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howto:hambasics:sections:test [2021/01/04 21:58] – [Cartesian vs Polar] va7fihowto:hambasics:sections:test [2021/01/04 22:00] – [Differential Equations] va7fi
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 What we've go so far says that our test function \$x(t) = e^{rt}\$ will satisfy the differential equation if \$r\$ is given by above equation.  There is still a lot to unpack however.  For example, since \$r\$ contains a square root, it could be real or complex depending on the values of \$a, b,\$ and \$c\$.  And as we saw above, if \$r\$ is real, then \$x(t)\$ will be a real exponential function.  But if \$r\$ is complex, then we can expect \$x(t)\$ to be some sort of sinusoidal function (recall the Euler Identity). What we've go so far says that our test function \$x(t) = e^{rt}\$ will satisfy the differential equation if \$r\$ is given by above equation.  There is still a lot to unpack however.  For example, since \$r\$ contains a square root, it could be real or complex depending on the values of \$a, b,\$ and \$c\$.  And as we saw above, if \$r\$ is real, then \$x(t)\$ will be a real exponential function.  But if \$r\$ is complex, then we can expect \$x(t)\$ to be some sort of sinusoidal function (recall the Euler Identity).
  
-To simplify the notation, let's define \$\alpha\$ and \$beta\$ as:+To simplify the notation, let's define \$\alpha\$ and \$\beta\$ as:
 \$$ \alpha = \dfrac{b}{2a} \qquad \text{and} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} \$$ \$$ \alpha = \dfrac{b}{2a} \qquad \text{and} \qquad \beta = \dfrac{\sqrt{|{b^2 - 4ac}|}}{2a} \$$
  
howto/hambasics/sections/test.txt · Last modified: 2021/02/13 19:14 by va7fi