Question

find the solution of the given initial value problem. $$ y^{\prime}-2 y=e^{2 x}, \quad y(0)=2 $$

Short answer

Answer: The solution for the given IVP is $y(x) = e^{2x}(x + 2)$.

Step-by-step solution

Identify the given differential equation and initial condition

The given differential equation is $$ y'(x)-2y(x)=e^{2x} $$ and the initial condition is $$ y(0) = 2 $$

Calculate the integrating factor

In our case, the function associated with y(x) is -2. Let's call this function p(x): $$ p(x) = -2 $$ Now, the integrating factor (IF) is given by the exponential of the integral of p(x) tensorwith respect to x: $$ \text{IF} = e^{\int p(x) \, dx} $$ Plugging the value of p(x) and simplify: $$ \text{IF} = e^{\int -2 \, dx} = e^{-2x}\ $$

Multiply the differential equation by the integrating factor

Now, multiply the given differential equation by the computed integrating factor: $$ e^{-2x} (y'(x) - 2y(x)) = e^{-2x} e^{2x} $$ which becomes: $$ (e^{-2x}y(x))' = 1 $$

Integrate both sides of the equation

Now, we integrate both sides of the equation with respect to x: $$ \int (e^{-2x}y(x))' \, dx = \int 1 \, dx $$ This results in $$ e^{-2x}y(x) = x + C $$

Rearrange the equation to find the general solution

Now, rearrange the equation above to find y(x): $$ y(x) = e^{2x}(x + C) $$ Now we have the general solution.

Apply the initial condition to find the particular solution

Using the initial condition y(0) = 2, we can find the particular solution: $$ 2 = e^{2\cdot0}(0 + C) $$ which gives us $$ C = 2 $$

Write down the final solution

Finally, plug the value of C back into the general solution to find the particular solution: $$ y(x) = e^{2x}(x + 2) $$ That is the solution of the given initial value problem.

Key concepts

Integrating Factor Method

The Integrating Factor Method is a technique used to solve first-order linear differential equations. This method transforms a given differential equation into one that can be more easily integrated. Here's how it works:

• Identify the linear differential equation in the standard form: \( y' + p(x)y = q(x) \).
• Calculate the integrating factor (IF), which is derived from the function \( p(x) \). The IF is expressed as \( e^{\int p(x) \, dx} \).
• Multiply every term in the original equation by the IF to rewrite the left side as the derivative of a product of functions.
• After rewriting, integrate both sides with respect to the variable to find a solution.

By applying the integrating factor, the left-hand side of the equation becomes an exact differential, making it easier to solve the problem.

Differential Equation

A differential equation is an equation that involves the derivatives of a function. These equations describe how a particular quantity changes relative to changes in another quantity. Differential equations can be used to model many real-world phenomena, such as:

• Growth rates (population, bacteria, etc.)
• Decay (radioactive decay, cooling laws)
• Motion (velocity, acceleration)

In the context of our exercise, we have the differential equation \( y' - 2y = e^{2x} \). This equation relates the rate of change of the function \( y(x) \) to a combination of \( y(x) \) and a given function of \( x \). Identifying the type of differential equation helps decide which method, such as the Integrating Factor Method, to use for solving it.

Particular Solution

A particular solution to a differential equation is one that satisfies the equation as well as any initial or boundary conditions provided. In contrast to the general solution, the particular solution is tailored to specific conditions.

For the given problem, after calculating the integrating factor and finding the general solution \( y(x) = e^{2x}(x + C) \), we used the initial condition \( y(0) = 2 \) to determine the specific constant \( C \). This involved substituting \( x = 0 \) and \( y = 2 \) into the general solution, solving for \( C \). The particular solution, \( y(x) = e^{2x}(x + 2) \), accurately reflects the unique scenario encapsulated by the initial value condition of the exercise.

General Solution

The general solution of a differential equation represents a family of functions that includes all possible solutions. It’s often expressed in terms of an arbitrary constant, which can be determined if additional conditions, such as initial values, are provided.

In solving the differential equation \( y' - 2y = e^{2x} \), the general solution was obtained as \( y(x) = e^{2x}(x + C) \), where \( C \) is an arbitrary constant. This solution accounts for any potential scenario by allowing \( C \) to take any real value. To derive a particular solution from the general solution, it is necessary to apply specific conditions (like \( y(0) = 2 \)). By doing so, we pinpoint a single value of \( C \), converting the general solution into a particular one that applies to the situation described by the initial condition.