How can I prove this function is a periodic phenomenom? I´ve been stuck with this math problem for a while and I would really appreciate it if someone could help me with this problem. Any tips or...

How can I prove this function is a periodic phenomenom?

I´ve been stuck with this math problem for a while and I would really appreciate it if someone could help me with this problem. Any tips or suggestions is fine, I would be really thankful.

So I´ve this differential equation here: 

Then I know that if the discriminant of the characteristic equation is negative, it´ll give me two complex solutions and it´ll describe a periodic phenomenon.

I´ve to prove that it´s the case with this equation: 

But I really don´t know how to start this? I´ve a general idea that if I get a two complex roots then I can find the modulus of the complex number, theta and also a point cos (x) and sin (y) on the plane. But how do I really prove this to be the case mathematically?

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

The periodic phenomena are very well modelled by the second order differential equations. In the provided case, you need to  start solving the characteristic equation for the given differential equation, such that:

`r^2 - alpha(beta+1)r + alpha*beta = 0`

Using quadratic formula, yields:

`r_(1,2) = (alpha(beta+1) +- sqrt(alpha^2(beta+1)^2 - 4alpha*beta))/2`

The quadratic equation has two different real solutions if the radicand `alpha^2(beta+1)^2 - 4alpha*beta > 0` , two equal real solutions if `alpha^2(beta+1)^2 - 4alpha*beta = 0` and complex conjugate solutions if `alpha^2(beta+1)^2 - 4alpha*beta < 0` .

You need to get a general solution to the given differential equation, such that:

`y = y_c + y_p`

`y_c` represents the complementary solution

`y_p` represents the particular solution

The complementary solution can be evaluated solving the characteristic equation `r^2 - alpha(beta+1)r + alpha*beta = 0` .

If the solutions `r_(1,2)` are different and real, then the complementary solution is the following:

`y_c = c_1*e^(r_1*t) + c_2*e^(r_2*t)`

If the solutions `r_(1,2) = a+-b*i` are complex, then the complementary solution is the following:

`y_c = c_1*e^(a*t)cos(b*t) + c_2*e^(a*t)sin(b*t) `

The problem does not provide the equation of the function `G` , but you may solve the problem to evaluate the constants `c_1` and `c_2` using the method of undetermined coefficients.

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user8940074's profile pic

user8940074 | (Level 1) eNoter

Posted on

But after I´ve found the complex root and decided the complementary solution, doesn´t I need to convert it into the typical periodic fucnction which is: f(x+P) = f(x)

to prove it to be periodic? 

I´m just confused, cause I´m not really sure what my teacher is out after with this problem. 

Anyway, thank you for taking your time answering my question.

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user8940074 | (Level 1) eNoter

Posted on

I didn´t get to include the picture files in my first post and I can´t edit my post. So here is the pictures of the math problem. 

Hope someone can help me.

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