Question

The given differential equation.$\mathit{x}\mathit{y}\mathbf{\text{'}}\mathbf{\text{'}}\mathbf{+}\mathit{y}\mathbf{\text{'}}\mathbf{=}\mathbf{0}$

Short answer

The solution of the differential equation is $y=c_1+c_2\text{ln}x$.

Step-by-step solution

Find the derivative

The starting differential equation can be written in the following form:

$x^2\frac{{\text{d}}^{2}y}{\text{d}{x}^2}+x\frac{\text{d}y}{\text{d}x}=0$

Assume that the solution is some power of $x,i.e.y=x^m.\mathrm{Calculate}\frac{dy}{dx}a\mathrm{nd}\frac{d^{2}y}{d{x}^2}:$

$$\begin{gathered} \frac{dy}{dx}=m{x}^{m-1} \\ \frac{d^{2}y}{d{x}^2}=m(m-1)x^{m-2} \end{gathered}$$

Deriving the given solution by differential equation

Substitute the values $\frac{dy}{dx}=m{x}^{m-1}and\frac{d^{2}y}{d{x}^2}=m(m-1)x^{m-2}$in the starting equation:

$\begin{array}{rcl}{x}^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}& =& 0\\ x^{2}m\left(m-1\right)x^{m-2}+xm{x}^{m-1}& =& 0\\ x^m\left(m\left(m-1\right)+m\right)& =& 0\\ x^m\left(m^2-m+m\right)& =& 0\\ x^m{m}^2& =& 0\\ & & \end{array}$

Therefore, the characteristic equation is:

$$\begin{gathered} m^2=0 \\ {m}_1=0and{m}_2=0 \end{gathered}$$

Therefore, take a look at "Case I: Distinct Real Roots" and conclude that the general solution is:

$y=c_1{x}^{m_1}+c_2{x}^{m_1}\text{ln}x$

where is repeated root of the characteristic equation. Finally, the general solution is then after substitution of$m_1:$$y=c_1+c_2\ln x$.