help me to solve this second order non-linear differential eq?

can anyone help me to solve this?

 y''+a*sign(y')*|y'|^1.8+by=c*sin(wt)
  y=y(t) ;
  y(0)=0 ;  y'(0)=0

[Jan]

Do you know, that you can convert an ODE system of degree 2 to one of degree 1?

[pedram]

yes it can be converted as this:

y(2)=y'(1)

y'(2)+a*sign*y'(1)*|y'(1)|^1.8+by(1)-c*sin(wt)=0

what should i do after this?

[Walter Roberson]

Where is the "x" in your revised ODE ?

[pedram]

i assumed x=c*sin(wt)

Answer 1

[Teja Muppirala]

Well you're almost there

Rewrite this:

    y(2)=y'(1)

    y'(2)+a*sign*y'(1)*|y'(1)|^1.8+by(1)-c*sin(wt)=0

Like this:

    y'(1) = y(2)

    y'(2) = -a*sign(y(2))*abs(y(2))^1.8 - b*y(1) + c*sin(w*t)

And then now it is in a form that you can use with MATLAB's many ODE solvers.

% Just picking some values for a,b,c,w
a = 1.2; 
b = 2.4;
c = 0.4;
w = 3;

dYdt = @(t,y) [y(2); -a*sign(y(2))*abs(y(2))^1.8 - b*y(1) + c*sin(w*t)]

ode45(dYdt, [0 30], [0 0])

[Greg Heath]

As Walter has stated, previously, the answer is y(t)= y'(t)=y''(t)= 0 for all t because at t=0 y, y' and y'' are all zero.

However if x = cos(w*t) there will be a solution that can be obtained numerically.

[Walter Roberson]

When I used Teja's formulation, then with non-negative "a" I get what appears to be a sine-convolved sine wave. With small negative "a" on the order of -1/10000 then the sine wave still looks more or less sine-convolved but with an expanding function, but once the wave gets to around +/- 1000 then the second derivative heads towards +/- infinity. Also, there are repeated small sections of the primary output which are not especially sine-convolved.

With cos() instead of sin() the output looks much the same.

Answer 2

[Walter Roberson]

The solution is y(t) = 0, x(t) = 0

[Greg Heath]

Only because this is an ill-posed problem with the silly condition x(0)=0. Typically. the assumption would be that x is an arbitrary known input, not necessarily zero at t=0. For example, your solution is correct for x(t) = sin(t), but not for x(t) = cos(t).

Then

y1' = 0*y1 + 1*y2

y2' = - b*y1 - a*y2*abs(y2)^0.8 + x

Which, as far as I can see, can only be solved numerically.

[Walter Roberson]

The original problem had x(t) on the right-hand side rather than c*sin(wt)