# Thread: Differential equation and initial condition

1. ## 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

2. 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

3. 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

4. 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

$\frac{dx}{\sqrt{3}-2\sin(x)}=dt$

5. thankyou.

you are correct, it isn't easy to integrate!!!

6. 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

$t = 2i\arctan{ \left[ i\sqrt{3}\tan{\left(\frac{x}{2}\right)} - 2i \right] } +C$

7. 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