# Doubts in Differential equations

1. Nov 6, 2004

### siddharth

If I have a second order differential equation (like in SHM) ,ie, (d^2x)/(dt^2)=-kx, (1) the solution is of the form Acos(wt+B) (2). This is obviously true because by substituting i get the result. But the answer could very well have been [-(kx^3)/6 +AX^2+BX + C]. Why do we not take this answer to solve the SHM problems?

Also, if I am solving the above differential equation, instead of equating (1) to -kx i could have equated it to zero and found the solution, and added to (2). It would still satisfy the equation. So I could keep adding many such solutions to (2). But then there would be an infinite number of solutions! How is that possible??

And finally what if I have (d^2x)/(dt^2)=-kx + C .What would be the general solution (I know that the frequency will not change).For example if I have a spring with two masses m1 and m2 on either side and I apply a force F1 and F2 respectivley on each of them, what would be the equation of motion of the center of mass?
It would be very helpful if anyone can tell me the equation or how to solve the Differential equation.

2. Nov 6, 2004

### arildno

"If I have a second order differential equation (like in SHM) ,ie, (d^2x)/(dt^2)=-kx, (1) the solution is of the form Acos(wt+B) (2). This is obviously true because by substituting i get the result. But the answer could very well have been [-(kx^3)/6 +AX^2+BX + C]. Why do we not take this answer to solve the SHM problems"
This is simply incorrect; where's your "t" dependence here, which is the one you differentiate with respect to???

3. Nov 6, 2004

### siddharth

OOPS! sorry i mean [-(kt^3)/6 +At^2+Bt + C].

4. Nov 6, 2004

### matt grime

Well, you could suggest that might be the answer, but if you put it into the equation, then it doesn't satisfy the differential equation.

We guess what the solution might be, because of experience, and then use some large result that states that this is the only answer.

d^2x/dt^2=-kx

tells us that the function x(t) is one that when differentiated twice gives us the negative of what we started with. Experience tells us that sin and cos are such functions, so we guess and answer, show it's ok, and then rely on the larger results to know these are the only ones.

writing L for the differntial operator of diffin twice wrt to t

we solve Lx=-kx+C by solving the homogenous part

Lx+kx=0, and then by solving for the particular solution involving C.

Last edited: Nov 6, 2004
5. Nov 6, 2004

### siddharth

Oh, i got it. I was just being plain stupid in the differentiation bit. But if I have
(d^2x)/(dt^2) + dx/dt = 0. Then I could have x(t) = A(e^Bt) + A1(e^b1t) + .......
such that B(B+1)=0 , B1(B1+1)=0, ......
So i would still have infinite soultions.
Solving Lx + kx =0 is fine. But if C is a constant, then how am I supposed to solve Lx=-kx+c?
Thanks for the help.

Last edited: Nov 6, 2004
6. Nov 6, 2004

### matt grime

if the coefficients B and B1 etc all satisfy y(y+1)=0 then there are only two possibilities, 0 and -1.

As for solving the particular bit, I'd guess a polynomial in t, possibly of the form x(t)=C/k, or possilby -C/k, your minus signs are changing somewhat randomly.
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