Solving a Differential Equation: Integrating and Finding the Correct Solution

[doroulla]

the question is to solve:
dx/dt -2t(2x-1)=0
when t=0, x=0..

I integrated and found:
ln(2x-1)=2t^2 + k
then : 2x-1 = e^2t^2 -2

then i put in the x=0 when t=0 and i found that k=-2

the solution i found is:

x= 0.5e(2t^2) - 0.5

is this correct? Thank you very much

 

[HallsofIvy]

What, again, is the anti-derivative of 2t dt?

 

[gb7nash]

k is incorrect. Also, a minor thing. You want:

ln|2x-1|=2t^2 + k

Now plug in x=0 and t=0 in this equation to see what k is.

What, again, is the anti-derivative of 2t dt?

I thought the same thing you did, but I think he just skipped a step. If you multiply out by the 1/2 from the integral of 1/(2x-1), it works out.

 

[doroulla]

yes i multiplied out.
If i put the conditions as it is then i have: ln(-1)=k
and this does not exist. You can't have ln(-1), right?

 

[gb7nash]

yes i multiplied out.
If i put the conditions as it is then i have: ln(-1)=k
and this does not exist. You can't have ln(-1), right?

The antiderivative of 1/(2x-1) is (1/2)ln|2x-1|. Notice the absolute value, so really you get ln 1

 

[doroulla]

i forgot about the absolute value. Thank you.

Now i get solution:
k=0
x= 0.5e2t^2 + 0.5

is this correct? :)

 

[SteamKing]

You can check this solution by plugging it back into the original differential equation.

 

[HallsofIvy]

the question is to solve:
dx/dt -2t(2x-1)=0
when t=0, x=0..

I integrated and found:
ln(2x-1)=2t^2 + k

You were correct up to here. Integrating both sides gives (1/2)ln|2x-1|= 2t^2+ k so that ln|2x-1|= 4t^2+2k and then
2x-1= e^{4t^2+ 2k}= Ce^{4t^2}\
where C= e^{2k}[/itex]<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> then : 2x-1 = e^2t^2 -2<br /> <br /> then i put in the x=0 when t=0 and i found that k=-2<br /> <br /> the solution i found is:<br /> <br /> x= 0.5e(2t^2) - 0.5 <br /> <br /> is this correct? Thank you very much </div> </div> </blockquote>

 

[doroulla]

when t=0 i get x=1

but when i put x=0 i get t=0


this should not make sense right?

But if i leave it as k=-2 as i found before as shown above i get x=0 when t=0. So that's is the correct solution?