Question

Verify, for any values \(c_{1}\) and \(c_{2}\), that the functions \(y_{1}(t)\) and \(y_{2}(t)\) satisfy the given system of linear differential equations. $$ \begin{aligned} &y_{1}^{\prime}=y_{1}+y_{2}, \quad y_{1}(t)=c_{1} e^{t} \cos t+c_{2} e^{t} \sin t \\ &y_{2}^{\prime}=-y_{1}+y_{2}, \quad y_{2}(t)=-c_{1} e^{t} \sin t+c_{2} e^{t} \cos t \end{aligned} $$

Short answer

Answer: Yes, the given functions \(y_1(t)\) and \(y_2(t)\) satisfy the provided system of linear differential equations for any coefficients \(c_1\) and \(c_2\).

Step-by-step solution

Find \(y_1'(t)\) and \(y_2'(t)\)

Firstly, let's find the derivatives of both functions with respect to \(t\): For \(y_1(t) = c_1 e^t \cos t + c_2 e^t \sin t\), we have: $$ y_1'(t) = \frac{d}{dt} (c_1 e^t \cos t + c_2 e^t \sin t) $$ For \(y_2(t) = -c_1 e^t \sin t + c_2 e^t \cos t\), we have: $$ y_2'(t) = \frac{d}{dt} (-c_1 e^t \sin t + c_2 e^t \cos t) $$

Apply product rule to find derivatives

Use the product rule (u(x)v(x))'= u'(x)v(x) + u(x)v'(x), where u(x) = \(c_1 e^t\) and v(x) = \(\cos t\) for \(y_1'(t)\), and u(x) = \(c_2 e^t\) and v(x) = \(\sin t\) for \(y_2'(t)\): For \(y_1'(t)\), we have: $$ y_1'(t) = (\frac{d}{dt} (c_1 e^t) )\cos t + (c_1 e^t )\frac{d}{dt}(\cos t) + (\frac{d}{dt}(c_2 e^t)) \sin t +(c_2 e^t) \frac{d}{dt}(\sin t) $$ For \(y_2'(t)\), we have: $$ y_2'(t) = (\frac{d}{dt} (-c_1 e^t) )\sin t + (-c_1 e^t )\frac{d}{dt}(\sin t) + (\frac{d}{dt}(c_2 e^t)) \cos t +(c_2 e^t) \frac{d}{dt}(\cos t) $$

Calculate the derivatives

Now, let's compute the derivatives and simplify the expressions: For \(y_1'(t)\), we get: $$ y_1'(t) = c_1 e^t \cos t - c_1 e^t \sin t + c_2 e^t \sin t + c_2 e^t \cos t $$ For \(y_2'(t)\), we get: $$ y_2'(t) = c_1 e^t \sin t - c_1 e^t \cos t - c_2 e^t \sin t + c_2 e^t \cos t $$

Check if functions satisfy differential equations

Now, we will check if \(y_1'(t)\) and \(y_2'(t)\) satisfy the given differential equations. For equation 1: \(y_1' = y_1 + y_2\) $$ c_1 e^t \cos t - c_1 e^t \sin t + c_2 e^t \sin t + c_2 e^t \cos t = (c_1 e^t \cos t + c_2 e^t \sin t) + (-c_1 e^t \sin t + c_2 e^t \cos t) $$ For equation 2: \(y_2' = - y_1 + y_2\) $$ c_1 e^t \sin t - c_1 e^t \cos t - c_2 e^t \sin t + c_2 e^t \cos t = - (c_1 e^t \cos t + c_2 e^t \sin t) + (-c_1 e^t \sin t + c_2 e^t \cos t) $$ Both expressions match, so we can conclude that the given functions \(y_1(t)\) and \(y_2(t)\) satisfy the provided system of linear differential equations for any coefficients \(c_1\) and \(c_2\).

Key concepts

Product Rule

The product rule is an essential mathematical tool that helps us find the derivative of a product of two functions. It states that if you have two functions, say, u(t) and v(t), the derivative of their product u(t)v(t) is given by u'(t)v(t) + u(t)v'(t). This rule is pivotal when differentiating complex expressions, such as those involving both exponential and trigonometric functions.

For example, if we take a function y(t) composed of an exponential function e^t and a trigonometric function cos(t), then the derivative y'(t) is found by applying the product rule: y'(t) = (d/dt)(e^t)cos(t) + e^t(d/dt)(cos(t)). In our exercise, we use this rule to find the derivatives for y_1(t) and y_2(t), both of which include products of exponential and trigonometric functions.

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. They are paramount in modeling situations where change is continuous, such as in physics, engineering, or economics. Linear differential equations, as shown in our exercise, are a specific type where the dependent variable and its derivatives appear to the first power and there are no products of the variable and its derivatives.

These equations can describe complex systems and their solutions, like the ones given for y_1(t) and y_2(t), represent the behavior of such systems under various conditions. Confirming that these solutions satisfy the given differential equations ensures they are valid for any constants c_1 and c_2, which represent initial conditions of the system.

Exponential Functions

Exponential functions are functions in the form f(t) = a^t, where 'a' is a positive constant and 't' is the variable. They are widely used in studies dealing with growth and decay processes, such as population growth, radioactive decay, and compound interest. Exponential growth or decay is characterized by the rate of change of a quantity being proportional to the quantity's current value.

In the context of our exercise, e^t is an exponential function with a growth rate proportional to its value at time t. This is why, when we differentiate e^t, the derivative is itself e^t, signifying constant growth rate. These types of functions are ideal for representing naturally occurring growth processes in differential equations.

Trigonometric Functions

Trigonometric functions like sine (sin(t)) and cosine (cos(t)) are fundamental in mathematics, especially when it comes to representing periodic phenomena such as waves and oscillations. These functions have properties that make them especially useful in solving differential equations that involve periodic behavior.

In our differential equations, trigonometric functions are paired with exponential functions. This combination leads to solutions that represent damped or amplified oscillations, depending on the context. Understanding how to differentiate these functions is crucial in our exercise, as we need to apply their derivatives correctly when using the product rule. For instance, the derivative of cos(t) is -sin(t), and the derivative of sin(t) is cos(t). These derivative functions also exhibit periodic behavior, which perfectly aligns with the properties of their original functions.