Differential equation problem

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November 18th 2012, 10:42 AM
entrepreneurforum.co.uk

Differential equation problem

Differential equation problem:

Question;

show any function $x=x(t)$ which satisfies $xe^{tx} = C$ is a solution of the differential equation $(1+tx)x' = x^2$

This is what I have so far;

$C' = x^2e^{tx}$ W.R.T. (t)
$x' = t$

so I have;

$C' = (xt * xt)e^{tx}$

$-x^2 = c'e^{tx}$

$-x^2=x^2 * e^{tx}$

therefore;

$-x^2 = [[xt*xt]e^{tx}]e^{tx}$

then I end up with

$-x^2 = xt(t)e^{2tx}$

$-x^2 = xt(x')$

What happens to the $e^{2tx}$ ???
November 18th 2012, 11:22 AM
topsquark

Re: Differential equation problem

Your troubles start way before your last line. C is a constant, not a function of x or t. See if that helps.

-Dan
November 18th 2012, 11:30 AM
entrepreneurforum.co.uk

Re: Differential equation problem

No it did not help, can you offer another tip to push me in the right direction?

Thanks
November 18th 2012, 04:51 PM
Prove It

Re: Differential equation problem

Quote:

Originally Posted by entrepreneurforum.co.uk

Differential equation problem:

Question;

show any function $x=x(t)$ which satisfies $xe^{tx} = C$ is a solution of the differential equation $(1+tx)x' = x^2$

This is what I have so far;

$C' = x^2e^{tx}$ W.R.T. (t)
$x' = t$

so I have;

$C' = (xt * xt)e^{tx}$

$-x^2 = c'e^{tx}$

$-x^2=x^2 * e^{tx}$

therefore;

$-x^2 = [[xt*xt]e^{tx}]e^{tx}$

then I end up with

$-x^2 = xt(t)e^{2tx}$

$-x^2 = xt(x')$

What happens to the $e^{2tx}$ ???

$\displaystyle \begin{align*} x\,e^{t\,x} &= C \\ \frac{d}{dt} \left( x\, e^{t\, x} \right) &= \frac{d}{dt} \left( C \right) \\ x \, \frac{d}{dt} \left( e^{t\, x} \right) + e^{t\, x} \, \frac{d}{dt} \left( x \right) &= 0 \\ x \left[ e^{t\, x} \, \frac{d}{dt} \left( t\, x \right) \right] + e^{t\, x} \, \frac{dx}{dt} &= 0 \\ x \, e^{t\, x} \left( t\, \frac{dx}{dt} + x \right) + e^{t\, x} \, \frac{dx}{dt} &= 0 \\ t\, x\, e^{t\, x} \, \frac{dx}{dt} + x^2\, e^{t\, x} + e^{t\, x}\, \frac{dx}{dt} &= 0 \\ e^{t\, x} \left( 1 + t\, x \right) \frac{dx}{dt} + x^2\, e^{t\, x} &= 0 \\ e^{t\, x} \left( 1 + t\,x \right) \frac{dx}{dt} &= -x^2\, e^{t\, x} \\ \left( 1 + t\,x \right) \frac{dx}{dt} &= -x^2 \end{align*}$

I assume this is what your original equation was supposed to be...
November 19th 2012, 07:38 PM
wallisonline

Re: Differential equation problem

Where can I find a good place with a lot of differential equation examples with solutions?

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