Question

Verify that the given function is a solution of the differential equation that follows it. $$u(t)=C_{1} e^{t}+C_{2} t e^{t} ; u^{\prime \prime}(t)-2 u^{\prime}(t)+u(t)=0$$

Short answer

Now, create a short answer. The given function \(u(t) = C_1 e^t + C_2te^t\) is indeed a solution to the differential equation \(u''(t) - 2u'(t) + u(t) = 0\) because, after finding its first and second derivatives and substituting them into the equation, it simplifies to 0.

Step-by-step solution

Find the first derivative of u(t)

To begin, we need to find the first derivative of the given function u(t) with respect to t. We apply the Sum Rule and Product Rule of differentiation here: \(\frac{d}{dt} (u(t)) = \frac{d}{dt} (C_1 e^t + C_2te^t)\) \(u'(t) = C_1 e^t + C_2 e^t + C_2te^t\)

Find the second derivative of u(t)

Next, we calculate the second derivative of u(t) by differentiating u'(t) with respect to t. Again, we apply the Sum Rule and Product Rule: \(\frac{d^2}{dt^2} (u(t)) = \frac{d}{dt} (C_1 e^t + C_2 e^t + C_2te^t)\) \(u''(t) = C_1 e^t + C_2 e^t + 2C_2 e^t + C_2 t e^t\)

Substitute the derivatives in the differential equation

Now that we have both the first and second derivatives, we can substitute them in the given differential equation and check if it holds true: \(u''(t) - 2u'(t) + u(t) = 0\) \((C_1 e^t + C_2 e^t + 2C_2 e^t + C_2 t e^t) - 2(C_1 e^t + C_2 e^t + C_2 te^t) + (C_1 e^t + C_2te^t) = 0\)

Simplify and check if the equation holds true

Next, we need to simplify the equation and check if it is zero: \(C_1e^t + C_2e^t + 2C_2e^t + C_2te^t - 2C_1e^t - 2C_2e^t - 2C_2te^t + C_1e^t + C_2te^t = 0\) \(e^t(2C_1 - 2C_1 + C_1) + te^t(C_2 - 2C_2 + C_2) = 0\) \(e^t(0) + te^t(0) = 0\) Since the equation holds true and equals zero, we can conclude that the given function u(t) is a solution of the given differential equation.

Key concepts

First Derivative

The first derivative of a function represents the rate of change of that function with respect to a variable. In the given exercise, we need to find the first derivative of \( u(t) = C_1 e^t + C_2 t e^t \) with respect to \( t \). This involves using two essential rules of differentiation: the Sum Rule and the Product Rule.

• Sum Rule: This rule states that the derivative of a sum is the sum of the derivatives. Mathematically, \( \frac{d}{dt}(f(t) + g(t)) = f'(t) + g'(t) \).
• Product Rule: When differentiating the product of two functions, we use this rule, which is given by \( \frac{d}{dt} (f(t)g(t)) = f'(t)g(t) + f(t)g'(t) \).

Applying these rules to \( C_1 e^t + C_2 t e^t \), we differentiate each component separately:- \( C_1 e^t \) simply becomes \( C_1 e^t \).- Using the Product Rule on \( C_2 t e^t \), - We treat \( C_2 t \) as one function and \( e^t \) as another, resulting in \( C_2 e^t + C_2 t e^t \).Thus, the first derivative is \( u'(t) = C_1 e^t + C_2 e^t + C_2 t e^t \).

Second Derivative

The second derivative provides information on how the rate of change (first derivative) itself changes. It offers insights into the curvature and concavity of the function. Continuing from the first derivative we calculated, the second derivative is found by differentiating \( u'(t) = C_1 e^t + C_2 e^t + C_2t e^t \) again with respect to \( t \).

Once more, we utilize the Sum Rule and Product Rule to find\( u''(t) \):

• \( C_1 e^t \) differentiates to \( C_1 e^t \), as \( e^t \) remains unchanged.
• \( C_2 e^t \) similarly differentiates to \( C_2 e^t \).
• Using the Product Rule on \( C_2 t e^t \), we find its derivative as \( 2C_2 e^t + C_2 t e^t \). Here, we differentiated \( C_2 t \) separately giving \( C_2 \), and \( e^t \)'s derivative remains \( e^t \).

Summing these results, we obtain the second derivative: \( u''(t) = C_1 e^t + C_2 e^t + 2C_2 e^t + C_2 t e^t \).

Product Rule

The Product Rule is vital when differentiating expressions where two functions are multiplied. The rule is particularly significant in this exercise for terms like \( C_2 t e^t \). It is applied to differentiate these products.

Here's how the Product Rule is applied:

• Identify the two functions in the product. For \( C_2 t e^t \), treat \( C_2 t \) as one function \( f(t) \) and \( e^t \) as another \( g(t) \).
• The derivative \( f'(t) \) of \( C_2 t \) is \( C_2 \), and \( g'(t) \) of \( e^t \) is \( e^t \).
• Apply the Product Rule: \( (C_2 t)' e^t + C_2 t (e^t)' = C_2 e^t + C_2 t e^t \).

This calculation ensures proper differentiation of products, leading to accurate results for both the first and second derivatives. Proper application of the Product Rule is essential in maintaining the integrity of your calculus operations.

Sum Rule

The Sum Rule allows us to break down the differentiation process by handling each term separately. This is particularly helpful when a function comprises multiple other functions added together, like \( u(t) = C_1 e^t + C_2 t e^t \).

To apply the Sum Rule, follow these steps:

• Take each function in the sum separately. Calculate the derivative for each function independently.
• Add all the individual derivatives together as the derivative of the entire function.

For example, with \( u(t) \), we have:- Derivative of \( C_1 e^t \) is \( C_1 e^t \), and - Derivative of \( C_2 t e^t \) involves the Product Rule resulting in \( C_2 e^t + C_2 t e^t \).
Add these derivatives using the Sum Rule to get the first derivative: \( u'(t) = C_1 e^t + C_2 e^t + C_2 t e^t \).Understanding the Sum Rule helps simplify and streamline the process of taking derivatives of complex functions efficiently.