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

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- Nov 29th 2008, 09:48 AMmathemanyakDif eqn.
dx/dt=ax+sinx => find the general solution of this eqn.

- Nov 29th 2008, 02:29 PMshawsend
I feel this is a perfectly fine representation for the general solution:

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

Or no? - Nov 29th 2008, 09:54 PMmathemanyakre:
no it is not

- Nov 29th 2008, 10:50 PMmr fantastic
Quite right. It should be

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

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. - Nov 30th 2008, 01:07 AMMathstud28
- Nov 30th 2008, 02:30 AMmr fantastic
- Nov 30th 2008, 04:28 AMmathemanyakre:
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)

- Nov 30th 2008, 05:05 AMshawsend
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:

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

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}]

- Nov 30th 2008, 09:43 AMshawsend
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: $\displaystyle \left(\int_1^x\frac{du}{x+\sin(u)},x\right)$ 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}}]