Question

Identify each of the differential equations as to type (for example, separable, linear first order, linear second order, etc.), and then solve it. \(x y^{\prime \prime}+y^{\prime}=4 x\)

Short answer

The differential equation is a linear second-order equation. The solution is \(y(x) = C + x^2\).

Step-by-step solution

Identify the Type of Differential Equation

First, consider the given differential equation: \(x y^{\backprime \backprime}+y^{\backprime}=4 x\). Notice that it involves the second derivative of \(y\) and the first derivative of \(y\), making it a second-order differential equation. It can be recast in standard linear form.

Rewrite the Equation in Standard Linear Form

The differential equation can be rewritten as: \[xy^{\backprime \backprime} + y^{\backprime} = 4x\]. This can be simplified to \[y^{\backprime \backprime} + \frac{1}{x} y^{\backprime} = 4\] by dividing both sides by \(x\). This is a linear non-homogeneous second-order differential equation.

Solve the Homogeneous Equation

Solve the homogeneous part of the equation: \[y^{\backprime \backprime} + \frac{1}{x} y^{\backprime} = 0\]. To do this, assume a solution of the form \(y = x^m\). Then, \(y^{\backprime} = mx^{m-1}\) and \(y^{\backprime \backprime} = m(m-1)x^{m-2}\). Substituting these into the homogeneous equation gives: \[m(m-1)x^{m-2} + \frac{m}{x}x^{m-1} = 0\]. Simplifying this, we get: \[m(m-1) + m = 0 \Rightarrow m^2 = 0 \Rightarrow m = 0\]. Thus, the homogeneous solution is \(y_h = C\) where \(C\) is a constant.

Solve the Non-Homogeneous Equation

For the non-homogeneous equation, we can use the method of undetermined coefficients. Assume a particular solution of the form \(y_p = Ax^2\). Then, \(y_p\backprime = 2Ax\) and \(y_p\backprime \backprime = 2A\). Substitute these into the non-homogeneous equation: \[2A + \frac{2Ax}{x} = 4\]. Simplifying, we get: \[2A + 2A = 4 \Rightarrow 4A = 4 \Rightarrow A = 1\]. So, the particular solution is \(y_p = x^2\).

Write the General Solution

Combine the homogeneous solution \(y_h\) with the particular solution \(y_p\) to form the general solution: \[y(x) = y_h + y_p = C + x^2\].

Key concepts

Linear Differential Equations

A linear differential equation is an equation involving derivatives of a function where each term is linear in the unknown function and its derivatives. In simpler terms, the dependent variable (like y) and its derivatives appear to the power of one and are not multiplied or divided by each other.
In our example, the differential equation \( x y^{\prime \prime} + y^{\prime} = 4x \) can be recast into standard linear form: \[ y^{\prime \prime} + \frac{1}{x} y^{\prime} = 4 \]. This makes it a second-order linear differential equation because it involves a second derivative, a first derivative, and powers of the variable x. Linear differential equations can be solved using various methods, like the method of undetermined coefficients.

Homogeneous Equations

Homogeneous differential equations are a type of differential equation where all terms involve derivatives of the same function, and the equation is set to zero. In essence, there is no external forcing function or non-homogeneous term.
For instance, consider the homogeneous part of our given equation: \( y^{\prime \prime} + \frac{1}{x} y^{\prime} = 0 \). The right-hand side is zero, indicating that it’s homogeneous. Solving a homogeneous differential equation usually results in finding the complementary function or the general solution to the homogeneous part. Here, we solved it by assuming a solution of the form \( y = x^m \), which led to the homogeneous solution \( y_h = C \), where C is a constant.

Method of Undetermined Coefficients

The method of undetermined coefficients is a technique used to find the particular solution of a non-homogeneous linear differential equation. It involves guessing a form of the particular solution based on the non-homogeneous term and determining the coefficients by substitution.
For our equation \( y^{\prime \prime} + \frac{1}{x} y^{\prime} = 4 \), we guessed that the particular solution might be of the form \( y_p = Ax^2 \), because the right-hand side is a polynomial. Substituting this guess into the differential equation, we solved for A, finding that \( A=1 \). Therefore, our particular solution is \( y_p = x^2 \).
The method of undetermined coefficients works well for cases where the non-homogeneous term is a polynomial, exponential, sine, or cosine function.

Particular Solution

The particular solution to a differential equation is a specific solution that satisfies the non-homogeneous equation. It incorporates the non-homogeneous term of the equation and complements the general solution obtained from the homogeneous equation.
Combining the solutions we obtained: the homogeneous solution \( y_h = C \) and the particular solution \( y_p = x^2 \), gives the general solution to the original non-homogeneous equation. Therefore, the final answer is \( y(x) = C + x^2 \). This general solution includes all possible solutions to the differential equation and satisfies the initial or boundary conditions provided in specific problems.