What is the maximum solution for the given differential equation?

[mathboy20]

Homework Statement

Dear Friends,

Given the differential equation

$$\frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)$$

with the condition that x(0) = 0

Then find the largest possible solution (this is how its stated) if either

$$\alpha_{1} = 1/2$$


$$\alpha_{2} = 1$$

The Attempt at a Solution

Don't I treat the above as a seperable differential equation?

Such that


$${(\frac{1}{1+x^2})}^{-1})} \frac{dx}{dt} = {({\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)})}^{-1}$$


$$1+x^2 dx = \frac{2}{\alpha \cdot \pi \cdot cos(t)} dt$$

$$\int x^2 dx = \int \frac{2}{\alpha \cdot \pi \cdot cos(t)} -1 \ dt + C$$

$$\frac{x^3}{3} = \frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi}$$

$$x^3 = 3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})$$


$$x(t) = (3 \cdot (\frac{2 \cdot ln(\frac{-cos(t)}{sin(t)-1}- \alpha \cdot \pi \cdot (t-C)}{\alpha \cdot \pi})^{\frac{1}{3}}$$

if I then insert x(0) = 0.

I get C = 0. Then this means that max solution is 0??

I just want to be sure. Can anybody please look at my result at then inform me. Have done this correctly?

Thank You in Advance

Fred

 

[rock.freak667]

$$\frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)$$

$$\Rightarrow \frac{1}{1+x^2}dx=\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t) dt$$

Now just integrate both sides.

 

[mathboy20]

$$\frac{1}{1+x^2} \frac{dx}{dt} = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t)$$

$$\Rightarrow \frac{1}{1+x^2}dx=\frac{\alpha \cdot 2 \pi}{4} \cdot cos(t) dt$$

Now just integrate both sides.

I then get

$$tan(x)^{-1} = \frac{\alpha \cdot 2 \pi}{4} \cdot sin(t)$$

I then take arctan on both sides of equation to arrive at a solution? with regards to the two alpha's?

 

[mathboy20]

I then get

$$tan(x)^{-1} = \frac{\alpha \cdot 2 \pi}{4} \cdot sin(t)$$

I then take arctan on both sides of equation to arrive at a solution? with regards to the two alpha's?

I meant to I take tan on each side of the equation, and thusly obtain x(t) = ??

and then by choosing either alpha1 or alpha2 see which of these gives largest possible solution?

 

[HallsofIvy]

I then get

$$tan(x)^{-1} = \frac{\alpha \cdot 2 \pi}{4} \cdot sin(t)$$

I then take arctan on both sides of equation to arrive at a solution? with regards to the two alpha's?

You mean tan^-1(x) or arctan(x). There is no need to solve for x(t). Just differentiate with respect to $\alpha$ to find a maximum.

 

[mathboy20]

You mean tan^-1(x) or arctan(x). There is no need to solve for x(t). Just differentiate with respect to $\alpha$ to find a maximum.

Dear Mister Hallsoft,

Just to be clear I find the derivative with respect to alpha of the original expression

$$\frac{1}{(x^2+1)} dx = \frac{\alpha \cdot 2 \pi}{4} \cdot cos(t) dt$$

$$\frac{-2x}{x^4 + 2x^2 +1} = \frac{\pi \cdot cos(t)}{4}$$

and then insert the two alpha values to obtain the maximum??

Best Regards

Mathboy