Question

Verifying solutions of initial value problems Verify that the given function \(y\) is a solution of the initial value problem that follows it. $$y(t)=16 e^{2 t}-10 ; y^{\prime}(t)-2 y(t)=20, y(0)=6$$

Short answer

Answer: Yes, y(t) = 16e^(2t) - 10 is a solution of the initial value problem as it satisfies both the differential equation and the initial condition.

Step-by-step solution

Find Derivative y'(t)

First, we need to find the derivative of y(t) with respect to t: y(t) = 16e^(2t) - 10 Differentiating with respect to t, we get: y'(t) = 32e^(2t)

Check y'(t) - 2y(t) = 20

Now, we need to substitute y(t) and y'(t) into the given equation and check if it is equal to 20: y'(t) - 2y(t) = 32e^(2t) - 2(16e^(2t) - 10) Simplify the equation: 32e^(2t) - 32e^(2t) + 20 = 20 As we can see, the left side of the equation is equal to the right side (20 = 20).

Check y(0) = 6

Now, we need to check if y(0) is equal to 6: y(0) = 16e^(2*0) - 10 y(0) = 16 - 10 y(0) = 6 As we can see, y(0) is equal to 6. As both conditions are satisfied, we can conclude that the given function y(t) is a solution of the initial value problem.

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They play a crucial role in describing the behavior of many physical systems. In the given exercise, a differential equation is provided:

• \( y'(t) - 2y(t) = 20 \)

This is a first-order linear differential equation, where \( y(t) \) is the function we are solving for, and \( y'(t) \) represents the derivative of \( y(t) \) with respect to \( t \).
We say this is an 'initial value problem' because it comes with an initial condition: \( y(0) = 6 \), which gives us a specific starting point.
To solve this type of problem, you typically find a general solution to the differential equation and then use the initial condition to solve for any constants. But in our exercise, we are verifying a provided solution, which simplifies our task.

Exponential Functions

Exponential functions are functions of the form \( f(x) = ae^{bx} \), where \( e \) is the base of the natural logarithms, \( a \) and \( b \) are constants. These functions are incredibly powerful in modeling growth and decay due to their unique properties.
In our problem, the function given is \( y(t) = 16e^{2t} - 10 \).
This represents an exponential component \((16e^{2t})\) combined with a constant shift of -10.

• The exponential part, \( 16e^{2t} \), indicates a rapid growth as \( t \) increases, because of the positive exponent \( 2t \).
• The subtraction by 10 shifts the entire curve downward.

The significance of exponential functions in differential equations stems from their properties under differentiation, making them ideal candidates for solutions to these equations.

Calculus

Calculus is the branch of mathematics that studies continuous change. It includes two main branches: differential calculus and integral calculus.

• **Differential calculus** focuses on finding derivatives, which represent rates of change.

In our exercise, we calculate the derivative of \( y(t) = 16e^{2t} - 10 \):

• Applying the differentiation rule for exponential functions, \( \frac{d}{dt}[e^{kt}] = ke^{kt} \), we find \( y'(t) = 32e^{2t} \).

• **Integral calculus**, on the other hand, involves the accumulation of quantities and the area under the curves, though not directly used in this exercise.

Understanding these fundamental calculus concepts is essential, especially when solving initial value problems, as they require the manipulation and interpretation of derivatives.