- anonymous

don't know how to solve this differential equation

- katieb

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- anonymous

##### 1 Attachment

- freckles

\[\frac{dy}{y}= 6x \sqrt{3x^2-1} dx\]

- freckles

integrate both sides

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## More answers

- anonymous

okay real quick wouldn't it be dy/dx on the left side?

- UsukiDoll

separation of variables.... in the form of h(y) dy = f(x) dx

- freckles

i multiplied both sides by dx

- UsukiDoll

you need all y's to the left and all x's to the right... then integrate both sides..

- freckles

i also divided both sides by y
and multiplied both sides by 6x

- anonymous

okay i see
one sec let me integrate

- anonymous

okay so im getting \[\frac{ 2(3x^2 -1)^3/2 }{ 3}+C\]

- anonymous

not sure how to do the left side

- freckles

\[\frac{2}{3}(3x^2-1)^\frac{3}{2}+C \text{ is \right for the \right hand side}\]

- freckles

the left hand hand is just ln|y|+k
but you only need one constant of integration in your equation

- freckles

\[\ln|y|=\frac{2}{3}(3x^2-1)^\frac{3}{2}+K\]

- anonymous

wait how did you get that from dy/y?

- 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

1/y dy the antiderivative of 1/y is just ln y

- freckles

oh not how I got dy/y

- freckles

but how you integrate it

- 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

the derivative of ln|y| w.r.t to y is y'/y=1/y

- anonymous

no i ment how did you integrate dy/y

- freckles

\[\int\limits \frac{du}{u}=\ln|u|+K \text{ since } \frac{d}{du}(\ln|u|+K)=\frac{1}{u}+0\]

- 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

well actually it is a definition in most calculus books the following:
\[\ln(x)=\int\limits_1^x \frac{1}{t} dt\]

- anonymous

okay i get it so now i just isolate y right?

- freckles

you can if you are asked to find an explicit form for y

- anonymous

it just says solve the differential equation ... so i assume i just solve for y

- 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

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

okay but then how do i get rid of the absolute value ?

- 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

\[e^{\ln|y|}=|y|\]

- anonymous

these are my answer choices by the way
and none of the above

##### 3 Attachments

- freckles

as long as y is not 0 of course

- freckles

e^K=C @Jdosio

- anonymous

ugh i thought your K was your way of saying Constant

- UsukiDoll

forgot that part... e^K = C substitution.

- 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

\[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}e^{K} \]
\[\large y =e^{\frac{2}{3}(3x^2-1)^\frac{3}{2}}C\]

- freckles

@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

okay so wtf is K then

- freckles

e^k is a constant

- freckles

C is a constant

- freckles

you are still replacing an unknown constant with another unknown constant

- anonymous

why did you use K instead of c then

- freckles

because you can

- freckles

you can mean any letter to be a constant

- 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

you can have any letters as a constant tbh. but it's usually C

- 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

e^K =C doesn't imply K=C

- UsukiDoll

I usually just leave the solution alone...unless it begs to be changed .

- anonymous

okay so then its just to keep track of which constant is which , okay that sort of makes sense

- freckles

it seems funny to me to replace e^C with C

- freckles

it is like replace e^5 with 5

- freckles

even though e^5 doesn't equal 5

- anonymous

i find it soothing to just be able to take out constants like that

- freckles

but yeah you could have just said e^C
then replace it with C
i think some people do that

- freckles

take out constants?

- anonymous

combine

- anonymous

anyway thank you very much for the help guys i wish i could give you both a medal :D

- anonymous

ugh who do i give it too D:

- UsukiDoll

I gave one to freckles already... so......?

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