Question

Verify that the given functions \(y_{1}\) and \(y_{2}\) satisfy the corresponding homogeneous equation; then find a particular solution of the given nonhomogeneous equation. In Problems 19 and \(20 g\) is an arbitrary continuous function. $$ t^{2} y^{\prime \prime}-t(t+2) y^{\prime}+(t+2) y=2 t^{3}, \quad t>0 ; \quad y_{1}(t)=t, \quad y_{2}(t)=t e^{t} $$

Short answer

The general solution of the given differential equation is: $$ y(t) = c_1t + c_2te^t + \frac{1}{2}t^3 $$ where \(c_1\) and \(c_2\) are arbitrary constants.

Step-by-step solution

Verify that \(y_1\) and \(y_2\) are solutions of the homogeneous equation

We will plug in \(y_1(t) = t\) and \(y_2(t) = te^t\) into the homogeneous equation $$ t^2y'' - t(t+2)y' + (t+2)y = 0 $$ and show that the equation is satisfied. First, plug in \(y_1(t) = t\): $$ t^2\left(\frac{d^2}{dt^2}(t)\right) - t(t+2)\left(\frac{d}{dt}(t)\right) + (t+2)(t) \stackrel{?}{=} 0 $$ The derivatives of \(t\) are: $$ y_1'(t) = 1, \quad y_1''(t) = 0 $$ Plugging these back into the equation: $$ t^2(0) - t(t+2)(1) + (t+2)(t) \stackrel{!}{=} 0 $$ Hence, \(y_1(t)=t\) is a solution of the homogeneous equation. Now, we'll plug in \(y_2(t) = te^t\): $$ t^2\left(\frac{d^2}{dt^2}(te^t)\right) - t(t+2)\left(\frac{d}{dt}(te^t)\right) + (t+2)(te^t) \stackrel{?}{=} 0 $$ The derivatives of \(te^t\) are: $$ y_2'(t) = e^t(t+1), \quad y_2''(t) = e^t(t+2) $$ Plugging these back into the equation: $$ t^2(e^t(t+2)) - t(t+2)(e^t(t+1)) + (t+2)(te^t) \stackrel{!}{=} 0 $$ Hence, \(y_2(t)=te^t\) is a solution of the homogeneous equation.

Find a particular solution

To find a particular solution, we'll use the method of undetermined coefficients. First, guess the form of the particular solution \(y_p(t)\): $$ y_p(t) = A t^3 $$ where \(A\) is an unknown constant. Now, compute the derivatives: $$ y_p'(t) = 3A t^2, \quad y_p''(t) = 6A t $$ Plug the expression for \(y_p(t)\) and its derivatives into the nonhomogeneous equation: $$ t^2(6A t) - t(t+2)(3A t^2) + (t+2)(At^3) = 2t^3 $$ Simplify and collect the terms: $$ (6At^3 - 3At^3 + At^3) = 2t^3 $$ Since the equation has to hold for all \(t>0\), we can equate the coefficients of the third-order terms: $$ (6A - 3A + A) t^3 = 2t^3 $$ Solve for \(A\): $$ 4A = 2 \implies A = \frac{1}{2} $$ Thus, the particular solution is: $$ y_p(t) = \frac{1}{2}t^3 $$

Obtain the general solution

Now that we have the homogeneous solutions and a particular solution, we can find the general solution as a linear combination of \(y_1(t), y_2(t),\) and \(y_p(t)\): $$ y(t) = c_1y_1(t) + c_2y_2(t) + y_p(t) = c_1t + c_2te^t + \frac{1}{2}t^3 $$ where \(c_1\) and \(c_2\) are arbitrary constants.

Key concepts

Homogeneous Differential Equations

Homogeneous differential equations are foundational to the study of differential equations and form the basis for understanding more complex, nonhomogeneous cases. In the simplest terms, these equations are characterized by having all terms containing the function or its derivatives, and equal to zero. Symbolically, a second-order homogeneous differential equation can be represented as
\[ a(t) y'' + b(t) y' + c(t) y = 0 \]
where \( a(t) \), \( b(t) \), and \( c(t) \) are functions of \( t \), and the right-hand side is consistently zero, signifying the absence of an external driving function.The solutions to homogeneous differential equations exhibit an important property known as the superposition principle, which allows individual solutions to be summed to form the general solution. In the context of our exercise, the provided functions \( y_1(t) = t \) and \( y_2(t) = te^t \) were verified to satisfy their corresponding homogeneous equation, illustrating how essential solutions can combine to address more complex equations.

Method of Undetermined Coefficients

The method of undetermined coefficients is a strategic approach for finding particular solutions to nonhomogeneous differential equations where the nonhomogeneous term is of a specific form—typically polynomial, exponential, or a sine/cosine function. This technique involves making an 'educated guess' about the form of the particular solution based on the nonhomogeneous term's nature.When applying this method, one initially assumes a solution structure with unknown coefficients (hence 'undetermined'). By differentiating this assumed solution and substituting back into the nonhomogeneous equation, one can determine the values of these coefficients by equating like terms. In our exercise, the particular solution guess was \( y_p(t) = At^3 \) for a nonhomogeneous term of \( 2t^3 \), which makes sense given the cubic nature of the term on the right-hand side.

Particular Solution

A particular solution to a nonhomogeneous differential equation is a specific solution that satisfies the entire nonhomogeneous equation, including the non-zero right-hand side. It reflects the influence of the external driving function on the system's behavior. Unlike the general solution, the particular solution does not contain arbitrary constants; it represents a single, unique function that accounts for this external influence.In our exercise, the particular solution was obtained using the method of undetermined coefficients, which allowed us to calculate the coefficient needed to satisfy the nonhomogeneous part of the equation. This solution \( y_p(t) = \frac{1}{2}t^3 \) contributes to the overall behavior of the system and is a crucial component in constructing the general solution for the nonhomogeneous differential equation.

General Solution

The general solution to a differential equation encompasses all possible solutions to the equation, accounting for all initial conditions or boundary values. It is typically composed of two parts: the homogeneous solution, which solves the equation without the nonhomogeneous term; and the particular solution, which specifically addresses the nonhomogeneous component.The general solution is often represented as a linear combination of the solutions to the homogeneous equation plus the particular solution. In the context of our exercise, the general solution for the nonhomogeneous differential equation is given by \( y(t) = c_1y_1(t) + c_2y_2(t) + y_p(t) \), where \( c_1 \) and \( c_2 \) are constants determined by the initial conditions of the problem. This expression means that, by appropriately selecting these constants, the general solution can adapt to fit a variety of specific scenarios, making it a powerful tool for understanding the system's dynamics under various circumstances.