# Thread: initial value problem for ODE's

1. ## initial value problem for ODE's

my question is on initial value problem for ODE's..

iv been asked to solve the following IVP:
dx/dt = (x^2)*(cos 2t) for x(pi/4)=0

Basically iv rearranged this and integrated to get

-1/x = 1/2*sin2t + c

But the problem came when i had to put in the values for x and t, when i plug x into the equation, i get -1/0, for which i cant do ....urm i dont know what to do next ...any help?? thankz

2. Originally Posted by dopi
my question is on initial value problem for ODE's..

iv been asked to solve the following IVP:
dx/dt = (x^2)*(cos 2t) for x(pi/4)=0

Basically iv rearranged this and integrated to get

-1/x = 1/2*sin2t + c

But the problem came when i had to put in the values for x and t, when i plug x into the equation, i get -1/0, for which i cant do ....urm i dont know what to do next ...any help?? thankz
You appear to be correct. But let's take this a step further. Take a quick look at the differential equation at the specific point x = 0. Then dx/dt = 0. So if we CAN get x to be 0 it will never change from this value. Thus there is a solution x(t) = 0. It's a "trivial" solution, but it does solve the differential equation.

-Dan

3. sorry i didnt quite understand what you mean

basically i want to find c in the equation

-1/x = 1/2*sin2t + c

for values of x=0 and t=pi/4

but u cant divide by 0 because 1/0 is is undefined...so what would i do...thankz

4. Originally Posted by dopi
sorry i didnt quite understand what you mean

basically i want to find c in the equation

-1/x = 1/2*sin2t + c

for values of x=0 and t=pi/4
No you don't; what you want to do is find a solution of the DE for which
x(t)=0 at t=pi/4.

Which as topsquark points out is the trivial solution x(t)=0 for all t.

RonL

5. I'm not certain why the standard solution method doesn't provide the x(t) = 0 solution. My best guess would be that the DE is non-linear. In any event, your problem with finding a C that fits the initial conditions is that the solution method you used has no solution that passes though the point given as the initial condition. Hence we have to look for other solutions using any method we can think of, a pretty hairy task in general. Fortunately the x(t) = 0 solution is obvious once it is looked for.

-Dan

6. Originally Posted by topsquark
I'm not certain why the standard solution method doesn't provide the x(t) = 0 solution.
I have no idea what you are talking about. And just joined the conversation. But let me say what my eyes see. I see that the equation is varaibles seperable. But you assume that $x^2\not =0$ in order to divide the equation. That assumption leads to non-trivial solutions. The other solutions are those that are $x^2=0$ are the trivial solutions. Hence the method does not work because it provides us with non-trivial solutions.

thankz