General solution of $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-y=0$ is

To find the general solution of the given differential equation, we first need to identify its type. The equation provided is $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-y=0$.

This equation is a special type of linear second-order homogeneous differential equation known as a Cauchy-Euler differential equation (also sometimes called an Euler-Cauchy equation). The general form of a homogeneous Cauchy-Euler equation of the second order is:

$a{x}^2\frac{d^{2}y}{d{x}^2}+bx\frac{dy}{dx}+cy=0$

By comparing our given equation $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-y=0$ with the general form, we can see that $a=1$, $b=1$, and $c=-1$.

Differential Equation Solution Approach

For Cauchy-Euler differential equations, we typically assume a solution of the form $y=x^m$, where $m$ is a constant that we need to determine.

- First, we find the first derivative of $y$ with respect to $x$:
- $\frac{dy}{dx}=\frac{d}{dx}(x^m)=m{x}^{m-1}$
- Next, we find the second derivative of $y$ with respect to $x$:
- $\frac{d^{2}y}{d{x}^2}=\frac{d}{dx}(m{x}^{m-1})=m(m-1)x^{m-2}$

Substitution and Auxiliary Equation Derivation

Now, we substitute $y$, $\frac{dy}{dx}$, and $\frac{d^{2}y}{d{x}^2}$ back into the original differential equation $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-y=0$:

$x^2\left(m(m-1)x^{m-2}\right)+x\left(m{x}^{m-1}\right)-x^m=0$

Let's simplify the terms by combining the powers of $x$:

$m(m-1)x^{m-2+2}+m{x}^{m-1+1}-x^m=0$

$m(m-1)x^m+m{x}^m-x^m=0$

Since we are looking for a non-trivial solution (where $x^m\ne 0$), we can factor out $x^m$ from the entire equation:

$x^m\left(m(m-1)+m-1\right)=0$

For this equation to hold true, the expression inside the parenthesis must be equal to zero. This expression is called the auxiliary equation or characteristic equation:

$m(m-1)+m-1=0$

Expand and simplify the equation:

$m^2-m+m-1=0$

$m^2-1=0$

Solving the Characteristic Equation

Our next step is to solve the characteristic equation $m^2-1=0$ for the values of $m$.

$m^2=1$

Taking the square root of both sides:

$m=\pm \sqrt{1}$

This gives us two distinct real roots:

- $m_1=1$
- $m_2=-1$

General Solution Formulation for Distinct Roots

For a Cauchy-Euler differential equation where the characteristic equation yields two distinct real roots, $m_1$ and $m_2$, the general solution is given by the formula:

$y=C_1{x}^{m_1}+C_2{x}^{m_2}$

Here, $C_1$ and $C_2$ are arbitrary constants determined by initial or boundary conditions.

Substitute the values of our roots, $m_1=1$ and $m_2=-1$, into this general formula:

$y=C_1{x}^1+C_2{x}^{-1}$

This can be rewritten as:

$y=C_{1}x+\frac{C_2}{x}$

This is the required general solution for the given differential equation.

Comparison with Options

Let's compare our derived general solution with the options provided:Our calculated general solution $y=C_{1}x+\frac{C_2}{x}$ perfectly matches Option 3.