Question

Consider the nonhomogeneous differential equation $$ y^{\prime \prime \prime}+a y^{\prime \prime}+b y^{\prime}+c y=g(t) \text {. } $$ In each exercise, the general solution of the differential equation is given, where \(c_{1}, c_{2}\), and \(c_{3}\) represent arbitrary constants. Use this information to determine the constants \(a, b, c\) and the function \(g(t)\). $$ y=c_{1} \sin 2 t+c_{2} \cos 2 t+c_{3} e^{t}+t^{2} $$

Short answer

Question: Determine the constants a, b, c, and the function g(t) for the given general solution: y = c1 * sin(2t) + c2 * cos(2t) + c3 * e^t + t^2. Solution: The constants are: a = 1, b = 0, and c = 1. The function g(t) is: g(t) = c3 * e^t + t^2.

Step-by-step solution

Calculate the first derivative

To find the first derivative (\(y'\)), we differentiate the given function with respect to \(t\): $$ y' = 2c_{1} \cos 2t - 2c_{2} \sin 2t + c_{3} e^{t} $$

Calculate the second derivative

To find the second derivative (\(y''\)), we differentiate the first derivative with respect to \(t\): $$ y'' = -4c_{1} \sin 2t - 4c_{2} \cos 2t + c_{3} e^{t} $$

Calculate the third derivative

To find the third derivative (\(y'''\)), we differentiate the second derivative with respect to \(t\): $$ y''' = -8c_{1} \cos 2t + 8c_{2} \sin 2t + c_{3} e^{t} $$

Substitute derivatives into the differential equation

Now we substitute the calculated derivatives into the given nonhomogeneous differential equation, replacing \(y\), \(y'\), \(y''\), and \(y'''\): $$ (-8c_{1} \cos 2t + 8c_{2} \sin 2t + c_{3} e^{t}) + a(-4c_{1} \sin 2t - 4c_{2} \cos 2t + c_{3} e^{t}) + b(2c_{1} \cos 2t - 2c_{2} \sin 2t + c_{3} e^{t}) + c(c_{1} \sin 2 t+c_{2} \cos 2 t+c_{3} e^{t}+t^{2}) = g(t) $$

Simplify the equation

By comparing coefficients, we can simplify the equation: $$ ((-a-2b+1) c_{1} \sin 2t + (-a+2b+1) c_{2} \cos 2t + (1+a+b+c) c_{3} e^{t} + ct^2) = g(t) $$

Determine the constants and \(g(t)\)

From the given general solution and the simplified equation, we can see that there is no sine or cosine terms in \(g(t)\). Thus, \(a = -2b+1\) and \(a = 2b+1\). Solving these two equations simultaneously, we get \(a = 1\) and \(b = 0\). The simplified equation now becomes: $$ (1+c) c_{3} e^{t} + ct^2 = g(t) $$ Thus, based on the given general solution and the simplified equation, we have \(c=1\). Therefore, the constants are: \(a=1\), \(b=0\), \(c=1\). And the function \(g(t)\) is: $$ g(t) = c_{3} e^{t} + t^2 $$

Key concepts

Differential Equation Solution

When faced with a nonhomogeneous differential equation like \( y^{\prime \prime \prime}+a y^{\prime \prime}+b y^{\prime}+c y=g(t) \) , it's essential to unravel the mystery in a coherent order. The first step is always to identify the type of differential equation we are dealing with. In this case, the presence of a term that does not depend on \( y \) or its derivatives (\( g(t) \) ) indicates it's a nonhomogeneous one. The method to find the solution can vary, but one effective way is to first seek the complementary solution (the solution of the homogenous equation) and then find a particular solution that satisfies the nonhomogeneous term.

The proposed general solution \( y=c_{1} \sin 2t+c_{2} \cos 2t+c_{3} e^{t}+t^{2} \) already suggests that it reflects the sum of the homogeneous solution (the sinusoidal and exponential terms) and a particular solution \( t^2 \) addressing the nonhomogeneous part. To find the solution, one must differentiate, substitute, and then align the results with the general solution to determine the constants \( a, b, \text{and} c \) and the function \( g(t) \). With careful analysis and comparison of coefficients, one can unveil these constants, which are the secretive keys to the solution’s complete form.

Third Derivative

Delving into the third derivative can be quite an expedition. In our differential equation \( y^{\prime \prime \prime}+a y^{\prime \prime}+b y^{\prime}+c y=g(t) \) , the third derivative \( y^{\prime \prime \prime} \) plays a pivotal role. It encapsulates how the acceleration (the second derivative) is changing; you might call it the 'acceleration of the acceleration.'

Understanding the Third Derivative

To grasp it, you think of it as the rate at which the curve's curvature is changing at any given point. Calculating the third derivative is a straightforward process of repeated differentiation. Each differentiation peels back a layer, revealing how the function’s rate of change is evolving. If the second derivative sheds light on the function's concavity or 'bendiness,' then the third derivative reveals the rate at which this concavity is intensifying or easing off.

It is crucial for understanding the behavior of solutions to differential equations beyond just the slope (first derivative) or concavity (second derivative) and it signifies the presence of a force or interaction that's impacting our system's state more complexly than just speed or direction.

Integrating Factor Method

Tackling a first-order linear nonhomogeneous differential equation could look like an uphill battle, but that’s where the integrating factor method comes into play, like a mathematical superhero. The method involves conjuring an 'integrating factor,' a cleverly chosen function, which when multiplied by the differential equation, transforms it into an exact differential, making it considerably easier to solve.

The Magic of the Integrating Factor

This handy function is generally represented as \( e^{\int P(t) \, dt} \), where \( P(t) \) is a function that depends on the particular differential equation at hand. The idea is to get rid of pesky variables that are entwined in a way that makes separation of variables not feasible. After finding the integrating factor, you multiply it through the entire differential equation which— as if by magic— allows the left side to be expressed as the derivative of the product of the integrating factor and the unknown function \( y \).

Although our exercise didn’t require this method for solution due to it being a higher-order equation, understanding the integrating factor method is a powerful tool in a mathematician's toolkit for dealing with various first-order scenarios, guaranteeing smoother sailing through differential equations that might initially seem intimidating.