Consider the differential equation $x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-4y=0$ with the boundary conditions of y(0) = 0 and y(1) = 1. The complete solution of the differential equation is

To find the complete solution of the given differential equation, we need to first solve the ordinary differential equation and then apply the given boundary conditions to determine the constants.

Differential Equation Analysis

The given differential equation is:

$x^2\frac{d^{2}y}{d{x}^2}+x\frac{dy}{dx}-4y=0$

This is a second-order, linear, homogeneous differential equation with variable coefficients. Specifically, it is a Cauchy-Euler (or Euler-Cauchy) differential equation because it has the form $a{x}^2\frac{d^{2}y}{d{x}^2}+bx\frac{dy}{dx}+cy=0$.

Solving the Cauchy-Euler Equation

For a Cauchy-Euler differential equation, we assume a solution of the form $y=x^m$. Let's find the derivatives:

- First derivative: $\frac{dy}{dx}=m{x}^{m-1}$
- Second derivative: $\frac{d^{2}y}{d{x}^2}=m(m-1)x^{m-2}$

Now, substitute these derivatives back into the original differential equation:

$x^2[m(m-1)x^{m-2}]+x[m{x}^{m-1}]-4x^m=0$

Simplify the terms:

$m(m-1)x^{m-2}{x}^2+m{x}^{m-1}{x}^1-4x^m=0$

$m(m-1)x^m+m{x}^m-4x^m=0$

Factor out $x^m$ (assuming $x\ne 0$):

$x^m[m(m-1)+m-4]=0$

Since $x^m\ne 0$, we must have the characteristic equation equal to zero:

$m(m-1)+m-4=0$

$m^2-m+m-4=0$

$m^2-4=0$

Solve for $m$:

$m^2=4$

$m=\pm \sqrt{4}$

$m_1=2$

$m_2=-2$

Since the roots $m_1$ and $m_2$ are real and distinct, the general solution of the differential equation is:

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

$y(x)=C_1{x}^2+C_2{x}^{-2}$

Applying Boundary Conditions

We are given two boundary conditions: $y(0)=0$ and $y(1)=1$.

**Boundary Condition 1: $y(0)=0$ **

Substitute $x=0$ into the general solution:

$y(0)=C_1(0{)}^2+C_2(0{)}^{-2}$

$y(0)=C_1\cdot 0+C_2\cdot \frac{1}{0^2}$

The term $C_2\cdot \frac{1}{0^2}$ is undefined. For the solution $y(x)$ to exist and be finite at $x=0$, the coefficient $C_2$ must be zero. If $C_2$ were non-zero, $y(0)$ would approach infinity, which contradicts the given condition $y(0)=0$.

Therefore, we must have $C_2=0$.

With $C_2=0$, the general solution simplifies to:

$y(x)=C_1{x}^2$

**Boundary Condition 2: $y(1)=1$ **

Now, substitute $x=1$ into the simplified solution $y(x)=C_1{x}^2$:

$y(1)=C_1(1{)}^2$

We are given $y(1)=1$, so:

$1=C_1\cdot 1$

$C_1=1$

Complete Solution of the Differential Equation

With $C_1=1$ and $C_2=0$, the complete solution of the differential equation that satisfies both boundary conditions is:

$y(x)=1\cdot x^2+0\cdot x^{-2}$

$y(x)=x^2$

Comparing this complete solution with the given options, we find that it matches option 1.