Dif eqn.

**mathemanyak** · November 29th 2008, 09:48 AM

dx/dt=ax+sinx => find the general solution of this eqn.

**shawsend** · November 29th 2008, 02:29 PM

I feel this is a perfectly fine representation for the general solution:

$\int_{x_0}^{x}\frac{du}{au+\sin(u)}=t$

Or no?

**mathemanyak** · November 29th 2008, 09:54 PM

no it is not

**mr fantastic** · November 29th 2008, 10:50 PM

Originally Posted by mathemanyak

no it is not

Quite right. It should be

$\int_{x_0}^{x}\frac{du}{au+\sin(u)} + t_0 = t$ .

If you have reason for thinking otherwise, please don't be backward in coming forward.

By the way, if this is meant to be some sort of challenge question, please say so clearly in the question itself. Members are happy to help people who are genuinely stuck and need help. However, some members have better things to do with their time if that's not the case.

**Mathstud28** · November 30th 2008, 01:07 AM

Originally Posted by mr fantastic

Quite right. It should be

$\int_{x_0}^{x}\frac{du}{au+\sin(u)} + t_0 = t$ .

If you have reason for thinking otherwise, please don't be backward in coming forward.

By the way, if this is meant to be some sort of challenge question, please say so clearly in the question itself. Members are happy to help people who are genuinely stuck and need help. However, some members have better things to do with their time if that's not the case.

Just curious, but isn't the constant that is gotten when integrating the integral sufficient? Or is the $+t_0$ also neccessary?

**mr fantastic** · November 30th 2008, 02:30 AM

Originally Posted by Mathstud28

Just curious, but isn't the constant that is gotten when integrating the integral sufficient? Or is the $+t_0$ also neccessary?

The boundary condition $x(t_0) = x_0$ has been built into this solution. There's no arbitrary constant of integration in this integral representation of the solution.

**mathemanyak** · November 30th 2008, 04:28 AM

thanks for advice but ı ask that can we do this solution integration factor, in this differantial equation? and ı solve this eqn with yours answer and ı get it (Arctan((1+axtan(x/2))/(x^2-1)+(1/2))/(x^2-1)+(1/2)

**shawsend** · November 30th 2008, 05:05 AM

Originally Posted by mathemanyak

thanks for advice but ı ask that can we do this solution integration factor, in this differantial equation? and ı solve this eqn with yours answer and ı get it (Arctan((1+axtan(x/2))/(x^2-1)+(1/2))/(x^2-1)+(1/2)

Hey mathemanyak. I do not feel that's an answer for one thing where's the t? Not sure what you did but allow me to make a suggestion which I have found very useful in helping me with differential equations: Solve it first numerically, then superimpose any analytical expression I get over the numerical solution. If they don't exactly agree within some tolerance, then it's very likely the analytical expression is wrong baring singularities or high oscillating problems or other non-linear effects. So I numerically solved the IVP:

$\frac{dx}{dt}=x+\sin(x);\quad x(0)=1$

using Mathematica's numerical solver "NDSolve":

Code:

sol = NDSolve[{Derivative[1][x][t] == 
     x[t] + Sin[x[t]], x[0] == 1}, x, 
   {t, 0, 5}]
Plot[Evaluate[x[t] /. sol], {t, 0, 5}]

and got the plot below. Now, any sort of analytical expression you get for this IVP very likely (see exceptions above) has to agree with this plot. You may wish to try yours.

**shawsend** · November 30th 2008, 09:43 AM

I was curious about this so I constructed a 10'th degree polynomial fit to the inverse data. That is, I numerically integrated the integral from 1 to 200 (increment by 1) and then formed a table of (t,x) pairs: $\left(\int_1^x\frac{du}{x+\sin(u)},x\right)$ and fit it to the polynomial. The plot below shows a overlay of the NDSolve results (in red) with the polynomial fit (green). I think red+green=brown.

Code:

f[x_] := NIntegrate[1/(u + Sin[u]), 
    {u, 1, x}]; 
tlist = Table[{f[x], x}, {x, 1, 200, 1}]; 
lp = ListPlot[tlist]
clist = Table[Subscript[a, n], {n, 0, 10}]
p[x_] = Sum[Subscript[a, n]*x^n, 
    {n, 0, 10}]; 
alist = FindFit[tlist, p[x], clist, x]; 
pic1 = Plot[p[x] /. alist, {x, 0, 5}, 
   PlotStyle -> Red]
sol = NDSolve[{Derivative[1][y][t] == 
      y[t] + Sin[y[t]], y[0] == 1}, y, 
    {t, 0, 5}]; 
pic2 = Plot[Evaluate[y[t] /. sol], 
   {t, 0, 5}, PlotStyle -> Green]
Show[{pic2, pic1}, PlotRange -> {{0, 5}, 
   {0, 200}}]

Thread: Dif eqn.

LinkBack

Thread Tools

Display

Dif eqn.

re:

re:

Search Tags