Question

Find the general solution of the given differential equation. $$ 4 y^{\prime \prime}-9 y=0 $$

Short answer

Question: Determine the general solution for the given second-order, linear, homogeneous differential equation. Differential Equation: $$ y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t} $$ Solution: The general solution for the given differential equation is $$ y(t) = C_1 e^{-t} + C_2te^{-t} - 2te^{-t} $$

Step-by-step solution

Find the complementary function

To find the complementary function, we first need to solve the homogeneous equation associated with the given differential equation: $$ y^{\prime \prime} + 2y^{\prime} + y = 0 $$ To solve this homogeneous equation, we will use the characteristic equation, which is formed by replacing the derivatives with powers of "r": $$ r^2 + 2r + 1 = 0 $$ This is a quadratic equation, and we can solve it using the quadratic formula: $$ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} $$ In this case, we have a = 1, b = 2, and c = 1, so the quadratic equation becomes: $$ r = \frac{-2 \pm \sqrt{2^2 - 4\cdot 1\cdot 1}}{2\cdot1} $$ Since the discriminant (b^2 - 4ac) is equal to 0, the equation has only one solution (a repeated root): $$ r = -1 $$ The general complementary function is given by: $$ y_c(t) = C_1 e^{r_1t} + C_2 e^{r_2t} $$ Since we have only one root, r = -1, the complementary function becomes: $$ y_c(t) = C_1 e^{-t} + C_2te^{-t} $$

Find a particular solution

Now, we need to find a particular solution to the original non-homogeneous equation: $$ y^{\prime \prime}+2 y^{\prime}+y=2 e^{-t} $$ We will assume a particular solution of the form: $$ y_p(t) = Ate^{-t} $$ Now, we need to find \(A\). First, find the first and second derivatives of \(y_p(t)\): $$ y^{\prime}_p(t) = Ae^{-t} - Ate^{-t} $$ $$ y^{\prime \prime}_p(t) = -Ae^{-t} + 2Ate^{-t} -Ae^{-t} $$ Now, substitute \(y_p(t)\), \(y^{\prime}_p(t)\), and \(y^{\prime \prime}_p(t)\) in the original non-homogeneous equation: $$ (-Ae^{-t} + 2Ate^{-t} -Ae^{-t}) + 2(Ae^{-t} - Ate^{-t}) + Ate^{-t} = 2e^{-t} $$ Simplify the equation and solve for A: $$ (1-2t)Ae^{-t} = 2e^{-t} $$ $$ A = \frac{2}{1-2t} $$ Since we want a particular solution that cancels the terms from the complementary function, the term \(te^{-t}\) should remain. We need to change the denominator to remove the terms with \(te^{-t}\): $$ A = \frac{2}{-1} $$ $$ A = -2 $$ So the particular solution is: $$ y_p(t) = -2te^{-t} $$

Find the general solution

Finally, we can determine the general solution by adding the complementary function and the particular solution: $$ y(t) = y_c(t) + y_p(t) $$ $$ y(t) = C_1 e^{-t} + C_2te^{-t} - 2te^{-t} $$ This is the general solution of the given differential equation.

Key concepts

Characteristic Equation

In the realm of ordinary differential equations (ODEs), the characteristic equation plays a significant role, particularly for linear differential equations with constant coefficients. This algebraic equation is derived from an ODE by assuming a solution of the form \( e^{rt} \). Here, \( r \) represents the roots that determine the behavior of the solution.

To illustrate this, let's look at the given differential equation \( y'' + 2y' + y = 2e^{-t} \). Firstly, we identify its associated homogeneous equation \( y'' + 2y' + y = 0 \). The characteristic equation is then formed by substituting \( y \) with \( e^{rt} \), \( y' \) with \( re^{rt} \), and \( y'' \) with \( r^2e^{rt} \) which leads to \( r^2 + 2r + 1 = 0 \). Solving this quadratic equation gives us the roots which signify the nature of the solutions, be they real, repeated, or complex.

General Solution

The general solution of a differential equation encompasses the entire family of possible solutions which include both the complementary function and any particular solution. For linear ODEs, it is expressed as the sum of these two functions. In the given example, the general solution is \( y(t) = y_c(t) + y_p(t) \), combining the homogeneous part \( y_c(t) \) that solves the associated homogeneous equation, and a \( y_p(t) \) that satisfies the non-homogeneous part.

This concept is crucial as it offers a complete picture of the behavior of the system described by the ODE. After finding the complementary function \( y_c(t) = C_1 e^{-t} + C_2te^{-t} \) and the particular solution \( y_p(t) = -2te^{-t} \) from the above exercise, the resulting general solution \( y(t) \) describes all possible trajectories depending on the constants \( C_1 \) and \( C_2 \) which are determined by the initial conditions.

Particular Solution

A particular solution to a differential equation is a single, specific solution that satisfies not just the differential equation, but also fits additional constraints or conditions, such as initial values. Unlike the complementary function, which represents a family of solutions, the particular solution does not contain arbitrary constants.

To find a particular solution for the non-homogeneous differential equation \( y''+2y'+y=2e^{-t} \), one must account for the non-homogeneous term, \( 2e^{-t} \). Guided by the method of undetermined coefficients, we choose \( y_p(t) = Ate^{-t} \) because the non-homogeneous part is a product of \( t \) and an exponential function. Through differentiation and substitution into the ODE, we can solve for the coefficient \( A \) and hence determine the particular solution.

Complementary Function

For solving homogeneous linear differential equations, where the right-hand side is zero, the complementary function \( y_c(t) \) forms the basis of the solution. It is composed of a linear combination of terms based on the roots of the characteristic equation. When those roots are real and distinct, they yield exponential solutions. In case of repeated roots, like \( r = -1 \) in our exercise, the complementary function includes \( t \) multiplied by an exponential term, leading to \( C_1 e^{-t} + C_2te^{-t} \).

The complementary function reflects the inherent dynamics of the system without external influences, symbolized by the homogeneous part of the differential equation. In our example, the complementary function addresses the solution to \( y'' + 2y' + y = 0 \) and represents all the trajectories that the system could follow due to its natural behavior.

Non-Homogeneous Differential Equation

A non-homogeneous differential equation incorporates an additional function on the right-hand side, which represents external forces or inputs that affect the system's behavior. This kind of ODE takes the form \( Ly = f(t) \) where \( L \) is a differential operator, \( y \) is the unknown function, and \( f(t) \) is a known non-zero function of \( t \).

In contrast to its homogeneous counterpart, the non-homogeneous equation requires finding a particular solution that accounts for the effect of \( f(t) \)—in our case, \( 2e^{-t} \). This particular solution \( y_p(t) \) is then combined with the complementary function \( y_c(t) \) found by setting \( f(t) = 0 \) to construct the general solution of the non-homogeneous equation. Thus, the end solution respects both the internal dynamics and the external factors influencing the system.