09 Euler
Euler’s Equidimensional DE¶
\[ x^2 y'' + px y' + qy = 0 \]
Transformation¶
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Let \(x = e^z \quad (z = \log x)\)
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Now, \(y\) is a function of \(z\), which in turn is a function of \(x\)
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Put the following substitutions; Refer to Custom Operators
\[ \begin{aligned} xD &= \theta \\ x^2 D^2 &= \theta(\theta - 1) \end{aligned} \]
- equation becomes
\[ \begin{aligned} \Big( \theta(\theta - 1) + p \theta + q \Big)y &= 0 \\ \theta(\theta - 1) + p \theta + q &= 0 & (y \ne 0) \end{aligned} \]
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Put \(\theta^2 \to m^2, \theta \to m\)
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Find gen solution in terms of \(z : y(z)\), using Constant Coefficient
- \(y = c_1 e^{m_1 z} + c_2 e^{m_2 z}\)
- \(y = e^{mz}(c_1 + c_2 z)\)
- \(y = e^{az}(c_1 \cos bz+ c_2 \sin bz)\)
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Find gen solution in terms of \(x\), by subbing \(z = \log x\)
Custom Operators¶
\[ \begin{aligned} D &= \frac{d}{dx} & D^2 &= \frac{d^2}{dx^2} \\ \theta &= \frac{d}{dz} & \theta^2 &= \frac{d^2}{dz^2} \end{aligned} \]
Formula¶
\[ x Dy = \theta y \implies xD = \theta \]