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Given an inhomogeneous linear complex differential equation of this form:

$z''(t)+a\cdot z'(t)+b \cdot z(t)=4\cdot \exp(i\cdot t)$ (*)

A particular solution to this differential equation is: $z_0=(4+6\cdot i)\exp(i\cdot t)$

Now observe a real differential equation with the same coefficients as above: $x''(t)+a\cdot x'(t)+b\cdot x(t)=4\cdot \cos(t)$ (**)

Now determine a particular solution to (**) of the form: $x_0(t)=c_1\cdot \cos(t)+c_2\cdot \sin(t)$

Determine $c_1$ and $c_2$.

Any hints to how I solve this? I have tried inserting the given particular in the equation, giving me:

$$(-6+4\cdot i)\exp(i\cdot t)=4\cdot \exp(i\cdot t)$$

I'm not sure how to progress from here.


I figured it out. My solution:

$$\Re((4+6i)\exp(it))=\Re((4+6i)(\cos(t)+i\sin(t)))=4\cos(t)-6\sin(t)$$

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As $a$ and $b$ have to be real numbers, take the real part of the first equation to find that $$x(t)=Re(z(t))$$ solves the real equation.


Because taking the real part commutes with multiplication with real factors and also with derivation. Thus $$x''+ax'+bx=Re(z''+az'+bz)=Re(4e^{it})=4\cos t.$$

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  • $\begingroup$ Could you elaborate? I'm aware the solution has to be real. $\endgroup$ – Steve Dec 2 '16 at 16:30
  • $\begingroup$ You just have to insert and use rules of complex number arithmetics. $\endgroup$ – LutzL Dec 2 '16 at 16:35

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