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Cauchy–Euler equation
(section)
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===Second order – solution through change of variables=== <math display="block">x^2\frac{d^2y}{dx^2} +ax\frac{dy}{dx} + by = 0 </math> We operate the variable substitution defined by <math display="block">t = \ln(x). </math> <math display="block">y(x) = \varphi(\ln(x)) = \varphi(t). </math> Differentiating gives <math display="block">\frac{dy}{dx}=\frac{1}{x}\frac{d\varphi}{dt}</math> <math display="block">\frac{d^2y}{dx^2}=\frac{1}{x^2}\left(\frac{d^2\varphi}{dt^2}-\frac{d\varphi}{dt}\right).</math> Substituting <math>\varphi(t)</math> the differential equation becomes <math display="block">\frac{d^2\varphi}{dt^2} + (a-1)\frac{d\varphi}{dt} + b\varphi = 0.</math> This equation in <math>\varphi(t)</math> is solved via its characteristic polynomial <math display="block">\lambda^2 + (a-1)\lambda + b = 0.</math> Now let <math>\lambda_1</math> and <math>\lambda_2</math> denote the two roots of this polynomial. We analyze the case in which there are distinct roots and the case in which there is a repeated root: If the roots are distinct, the general solution is <math display="block">\varphi(t)=c_1 e^{\lambda_1 t} + c_2 e^{\lambda_2 t},</math> where the exponentials may be complex. If the roots are equal, the general solution is <math display="block">\varphi(t)=c_1 e^{\lambda_1 t} + c_2 t e^{\lambda_1 t}.</math> In both cases, the solution <math>y(x)</math> can be found by setting <math>t = \ln(x)</math>. Hence, in the first case, <math display="block">y(x) = c_1 x^{\lambda_1} + c_2 x^{\lambda_2},</math> and in the second case, <math display="block">y(x) = c_1 x^{\lambda_1} + c_2 \ln(x) x^{\lambda_1}.</math>
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