Question

For the given differential equation, (a) Determine the complementary solution, \(y_{c}(t)=c_{1} y_{1}(t)+c_{2} y_{2}(t)\). (b) Use the method of variation of parameters to construct a particular solution. Then form the general solution. $$ y^{\prime \prime}-(2 / t) y^{\prime}+\left(2 / t^{2}\right) y=t /\left(1+t^{2}\right) $$

Short answer

Question: Determine the general solution of the given second-order linear non-homogeneous differential equation: $$ y^{\prime \prime}-(2 / t) y^{\prime}+\left(2 / t^{2}\right) y=t /\left(1+t^{2}\right) $$ Answer: The general solution of the given differential equation is: $$ y(t) = c_1t^2 + c_2\frac{1}{t} + u_1(t)t^2 + u_2(t)\frac{1}{t} $$ where \(c_1\) and \(c_2\) are constants, and \(u_1(t)\) and \(u_2(t)\) are functions obtained through the integration of \(u_1'(t)\) and \(u_2'(t)\).

Step-by-step solution

Find the Complementary Solution

For finding the complementary solution, first consider the homogeneous equation: $$ y_h^{\prime \prime}-(2 / t) y_h^{\prime}+\left(2 / t^{2}\right) y_h = 0 $$ This is an Euler equation, so we can use the substitution \(y_h(t)=t^r\) to transform it into a polynomial equation. The derivatives of \(y_h(t)\) are: - \(y_h^{\prime}(t) = rt^{r-1}\) - \(y_h^{\prime \prime}(t) = r(r-1)t^{r-2}\) Now, substitute these in the homogeneous equation: $$ r(r-1)t^{r-2}-\frac{2}{t}rt^{r-1}+\frac{2}{t^2}t^r=0 $$ Cancel the common term \(t^{r-2}\): $$ r(r-1)-2r+\frac{2}{t}=0 $$ This results in a quadratic equation in r: $$ r^2-r-2=0 $$ Solve the quadratic equation to get the roots: \(r_1=2\) and \(r_2=-1\). Hence, the complementary solution is: $$ y_c(t)=c_1t^2+c_2\frac{1}{t} $$

Variation of Parameters

Now we will use the method of variation of parameters to find a particular solution. Let \(Y(t)\) be the particular solution. We assume that \(Y(t)=u_1(t)y_1(t) + u_2(t)y_2(t)\), where \(y_1(t)=t^2\) and \(y_2(t)=\frac{1}{t}\) are the complementary functions and \(u_1(t)\) and \(u_2(t)\) are unknown functions. The first derivatives \(Y^{\prime}(t)\) and \(Y^{\prime\prime}(t)\) are: - \( Y'(t) = u_1't^2 + 2u_1t + u_2'\frac{1}{t} - u_2\frac{1}{t^2} \) - \( Y''(t) = u_1''t^2 + 4u_1't + 2u_1 + u_2''\frac{1}{t} - 2u_2'\frac{1}{t^2} + 2u_2\frac{1}{t^3} \) Now substitute \(Y(t)\), \(Y^{\prime}(t)\), and \(Y^{\prime\prime}(t)\) into the original non-homogeneous differential equation and simplify: $$ u_1''t^2 + 4u_1't + 2u_1 + u_2''\frac{1}{t} - 2u_2'\frac{1}{t^2} + 2u_2\frac{1}{t^3} - (2/t)(u_1't^2+2u_1t+u_2'\frac{1}{t} - u_2\frac{1}{t^2}) + \frac{2}{t^2}(u_1t^2+u_2\frac{1}{t})= \frac{t}{1+t^2} $$ Simplify the equation by canceling out terms and collect like terms: $$ u_1''t^2 + 2u_1 + u_2''\frac{1}{t} - 2u_2'\frac{1}{t^2} = \frac{t}{1+t^2} $$ Now, we use the condition that \(u_1''t^2 + u_2''\frac{1}{t} = 0\). Thus, we have: - \(u_1''t^2 = -u_2''\frac{1}{t}\) Integrate both sides with respect to \(t\): - \(u_1't = \int -u_2''\frac{1}{t}dt\) Now we get a new condition: \(2u_1 + 2u_2\frac{1}{t^3}- 2u_2'\frac{1}{t^2}= \frac{t}{1+t^2}\). Substitute the previous result: $$ 2u_1 + \frac{1}{t^2}\left(2u_2-\int u_1'tdt \right)= \frac{t}{1+t^2} $$ Now solve for \(u_1'(t)\) and \(u_2'(t)\): - \(u_1'(t) = \int \left(\frac{t}{2(1+t^2)}\right)dt\) - \(u_2'(t) = \int -u_1'(t) dt\) Integrate these two equations to find the unknown functions \(u_1(t)\) and \(u_2(t)\). Then, substitute these functions into the particular solution \(Y(t)\).

