# Differential equation and initial condition

• October 25th 2009, 07:48 AM
ben.mahoney@tesco.net
Differential equation and initial condition
Q) The question is for me to sketch the graph of x(t) for different initial conditions.

dx/dt = sqrt(3) - 2sin(x)

A) i cant see how this is separable variables or integrating factor method.

So do I do: x(t) = sqrt(3)t - 2tsin(x) + C

Saying x = x0 at t=0 just gives me c = x0

Does this look correct?

Thanks

• October 25th 2009, 09:19 AM
CaptainBlack
Quote:

Originally Posted by ben.mahoney@tesco.net
Q) The question is for me to sketch the graph of x(t) for different initial conditions.

dx/dt = sqrt(3) - 2sin(x)

A) i cant see how this is separable variables or integrating factor method.

So do I do: x(t) = sqrt(3)t - 2tsin(x) + C

Saying x = x0 at t=0 just gives me c = x0

Does this look correct?

Thanks

1. Have you checked that that is a solution by differentiating it and seeing if it satisfies the equation?

2. Are you sure that you have typed the equation correctly?

CB
• October 25th 2009, 09:34 AM
ben.mahoney@tesco.net
if you differentiate it you just take away the t's on each part so in that respect it does satisfy equation.

Yes that is definitly the question
• October 25th 2009, 10:28 AM
TheEmptySet
Quote:

Originally Posted by ben.mahoney@tesco.net
if you differentiate it you just take away the t's on each part so in that respect it does satisfy equation.

Yes that is definitly the question

That is false....

x is a function of t. You need to use implict differentation.

$x(t)=\sqrt{3}t-2t\sin(x)+c$

$\frac{dx}{dt}=\sqrt{3}\underbrace{-2\sin(x)-2t\cos(x)\frac{dx}{dt}}_{\text{Product rule}}$

Your equation does seperate but is not easy to integrate

$\frac{dx}{\sqrt{3}-2\sin(x)}=dt$
• October 25th 2009, 11:14 AM
ben.mahoney@tesco.net
thankyou.

you are correct, it isn't easy to integrate!!!
• October 25th 2009, 02:18 PM
Prove It
Quote:

Originally Posted by ben.mahoney@tesco.net
Q) The question is for me to sketch the graph of x(t) for different initial conditions.

dx/dt = sqrt(3) - 2sin(x)

A) i cant see how this is separable variables or integrating factor method.

So do I do: x(t) = sqrt(3)t - 2tsin(x) + C

Saying x = x0 at t=0 just gives me c = x0

Does this look correct?

Thanks

According to Wolfram Integrator, the solution is

$t = 2i\arctan{\left\{\sec{\left(\frac{x}{2}\right)}\le ft[i\sqrt{3}\sin{\left(\frac{x}{2}\right)} - 2i\cos{\left(\frac{x}{2}\right)}\right]\right\} } + C$

Now you can use your initial condition to find $C$.

Edit: Upon further inspection, it appears that this can be simplified to

$t = 2i\arctan{ \left[ i\sqrt{3}\tan{\left(\frac{x}{2}\right)} - 2i \right] } +C$
• October 26th 2009, 01:44 PM
ben.mahoney@tesco.net
The question that this equation involves is:
Use a geometric approach and qualitatively analyse the following equations (in each case find fixed points, sketch the vector field and classify stability of each fixed point, sketch the graph of x(t) for different initial conditions).

Does this mean that I actually do have to integrate dx/dt.

On asking my lecturer he said that the hint is in the brackets.

Im so confused :(