Question

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

Question: Verify if the function \(y(t) = 16e^{2t} - 10\) is a solution to the initial value problem \(y'(t) - 2y(t) = 20\) with the initial condition \(y(0) = 6\). Answer: Yes, the given function \(y(t) = 16e^{2t} - 10\) is a solution to the initial value problem, as it satisfies both the given differential equation and the initial condition.

Step-by-step solution

Find the derivative of y(t) with respect to t

First, we need to find the derivative of \(y(t) = 16e^{2t} - 10\) with respect to t. This can be done using the chain rule for the exponential term and the constant rule for the constant term. $$ \frac{d}{dt}(y(t)) = \frac{d}{dt}(16 e^{2 t}) - \frac{d}{dt}(10) = 16 \cdot 2 e^{2 t} - 0 = 32 e^{2 t} $$

Replace y(t) and y'(t) in the given differential equation

Now, we substitute the function \(y(t)\) and its derivative \(y'(t)\) into the given differential equation \(y'(t) - 2y(t)=20\): $$ 32 e^{2 t} - 2(16 e^{2 t} - 10) = 20 $$

Simplify the equation

Now we have to simplify this equation to see if it holds true: $$ 32 e^{2 t} - 32 e^{2 t} + 20 = 20 $$ $$ 20 = 20 $$ Since the left side is equal to the right side, this means that the given function \(y(t)\) satisfies the differential equation.

Check the initial condition

Now we have to check if the function \(y(t)\) satisfies the initial condition \(y(0) = 6\). To do this, we'll plug in \(t = 0\) into the function \(y(t)\): $$ y(0) = 16 e^{2 \cdot 0} - 10 = 16 \cdot 1 - 10 = 6 $$ Since the function \(y(0) = 6\), the initial condition is also satisfied. In conclusion, since the function \(y(t)\) satisfies both the given differential equation and the initial condition, it is indeed a solution to the initial value problem.

Key concepts

Differential Equation

A differential equation involves derivatives of a function. They express relationships between functions and their rates of change. In this problem, you encounter a differential equation given as \(y'(t) - 2y(t) = 20\). Here, \(y(t)\) is the original function and \(y'(t)\) is its derivative.

Differential equations are essential in mathematics and applied sciences. They help us understand how things change over time or space. In this case, the differential equation is linear because it only involves linear combinations of \(y(t)\) and \(y'(t)\).

• Linear differential equations have the form \(a(t)y' + b(t)y = c(t)\), where \(a(t)\), \(b(t)\), and \(c(t)\) are functions of \(t\).
• This particular equation is simpler because it has constant coefficients and the term \(c(t)\) is a constant (20).

Once a function meets all conditions set by the differential equation and the initial conditions, you have a solution to the problem.

Exponential Function

Exponential functions are critical when solving differential equations, especially when the variable is in the exponent. In this exercise, the function \(y(t) = 16e^{2t} - 10\) involves an exponential function, \(e^{2t}\).

Exponential functions like \(e^{x}\) grow rapidly. They are used to model growth processes like populations or interest over time. The constant \(e\) is an irrational number, approximately 2.718, and it's the base for natural logarithms.

• In \(16e^{2t} - 10\), 16 is a coefficient affecting the amplitude of the exponential curve, and 10 shifts the curve vertically.
• The expression \(2t\) specifies the rate at which the function grows. The larger the value of 2, the steeper the curve.

This exponential component is key to solving the differential equation, as it reflects how the solution behaves over time.

Chain Rule

The chain rule is a fundamental tool in calculus that helps us differentiate composite functions. To calculate the derivative of \(y(t) = 16e^{2t} - 10\), we apply the chain rule to the term \(16e^{2t}\).

A composite function is a function within another function, like \(e^{2t}\) in this context. When differentiating such functions, we first take the derivative of the outer function, and then multiply it by the derivative of the inner function.

• Here, the outer function is \(e^x\) and its derivative stays the same, \(e^x\).
• The inner function is \(2t\), and its derivative is 2.
• So, applying the chain rule, \(\frac{d}{dt}(16e^{2t}) = 16 \cdot e^{2t} \cdot 2 = 32e^{2t}\).

By understanding and using the chain rule, you can differentiate more complex functions effectively, which is crucial for solving differential equations.