Solve differential equation still confused

[doroulla]

hi. I have asked for help on this again but i got even more confused. The question says solve:
dx/dt - 2t(2x-1) = 0 when t=0 when x=0.

I did:
0.5 ln(2x-1) = 2t^2 / 2 + K

then i have 2x-1 = 0.5(e^t^2) + K

putting x=0 and t=0 i get:
K=0

thus i have 2x-1 = e^(2t^2)

thus: x=0.5e^(2t^2) + 1

but i know it is wrong because when i put t=0 i do not get x=0

someone please help me. I've been stuck on this coursework for a week! Thank you

 

[Sethric]

0.5 ln(2x-1) = 2t^2 / 2 + K

then i have 2x-1 = 0.5(e^t^2) + K

There is your problem. You can't manipulate the right side and forget the constant. In this case, it will become a constant in front of the exponential.

 

[LCKurtz]

hi. I have asked for help on this again but i got even more confused. The question says solve:
dx/dt - 2t(2x-1) = 0 when t=0 when x=0.

I did:
0.5 ln(2x-1) = 2t^2 / 2 + K

This should be (1/2)ln|2x-1| =t² + k, with absolute value signs.

ln|2x-1| = 2t²+2K

|2x-1| = e^{2t^2+2K}=e^{2t^2}e^{2K}

Best not to evaluate the constant yet. Notice that with the absolute value signs, both sides are positive. We can drop the absolute value signs and write:

2x-1 = \pm e^{2K}e^{2t^2} = Ce^{2t^2}

where C is the new constant which can be positive or negative. Now put in your x=0 when t = 0 to evaluate the constant and see if it works.

 

[doroulla]

thank you very much. I got it correct now. I found C=-1
so i get: 2x-1 = -e^(2t^2)

thus i get : x= 0.5 - 0.5e^(2t^2)

when i put t=0 in this equation, i get x=0 thus this is correct right? Thank you

 

[LCKurtz]

thank you very much. I got it correct now. I found C=-1
so i get: 2x-1 = -e^(2t^2)

thus i get : x= 0.5 - 0.5e^(2t^2)

when i put t=0 in this equation, i get x=0 thus this is correct right? Thank you

You don't have to ask me. You know it satisfies the initial conditions. You can always check if you have it correct by substituting it back in the original DE to see if it works. That's a good check against arithmetic errors.