General Solution

The general solution is the sum of the complementary solution and the particular solution: $$ y(t) = y_c(t) + Y(t) $$ Substitute the complementary and particular solutions to get the final answer: $$ y(t) = c_1t^2 + c_2\frac{1}{t} + u_1(t)t^2 + u_2(t)\frac{1}{t} $$

Key concepts

Complementary Solution

When dealing with non-homogeneous differential equations, the first step is usually to find the complementary solution, also known as the homogeneous solution or the associated homogeneous equation's solution. This solution addresses the system's response that naturally diminishes over time without any external forces or inputs.

The complementary solution is denoted as \(y_c(t)\) and is a linear combination of solutions to the corresponding homogeneous equation. In the case of our example, we turn our attention to the equation without the non-homogeneous part (the right side of the equation).

This process often utilizes standard methods for solving homogeneous differential equations, such as characteristic equations, undetermined coefficients, or, more specifically in this case, an approach suitable for Euler equations. We use the Euler method by making a power substitution, leading to a characteristic equation that can be solved for the roots which, in turn, provide us with the fundamental solutions, \(y_1(t)\) and \(y_2(t)\). These solutions depend on arbitrary constants \(c_1\) and \(c_2\), which reflect the initial conditions of the problem and make up the complementary solution.

Variation of Parameters

The Variation of Parameters technique allows us to find a particular solution to the non-homogeneous differential equation. Starting from the complementary solution, we admit the possibility that the constants \(c_1\) and \(c_2\) can be functions of \(t\), namely \(u_1(t)\) and \(u_2(t)\).

By allowing these parameters to vary, we accommodate the external inputs represented by the non-homogeneous part of the equation. This method requires the calculation of the derivatives of the assumed particular solution, then substituting back into the original equation to find conditions that the varying parameters must satisfy.

After setting up the system of equations, we solve for the derivatives of the parameters, which often involves integration. Determining the functions \(u_1(t)\) and \(u_2(t)\) gives us the particular solution that is then added to the complementary solution to form the general solution.

Euler Equation

An Euler Equation, sometimes also referred to as the Cauchy-Euler equation, is a type of linear homogeneous differential equation characterized by its variable coefficients being powers of the independent variable, typically denoted as \(t\).

The general form of a second-order Euler equation is \(a_2t^2y'' + a_1ty' + a_0y = 0\), where \(a_i\) are constants. As in our exercise, to solve it, you typically make the substitution \(y(t) = t^r\), which in turn transforms the differential equation into an algebraic equation in terms of \(r\). This method exploits the power-law nature of the equation to simplify it significantly, often resulting in a polynomial equation whose roots determine the specific form of the complementary solution.

General Solution

The general solution of a differential equation encapsulates all possible solutions, embodying both the system’s inherent behavior and its response to external influences. It is composed of two parts: the complementary (or homogeneous) solution, which resolves the differential equation without the non-homogeneous component, and the particular solution, which is a specific solution that accounts for the non-homogeneous part of the equation.

In forming the general solution, one combines the complementary solution, which includes arbitrary constants and reflects the general behavior under initial conditions, with the particular solution, adjusted for the given non-homogeneity. The general solution is especially meaningful because it provides a comprehensive formula that can adapt to various initial conditions, giving it broad applicability across different scenarios.