Differential equation what going on?

[jamesd2008]

Hi could someone explain why this result occurs when solving this differential equation.

dx/dt= xt

1/xdx/dt=t then intergrating both side we get with respect to dt we get,

Inx=t^2/2 + c now this is the bit i don't understand why does the answer then become,

x=Ce^2/t^2

I get really confused how the equation gets re-arranged into the above form after the intergration has occured. Any help please?

Thanks
James

 

[sutupidmath]

Inx=t^2/2 + c now this is the bit i don't understand why does the answer then become,

x=Ce^2/t^2

I get really confused how the equation gets re-arranged into the above form after the intergration has occured. Any help please?

Thanks
James

You are just exponentiating both sides of the equation. But remember that exponential functions and log's are inverse functions so e^lnx=x

 

[jamesd2008]

Thanks but how come it is c*the e and then raised to the power of 2/t^2. Is that just a rule of exponentials?
Please help doing my head in!

Thanks
James

 

[tiny-tim]

Inx=t^2/2 + c now this is the bit i don't understand why does the answer then become,

x=Ce^2/t^2

Hi James!

No, x = Ce^t2/2 …

C is e^c.

 

[EliotHijano]

Hey James
dx/dt= xt
dx/x = t·dt
Ln(x) = (t^2)/2 + K
x=e ^{t2 /2 +K} = e ^K · e ^{t2 /2}
If we say that e^K = C then;
x= C·e ^{t2 /2}

I hope it helps

 

[jamesd2008]

Thanks all for your help got it now! thanks again james : -)