y dy the antiderivative of 1/y is just ln y
freckles
  • freckles
oh not how I got dy/y
freckles
  • freckles
but how you integrate it
UsukiDoll
  • UsukiDoll
but we're still not done because we need a y by itself... we have to take e^ {} on everything so that we have y = (all the contents)
freckles
  • freckles
the derivative of ln|y| w.r.t to y is y'/y=1/y
anonymous
  • anonymous
no i ment how did you integrate dy/y
freckles
  • freckles
\[\int\limits \frac{du}{u}=\ln|u|+K \text{ since } \frac{d}{du}(\ln|u|+K)=\frac{1}{u}+0\]
UsukiDoll
  • UsukiDoll
first rewrite this as 1/y dy then the antiderivative of 1/y is ln y it's a logarithm the derivative is 1/y (1) because we also have to take the derivative of the denominator but it's just 1, so it's 1/y only
freckles
  • freckles
well actually it is a definition in most calculus books the following: \[\ln(x)=\int\limits_1^x \frac{1}{t} dt\]
anonymous
  • anonymous
okay i get it so now i just isolate y right?
freckles
  • freckles
you can if you are asked to find an explicit form for y
anonymous
  • anonymous
it just says solve the differential equation ... so i assume i just solve for y
freckles
  • freckles
you could stop at \[\ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K\] then but solving this one for y is doable
UsukiDoll
  • UsukiDoll
e^{ln y} e and ln are inverses of each other so it cancels out for y we're you given initial conditions in your problem ? if you were, then we can solve... if not we can just take e^{} on both sides
anonymous
  • anonymous
okay but then how do i get rid of the absolute value ?
UsukiDoll
  • UsukiDoll
\[\large \ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K\] \[\large e^{ln|y|}=e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K}\] \[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K}\]
freckles
  • freckles
\[e^{\ln|y|}=|y|\]
anonymous
  • anonymous
these are my answer choices by the way and none of the above
freckles
  • freckles
as long as y is not 0 of course
freckles
  • freckles
e^K=C @Jdosio
anonymous
  • anonymous
ugh i thought your K was your way of saying Constant
UsukiDoll
  • UsukiDoll
forgot that part... e^K = C substitution.
freckles
  • freckles
\[\text{ assuming } y>0 \text{ we do have } \\ y=e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}+K} \\ \\ \text{ but we can also rewrite this } \\ y=e^{k} e^{\frac{2}{3} (3x^2-1)^\frac{3}{2}} \text{ by law of exponents } \\ \text{ so now \it doesn't matter much about the } y>0 \text{ since } C \\ \text{ could be negative or positive } \\ \\ y=C e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}\]
UsukiDoll
  • UsukiDoll
\[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}e^{K} \] \[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}C\]
freckles
  • freckles
@UsukiDoll is not really part you forgot. it is just a little something extra that isn't really needed but people write because it does look prettier
anonymous
  • anonymous
okay so wtf is K then
freckles
  • freckles
e^k is a constant
freckles
  • freckles
C is a constant
freckles
  • freckles
you are still replacing an unknown constant with another unknown constant
anonymous
  • anonymous
why did you use K instead of c then
freckles
  • freckles
because you can
freckles
  • freckles
you can mean any letter to be a constant
freckles
  • freckles
I used K as a constant above we ended up with e^K which still an unknown constant you can replace that with another unknown constant call it C
UsukiDoll
  • UsukiDoll
you can have any letters as a constant tbh. but it's usually C
anonymous
  • anonymous
okay i understand that the e^anyunknown constant equals another constant my question was why you used two different symbols for constant in the same equation
freckles
  • freckles
e^K =C doesn't imply K=C
UsukiDoll
  • UsukiDoll
I usually just leave the solution alone...unless it begs to be changed .
anonymous
  • anonymous
okay so then its just to keep track of which constant is which , okay that sort of makes sense
freckles
  • freckles
it seems funny to me to replace e^C with C
freckles
  • freckles
it is like replace e^5 with 5
freckles
  • freckles
even though e^5 doesn't equal 5
anonymous
  • anonymous
i find it soothing to just be able to take out constants like that
freckles
  • freckles
but yeah you could have just said e^C then replace it with C i think some people do that
freckles
  • freckles
take out constants?
anonymous
  • anonymous
combine
anonymous
  • anonymous
anyway thank you very much for the help guys i wish i could give you both a medal :D
anonymous
  • anonymous
ugh who do i give it too D:
UsukiDoll
  • UsukiDoll
I gave one to freckles already... so......?

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