Question

Identify each of the differential equations as type (for example, separable, linear first order, linear second order, etc.), and then solve it. $$x^{2} y^{\prime \prime}-x y^{\prime}+y=x$$

Short answer

The differential equation is a linear second-order differential equation. The general solution is: \( y = C_1 x + C_2 x \ln(x) - x \)

Step-by-step solution

Identify the Type of Differential Equation

The given differential equation is \(x^{2} y^{\text{''}} - x y^{\text{'}} + y = x\). Notice that the highest derivative of the function is \(y^{\text{''}}\), indicating it is a second-order differential equation. Because all terms involve either \(y\), its derivatives, or \(x\) (the independent variable), and they follow a linear relationship, this is a linear second-order differential equation.

Rewrite the Equation in Standard Form

Rewrite the differential equation in standard linear second-order form: \[ y^{\text{''}} + p(x)y^{\text{'}} + q(x)y = g(x) \] First, divide the entire equation by \(x^2\): \[ y^{\text{''}} - \frac{1}{x} y^{\text{'}} + \frac{1}{x^2} y = \frac{1}{x} \]

Solve the Homogeneous Equation

First, solve the homogeneous part: \[ y^{\text{''}} - \frac{1}{x} y^{\text{'}} + \frac{1}{x^2} y = 0 \] Assume a solution of the form \(y = x^r\). Taking derivatives, we get \[ y^{'} = r x^{r-1} \]\[ y^{\text{''}} = r (r-1) x^{r-2} \] Substitute these into the homogeneous equation: \[ r (r-1) x^{r-2} - \frac{1}{x} (r x^{r-1}) + \frac{1}{x^2} x^r = 0 \] Simplifying, \[ r (r-1) x^{r-2} - r x^{r-2} + x^{r-2} = 0 \] Factor out \(x^{r-2}\): \[ (r^2 - 2r + 1) x^{r-2} = 0 \] This simplifies to \[ (r-1)^2 = 0 \] Hence, \(r = 1\) is a double root.

General Solution to the Homogeneous Equation

Because we have a double root, the general solution for the homogeneous equation is \[ y_h = C_1 x + C_2 x \ln(x) \]

Find a Particular Solution

Now, consider a particular solution to the non-homogeneous equation \[ y^{\text{''}} - \frac{1}{x} y^{\text{'}} + \frac{1}{x^2} y = \frac{1}{x} \] We use undetermined coefficients and try a solution of the form \[ y_p = Ax \]. Then \[ y_p^{\text{'}} = A \] and \[ y_p^{\text{''}} = 0 \]. Substitute into the left-hand side of the equation: \[ 0 - \frac{1}{x}A + \frac{1}{x^2}Ax = \frac{1}{x} \]. Simplifying, \[ -\frac{A}{x} = \frac{1}{x} \] Hence, \(A = -1\), so : \[ y_p = -x \]

Write the General Solution

Combine the homogeneous and particular solutions to get the full general solution: \[ y = y_h + y_p \]\[ y = C_1 x + C_2 x \ln(x) - x \]

Key concepts

Differential Equations

A differential equation involves variables and their rates of change with respect to each other. They play a crucial role in fields like physics, engineering, and economics because they describe how quantities evolve. Understanding the type of a differential equation helps in choosing the right method to solve it.
For instance, consider the differential equation from our exercise, \(x^{2} y^{\text{''}} - x y^{\text{'}} + y = x\). It shows how the function \(y(x)\) and its derivatives interact. Recognizing that this is a second-order differential equation is key to applying the correct solution method.

Second-Order

Second-order differential equations involve the second derivative of the unknown function. These equations can model scenarios where acceleration or concavity is important, such as in motion or oscillations. The form of a second-order linear differential equation is \(y^{\text{''}} + p(x)y^{\text{'}} + q(x)y = g(x)\).
In our exercise, the given equation, \(x^{2} y^{\text{''}} - x y^{\text{'}} + y = x\), needs to be rewritten in this standard form. By dividing by \(x^2\), we get \(y^{\text{''}} - \frac{1}{x} y^{\text{'}} + \frac{1}{x^2} y = \frac{1}{x}\). This makes it easier to identify and solve the homogeneous and particular solutions.

Homogeneous Equations

A homogeneous differential equation has the form \(y^{\text{''}} + p(x)y^{\text{'}} + q(x)y = 0\), where the right-hand side is zero. Solving the homogeneous part provides insight into the general shape of solutions.
For our exercise, solving \(y^{\text{''}} - \frac{1}{x} y^{\text{'}} + \frac{1}{x^2} y = 0\) involves finding the characteristic equation. Assuming \(y = x^r\), we find \(r\) values that satisfy the equation. Simplifying, we get \((r-1)^2 = 0\), giving a double root \(r = 1\). This leads to a general solution for the homogeneous part: \(y_h = C_1 x + C_2 x \text{ln}(x)\).

Particular Solutions

Particular solutions address the non-homogeneous part of a differential equation. For \(y^{\text{''}} - \frac{1}{x} y^{\text{'}} + \frac{1}{x^2} y = \frac{1}{x}\), we identify a specific solution that satisfies the entire equation, not just the homogeneous part.
We use a trial solution, \(y_p = Ax\), for simplicity. Substituting \(y_p\) and its derivatives into the equation, we find that \(A = -1\), resulting in \(y_p = -x\).
Combining the homogeneous and particular solutions gives the complete solution: \(y = C_1 x + C_2 x \text{ln}(x) - x\). This approach ensures we account for both the structure of the equation and the external forces affecting it.