Question

Suppose ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{,}\mathbf{}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{5}\mathbf{}\mathit{a}\mathit{n}\mathit{d}\mathbf{}{\mathit{m}}_{\mathbf{3}}\mathbf{=}\mathbf{1}$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:

$y=c_1{x}^3+c_2{x}^{-5}\ln x+c_{4}x+c_{5}x\ln x+c_{6}x{\left(\ln x\right)}^2$

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

$$\begin{gathered} y=c_1{x}^{m_1}+c_2{x}^{m_2}+c_3{x}^{m_3}+c_4{x}^{m3}+c_5{x}^{m_3}\ln x+c_6{x}^{m_3}{\left(\ln x\right)}^2 \\ =c_1{x}^3+c_2{x}^{-5}+c_3{x}^{-5}\ln x+c_{4}x+c_{5}x\ln x+c_{6}x{\left(\ln x\right)}^2 \end{gathered}$$

The general solution of the homogeneous linear DE if it is Cauchy-Euler Equation is:

$y=c_1{x}^3+c_2{x}^{-5}\ln x+c_{4}x+c_{5}x\ln x+c_{6}x{\left(\ln x\right)}^2$