# Thread: Dif eqn.

1. ## Dif eqn.

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

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

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

Or no?

3. ## re:

no it is not

4. 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.

5. 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?

6. 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.

7. ## re:

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)

8. 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.

9. 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}}]