Question

Does the function \(y(t)=6 e^{-3 t}\) satisfy the initial value problem \(y^{\prime}(t)-3 y(t)=0, y(0)=6 ?\)

Short answer

Answer: No, the function does not satisfy the initial value problem, although it does satisfy the given initial condition.

Step-by-step solution

Differentiate the given function

First, we find the derivative of \(y(t)=6 e^{-3 t}\). To do this, we will use the chain rule: \(\frac{d}{dt}(e^{-3 t}) = -3 e^{-3 t}\). So, \(y^{\prime}(t) = -18 e^{-3 t}\). Now we have the function and its derivative: \(y(t) = 6 e^{-3t}\) and \(y^{\prime}(t) = -18 e^{-3t}\).

Substitute the given function and its derivative into the initial value problem equation

Next, we substitute the function and its derivative into the given equation: \(y^{\prime}(t) - 3y(t) = 0\). \((-18 e^{-3t}) - 3(6 e^{-3t}) = (-18 e^{-3t}) - (18 e^{-3t})\). Adding the two terms together, we get \(-36 e^{-3t}\). Since it is not equal to 0, the function does not satisfy the given equation.

Check the initial conditions

Lastly, let's check the initial conditions given by \(y(0) = 6\). We have: \(y(0) = 6 e^{-3\cdot0} = 6 e^0 = 6.\) As the initial condition is satisfied, it implies that the function satisfies the initial value problem for the initial condition, but not for the differential equation.

Key concepts

Differential Equations

Differential equations are mathematical tools used to relate functions to their derivatives. They describe how a quantity changes over time or space and are fundamental in expressing the laws of nature, such as Newton's laws of motion or the heat equation.

When presented with a differential equation, the goal is to find a function or set of functions that solve the equation. An initial value problem (IVP) is a type of differential equation that is accompanied by specific values the function must take at certain points, referred to as initial conditions. The IVP is considered solved when a function satisfies both the differential equation and the initial conditions.

In the context of the given exercise, the problem is defining an equation involving the derivative of a function, denoted as \(y'(t)\), and the function itself, \(y(t)\). The challenge arises in verifying if the proposed function not only satisfies the equation, but also agrees with the initial value provided.

Chain Rule Differentiation

The chain rule is a principle in calculus that's used when differentiating composite functions. Intuitively, it allows us to understand how the rate of change of a function is affected by another function nested within it.

When faced with a function like \(e^{-3t}\), where the variable \(t\) is multiplied by a constant within the exponent, the chain rule becomes vital. To apply the rule, one identifies the 'outside' function (in this case \(e^x\)) and the 'inside' function (in this case \(-3t\)), then differentiates each part separately before multiplying the results together.

However, it seems there has been a mistake in the original solution. The derivative of \(6e^{-3t}\) with respect to \(t\) should actually be \(6 \times (-3) \times e^{-3t} = -18e^{-3t}\), not \(-18e^{-3t}\) as stated in the exercise improvement advice. Such an error can fundamentally change the outcome when verifying if the function satisfies the differential equation.

Exponential Functions

Exponential functions are characterized by a constant base raised to a variable exponent. They are widely used across multiple disciplines, from population models in biology to radioactive decay in physics.

The function \(y(t) = 6e^{-3t}\) is an exponential function where \(e\) (approximately 2.71828) is the base and \(-3t\) is the exponent. Exponential functions have unique properties, such as the fact that the derivative of \(e^x\) with respect to \(x\) is itself, \(e^x\). This self-derivative property is specific to the base \(e\), and it helps in solving many calculus problems.

At the initial condition, \(t=0\), the exponential term \(e^{-3t}\) simplifies to \(e^0\), which equals 1. Therefore, the function value is \(y(0) = 6e^0 = 6\), which confirms that the initial condition is satisfied. This illustrates how exponential functions can be analyzed and simplified at particular points to check against given conditions.