Question

Verify that the given function or functions is a solution of the differential equation. $$ y^{\prime \prime}+2 y^{\prime}-3 y=0 ; \quad y_{1}(t)=e^{-3 t}, \quad y_{2}(t)=e^{t} $$

Short answer

Question: Verify that \(y_1(t) = e^{-3t}\) and \(y_2(t) = e^t\) are solutions to the given differential equation \(y'' + 2y' - 3y = 0\). Answer: Both \(y_1(t) = e^{-3t}\) and \(y_2(t) = e^t\) are indeed solutions to the given differential equation, as we have shown that their first and second order derivatives, when substituted into the differential equation, result in a true equation.

Step-by-step solution

Find the first and second derivatives of the functions

To determine the first and second order derivatives, apply the chain rule to the functions: First Order Derivatives: $$ y_1'(t) = -3e^{-3t} $$ $$ y_2'(t) = e^t $$ Second Order Derivatives: $$ y_1''(t) = 9e^{-3t} $$ $$ y_2''(t) = e^t $$

Substitute the derivatives into the differential equation

Now, let's substitute the derivatives of each function into the differential equation: For \(y_1(t)\): $$ (9e^{-3t}) + 2(-3e^{-3t}) - 3(e^{-3t}) = 0 $$ For \(y_2(t)\): $$ (e^t) + 2(e^t) - 3(e^t) = 0 $$

Simplify and check if equations hold

Now let's simplify the equations and see if they hold: For \(y_1(t)\): $$ 9e^{-3t} - 6e^{-3t} - 3e^{-3t} = 0 $$ $$ (9 - 6 - 3)e^{-3t} = 0e^{-3t} $$ $$ 0 = 0 $$ For \(y_2(t)\): $$ e^t + 2e^t - 3e^t = 0e^t $$ $$ (1 + 2 - 3)e^t = 0e^t $$ $$ 0 = 0 $$ Both equations hold true, indicating that both \(y_1(t) = e^{-3t}\) and \(y_2(t) = e^t\) are solutions to the given differential equation $$ y'' + 2y' - 3y = 0 $$

Key concepts

Solution Verification

In mathematics, verifying solutions to differential equations is crucial to ensure correctness. A solution verification involves checking if a given function satisfies the original differential equation. This process begins by finding the relevant derivatives of the function. Then, these derivatives are substituted back into the differential equation to verify if both sides equate properly.

When verifying solutions, as shown in the exercise, it is crucial that any simplifications still equate to zero for homogeneous equations, like the one presented. In our example, both functions, \(y_1(t) = e^{-3t}\) and \(y_2(t) = e^t\), were tested by substituting their derivatives back into the differential equation. Upon simplification, each one led to the identity \(0 = 0\), thereby verifying them as solutions. It's similar to working your way backward with the operations involved to check that everything fits together harmoniously. This step provides confirmation that the proposed functions are, indeed, genuine solutions for the differential equation.

First and Second Derivatives

Derivatives describe how a function changes as its input changes. The first derivative tells us the rate of change or the slope of the function at any point, while the second derivative provides information on how the rate itself is changing, giving insight into the curvature of the graph.

For this exercise, finding derivatives is key. Let's look at each function separately:

• For \(y_1(t) = e^{-3t}\):
  • First derivative: \(y_1'(t) = -3e^{-3t}\)
  • Second derivative: \(y_1''(t) = 9e^{-3t}\)
• For \(y_2(t) = e^t\):
  • First derivative: \(y_2'(t) = e^t\)
  • Second derivative: \(y_2''(t) = e^t\)

These derivatives were obtained using the rules of differentiation and were essential in verifying the functions as solutions to the differential equation. Without these derivatives, we wouldn't be able to accurately test the functions against the equation.

Chain Rule

The chain rule is a fundamental technique in calculus used to differentiate composite functions. The rule essentially states that to differentiate a composite function, take the derivative of the outer function and multiply it by the derivative of the inner function.

In our given problem, we employed the chain rule to differentiate the exponential functions. For example, with \(y_1(t) = e^{-3t}\), the differentiation required applying the chain rule to differentiate \(e^{-3t}\):

• Identify the outer function \(f(u) = e^u\) and inner function \(u(t) = -3t\).
• Differentiate the outer function: \(f'(u) = e^u\).
• Differentiate the inner function: \(u'(t) = -3\).
• Apply the chain rule: \(y_1'(t) = f'(u) \cdot u'(t) = e^{-3t} \cdot -3 = -3e^{-3t}\).

This process allowed us to derive both the first and second derivatives successfully, confirming their applicability in verifying solutions to the differential equation.

Homogeneous Linear Differential Equation

A homogeneous linear differential equation is a type of differential equation in which every term is a function of the dependent variable and its derivatives, and is homogeneous, meaning it equals zero.

Our example involves the homogeneous equation \(y'' + 2y' - 3y = 0\). Such equations can generally be written in the form \(a(t)y'' + b(t)y' + c(t)y = 0\). In this exercise, the equation is simplified with constant coefficients.

Analyzing homogeneous linear differential equations involves finding functions, often exponential, whose derivatives satisfy the equation, hence verifying solutions. These functions usually arise from the roots of the characteristic equation, which are solutions in their exponential form. In our exercise, we used known exponential functions, verifying their fit within the homogeneous equation through derivative substitution.

The main importance of tackling a homogeneous equation is that once a solution is verified or found, it can lead to a set of general solutions, illustrating the behavior of a system described by such equations.