Question

Find the general solution of $$ x^{2} \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+y=x $$ given that \(y(1)=1\) and \(y(e)=2 e\).

Short answer

The general solution is \( y = x + x \, \ln(x) \).

Step-by-step solution

Identify the type of differential equation

Recognize that the given equation is a Cauchy-Euler differential equation: \[ x^{2} \frac{d^{2} y}{d x^{2}} - x \frac{d y}{d x} + y = x \].

Rewrite the equation

For a Cauchy-Euler differential equation, a substitution is often used. Let \(y = x^m\). Then, \(y’ = mx^{m-1}\) and \(y'' = m(m-1)x^{m-2}\). Substitute these into the equation.

Substitute and simplify

Substitute \(y, y’\) and \(y''\) into the original equation: \[ x^{2}(m(m-1)x^{m-2}) - x(mx^{m-1}) + x^m = x \]. Simplifying, \[ m(m-1)x^m - mx^m + x^m = x \].

Solve the characteristic equation

Collect terms involving \(x^m\): \[ (m^2 - m - m + 1)x^m = x \]. The characteristic equation is \[ m^2 - 2m + 1 = 0 \], which simplifies to \[ (m-1)^2 = 0 \]. So, \(m = 1\) (a repeated root).

Form the general solution

Since we have a repeated root, the general solution involves \(m=1\) and an additional term: \[ y_c = C_1 x + C_2 x \, \ln(x) \].

Find particular solution

Assume a particular solution of the form \(y_p = A x\). Substitute into the original equation to find \(A\): \[ x^{2}(0) - x(A) + (A x) = x \]. This leads to \( -Ax + Ax = x \), giving \(A=1\), so \(y_p = x\).

Complete general solution

Combine the general solution with the particular solution: \[ y = y_c + y_p = C_1 x + C_2 x \, \ln(x) + x \], which simplifies to \[ y = (C_1+1) x + C_2 x \, \ln(x) \].

Apply boundary conditions

Use the boundary conditions \(y(1) = 1\) and \(y(e) = 2e\). Substituting these into the general solution to solve for \(C_1\) and \(C_2\): \[ y(1) = (C_1 + 1) \cdot 1 + C_2 \cdot 1 \cdot \ln(1) = 1 \Rightarrow C_1 + 1 = 1 \Rightarrow C_1 = 0 \] and \[ y(e) = (C_1 + 1)e + C_2 e \, \ln(e) = 2e \Rightarrow e + C_2 e \cdot 1 = 2e \Rightarrow 1 + C_2 = 2 \Rightarrow C_2 = 1 \].

Write final solution

Substitute \(C_1 = 0\) and \(C_2 = 1\) back into the general solution to get: \[ y = (0+1)x + 1x \, \ln(x) = x + x \, \ln(x) \]. Therefore, the final solution is \[ y = x + x \, \ln(x) \].

Key concepts

general solution

The general solution of a differential equation reflects the complete set of solutions that can be found by solving the equation. For a Cauchy-Euler differential equation, the form of the solution often takes advantage of specific patterns in the equation. In this case, we start by recognizing the equation as a Cauchy-Euler type and proceed to solve the associated characteristic equation. The characteristic equation reveals the roots which help us form the structure of the general solution. When these roots are repeated, it modifies the general solution to include a term involving a logarithmic function.

For example, in our problem, the characteristic equation yielded a repeated root. This means our general solution is of the form: \[ y_c = C_1 x + C_2 x \, \ln(x) \] These terms collectively cover the most general form of the solution for the given type of differential equation.

boundary conditions

Boundary conditions allow us to find specific values for the constants in the general solution. These conditions are given values that the solution must satisfy at specific points. They help narrow down the infinite range of possible solutions to a single, unique solution.

In our problem, the boundary conditions are given as \( y(1)=1 \) and \( y(e)=2e \). By substituting these points into the general solution that we derived, we can solve for the unknown constants \( C_1 \) and \( C_2 \).

For instance, plugging in the boundary conditions results in:
- \( y(1) = (C_1 + 1) \cdot 1 + C_2 \cdot 1 \cdot ln(1) = 1 \rightarrow C_1 + 1 = 1 \rightarrow C_1 = 0 \)
- \( y(e) = (C_1 + 1)e + C_2 e \cdot ln(e) = 2e \rightarrow e + C_2 e \cdot 1 = 2e \rightarrow 1 + C_2 = 2 \rightarrow C_2 = 1 \)

characteristic equation

The characteristic equation is a crucial part of solving a Cauchy-Euler differential equation. It's derived by assuming a solution of the form \( y = x^m \) and substituting it into the differential equation. This transforms the differential equation into an algebraic equation in terms of \( m \), which we call the characteristic equation.

In our problem, substituting \( y = x^m \) leads us to: \[ x^{2}(m(m-1)x^{m-2}) - x(mx^{m-1}) + x^m = x \] Simplifying yields the characteristic equation: \[ m^2 - 2m + 1 = 0 \] or, factoring, \[ (m-1)^2 = 0 \] This means we have a repeated root \( m = 1 \).

These roots dictate the form of the solution where repeated roots result in an additional term involving \( \ln(x) \).

particular solution

A particular solution is a solution to a nonhomogeneous differential equation that specifically matches the nonhomogeneous part. After finding this solution, it is added to the general solution of the associated homogeneous equation.

For our differential equation \[ x^{2} \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}+y=x \] we assume a particular solution of the form \( y_p = Ax \).

Substituting \( y_p = Ax \) into the original equation, we get: \[ x^{2}(0) - x(A) + A x = x \] Simplifying, we deduce that \( A = 1 \). Hence, our particular solution is \( y_p = x \).

Combining this with the general solution of the homogeneous equation gives us the complete general solution: \[ y = (C_1 + 1)x + C_2 x \ln(x) \]