Complex differential equation: d^2 z/ d^2 z = -z for z

[Mondo]

I try to solve this equation [imath]\frac{d^2}{d^2z}z = -z[/imath] for z. I integrated the left hand side twice to get [imath]\frac{-z^3}{6} + C[/imath] but it doesn't seem to be the correct solution which is [imath]z = e^{\pm it}[/imath]. What do I do wrong?

 

[topsquark]

I try to solve this equation [imath]\frac{d^2}{d^2z}z = -z[/imath] for z. I integrated the left hand side twice to get [imath]\frac{-z^3}{6} + C[/imath] but it doesn't seem to be the correct solution which is [imath]z = e^{\pm it}[/imath]. What do I do wrong?

Looks like your given answer is a solution to
[imath]\dfrac{d^2z}{dt^2} = -z(t)[/imath].

Your equation makes no sense because
[imath]\dfrac{d^2z}{dz^2} = 0 \: \forall z \in \mathbb{C}[/imath]
so [imath]z \equiv 0[/imath].

Are you trying to find the solution for
[imath]\dfrac{d^2f(z)}{dz^2} = - f(z)[/imath]

-Dan

 

[Mondo]

Right, I meant to say [imath]\frac{d^2}{d^2t}z = -z[/imath]. Thanks for correcting my incorrect notation. But still, when I calculate it I get [math]\int\frac{d^2z}{d^2t} = \int -z = ie^{it} + C = \frac{d}{dt}z[/math] so now we have [imath]\frac{d}{dt}z = ie^{it} + C \rightarrow z = -e^{it} + ct + D[/imath] but it is not equal to expected result [imath]e^{\pm it}[/imath]

 

[blamocur]

Can you provide more context? Are you solving an ordinary differential equation [imath]\frac{d^2}{dt^2} z(t) = -z[/imath] ? If yes, do you have initial conditions specified? Do you have a complete statement of the problem?
Thank you.

 

[Mondo]

It is supposed to be a differential equation for the motion of an object going round elliptical trajectory. I have no initial conditions.
Yes I think the differential equation is as you wrote [imath]\frac{d^2}{d^2t}z(t) = -z[/imath]

 

[Dr.Peterson]

Right, I meant to say [imath]\frac{d^2}{d^2t}z = -z[/imath]. Thanks for correcting my incorrect notation. But still, when I calculate it I get [math]\int\frac{d^2z}{d^2t} = \int -z = ie^{it} + C = \frac{d}{dt}z[/math] so now we have [imath]\frac{d}{dt}z = ie^{it} + C \rightarrow z = -e^{it} + ct + D[/imath] but it is not equal to expected result [imath]e^{\pm it}[/imath]

Please show how you did that integral. It is not at all clear.

It is supposed to be a differential equation for the motion of an object going round elliptical trajectory. I have no initial conditions.
Yes I think the differential equation is as you wrote [imath]\frac{d^2}{d^2t}z(t) = -z[/imath]

Please show us the actual problem as given to you, and explain where it came from. What topic are you studying?

It appears that you are using complex-valued functions of a real variable. What has been said about this?

 

[topsquark]

Right, I meant to say [imath]\frac{d^2}{d^2t}z = -z[/imath]. Thanks for correcting my incorrect notation. But still, when I calculate it I get [math]\int\frac{d^2z}{d^2t} = \int -z = ie^{it} + C = \frac{d}{dt}z[/math] so now we have [imath]\frac{d}{dt}z = ie^{it} + C \rightarrow z = -e^{it} + ct + D[/imath] but it is not equal to expected result [imath]e^{\pm it}[/imath]

It is supposed to be a differential equation for the motion of an object going round elliptical trajectory. I have no initial conditions.
Yes I think the differential equation is as you wrote [imath]\frac{d^2}{d^2t}z(t) = -z[/imath]

So take
[imath]z^{\prime \prime} (t) = -z(t), \quad z(0) = a, \quad z^{\prime}(0) = b[/imath]

The symbol
[imath]\displaystyle \int \dfrac{d^2z}{dt^2}[/imath]

has no meaning. You would need to write something like
[imath]\displaystyle \int \dfrac{d^2z}{dt^2} \, dt[/imath]

But to do it this way would require the steps:
[imath]\dfrac{d^2z}{dt^2} = -z(t)[/imath]

[imath]\displaystyle \int \dfrac{d^2z}{dt^2} \, dt = - \int z(t) \, dt[/imath]

[imath]\displaystyle \dfrac{dz}{dt} = - \int z(t) \, dt[/imath]

but what is the integral of the unknown function z(t)?

See here for a suggestion how to solve this equation. Look for the section on "Homogeneous Equation with Constant Coefficients."

-Dan

 

[blamocur]

You can verify that [imath]z=e^{\pm t}[/imath] is a solution by plugging it in into your equation. More generally, the solution can be represented as [imath]z = Ae^{it}+Be^{-it}[/imath]. Can you see that this is an ellipse? Can you figure out its axes?

Another note on the topic: while there exists a very elegant technique for solving linear ODEs with constant coefficients there is also a simpler trick for solving your case:
[math]\ddot z = -z \;\;\longleftrightarrow\;\; \ddot z + z = 0 \;\;\longrightarrow\;\; \ddot z \dot z + \dot z z = 0[/math]