"amadeo constantine" <amadeoconstantine@gmail.com> wrote in message <icgpfs$r41$1@fred.mathworks.com>...
> Hi Matlabers,
>
> I have problem on second order differential equation.
> D2(y) = 20*sin(y)  200
>
> where D2(y) is second order diff of Y where Dy = dy/dt
> I try to use this:
> Q = dsolve('D2y = 20*sin(y)  200, 'y(0) = 0','Dy(0) = 0','t')
> but it doesn't work because of sin(y).
>
> Please give enlightment, Thanks.
>
> Amadeo
         
There are some tricks you can play with this differential equation. Call v = dy/dt. Then
d^2y/dt^2 = dv/dt = dv/dy*dy/dt = dv/dy*v = 1/2*d(v^2)/dy = 20*sin(y)200
v^2 = int(40*sin(y)400,'y') = 40*cos(y)400*y+C
The initial conditions imply that C = 40, giving
v^2 = 40*(1cos(y))400*y = 80*sin(y/2)^2400*y
It is clear that with increasing t the quantity y must become negative and therefore v = dy/dt must initially become negative. Hence
v = dy/dt = sqrt(80*sin(y/2)^2400*y) for t >= 0
dt/dy = 1/sqrt(80*sin(y/2)^2400*y)
t = int(1/sqrt(80*sin(yy/2)^2400*yy),'yy',0,y) =
int(1/sqrt(80*sin(yy/2)^2400*yy),'yy',y,0)
where yy is the dummy integration variable for y, and where the y lower limit of integration is negative.
Thus the elapsed t can be obtained as a function of y provided this integral can be evaluated. It has a square root type singularity at y = 0, but the integral would nevertheless be finite. An appropriate change of variable could avoid the singularity. I don't know an explicit form for this integral but presumably it can be evaluated numerically for various values of y (with perhaps the aforementioned change of variable.)
Roger Stafford
