Question

Suppose ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{5}$, and${\mathit{m}}_{\mathbf{3}}\mathbf{=}\mathbf{1}$are roots of multiplicity one, two, and three, respectively, of an auxiliary equation. Write down the general solution of the corresponding homogeneous linear DE if it is a Cauchy-Euler equation.

Short answer

The general solution of the homogeneous linear DE if it is Cauchy-Euler Equation is:$y=c_1{x}^3+c_2{x}^{-5}+c_3{x}^{-5}lnx+c_{4}x+c_{5}xlnx+c_{6}x(lnx{)}^2$

Step-by-step solution

Define

In mathematics, an Euler–Cauchy equation, or Cauchy–Euler equation, or simply Euler's equation is a linear homogeneous ordinary differential equation with variable coefficients. It is sometimes referred to as an equidimensional equation

Solve the equations.

For Cauchy Euler equation, we have

$y=c_1{x}^{m_1}+c_2{x}^{m_2}+c_3{x}^{m_2}lnx+c_4{x}^{m_3}+c_5{x}^{m_3}lnx+c_6{x}^{m_3}(lnx{)}^2$

$=c_1{x}^3+c_2{x}^{-5}+c_3{x}^{-5}lnx+c_{4}x+c_{5}xlnx+c_{6}x(lnx{)}^2$

The general solution of the homogeneous linear DE if it is Cauchy-Euler Equation is:$y=c_1{x}^3+c_2{x}^{-5}+c_3{x}^{-5}lnx+c_{4}x+c_{5}xlnx+c_{6}x(lnx{)}^2$