To solve the differential equation \( y'' - 2y' + 5y = 0 \), we start by finding the characteristic equation associated with the homogeneous equation. The characteristic equation is obtained by substituting \( y = e^{rt} \), where \( r \) is a constant, into the differential equation.
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Substitute \( y = e^{rt} \): \[ y' = re^{rt}, \quad y'' = r^2 e^{rt} \]
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Plug these into the differential equation: \[ r^2 e^{rt} - 2re^{rt} + 5e^{rt} = 0 \]
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Factor out \( e^{rt} \) (which is never zero): \[ e^{rt}(r^2 - 2r + 5) = 0 \]
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Thus, we need to solve the characteristic equation: \[ r^2 - 2r + 5 = 0 \]
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Use the quadratic formula to find the roots: \[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{2 \pm \sqrt{(-2)^2 - 4 \cdot 1 \cdot 5}}{2 \cdot 1} = \frac{2 \pm \sqrt{4 - 20}}{2} = \frac{2 \pm \sqrt{-16}}{2} \]
Simplifying this gives: \[ r = \frac{2 \pm 4i}{2} = 1 \pm 2i \]
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The roots of the characteristic equation are \( r = 1 + 2i \) and \( r = 1 - 2i \). These are complex conjugates.
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The general solution to the differential equation can be written using the standard form for complex roots: \[ y(t) = e^{\alpha t} (C_1 \cos(\beta t) + C_2 \sin(\beta t)) \] where \( \alpha = 1 \) and \( \beta = 2 \).
Thus, the general solution is: \[ y(t) = e^{t} (C_1 \cos(2t) + C_2 \sin(2t)) \]
where \( C_1 \) and \( C_2 \) are constants determined by initial conditions.
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