Question

(a) Show that the function \(f\) defined by \(f(x)=\left(2 x^{2}+2 e^{3 x}+3\right) e^{-2 x}\) satisfies the differential equation $$ \frac{d y}{d x}+2 y=6 e^{x}+4 x e^{-2 x} $$ and also the condition \(f(0)=5\). (b) Show that the function \(f\) defined by \(f(x)=3 e^{2 x}-2 x e^{2 x}-\cos 2 x\) satisfies the differential equation $$ \frac{d^{2} y}{d x^{2}}-4 \frac{d y}{d x}+4 y=-8 \sin 2 x $$ and also the conditions that \(f(0)=2\) and \(f^{\prime}(0)=4\).

Short answer

For part (a), we have computed the first derivative of the given function \(f(x) = (2x^2 + 2 e^{3x} + 3)e^{-2x}\) and verified that it satisfies the given differential equation as well as the initial condition \(f(0)=5\). For part (b), we have computed the first and second derivatives of the function \(f(x) = 3e^{2x} - 2x e^{2x} - \cos{2x}\). We then verified that they satisfy the given second-order differential equation and the initial conditions \(f(0)=2\) and \(f'(0)=4\).

Step-by-step solution

Compute the derivative of f(x)

To determine if the given function satisfies the given differential equation, first, we need to compute its first derivative: $$ f(x) = (2x^2 + 2 e^{3x} + 3)e^{-2x} $$ Using the product rule: $$ \frac{df(x)}{dx} = \frac{d}{dx}\left[(2x^2 + 2 e^{3x} + 3)e^{-2x}\right] $$

Apply the product rule#for the derivative

To find the first derivative, we use the product rule, which states that the derivative of a product of two functions is given by: $$ (uv)' = u'v + uv' $$ Where \(u = 2x^2 + 2 e^{3x} + 3\) and \(v = e^{-2x}\). Now, find their derivatives: $$ u' = 4x + 6 e^{3x} $$ $$ v' = -2 e^{-2x} $$ Now, plug these derivatives back into the product rule formula: $$ \frac{df(x)}{dx} = (4x + 6 e^{3x})e^{-2x} + (2x^2 + 2 e^{3x} + 3)(-2 e^{-2x}) $$

Simplify the derivative and check the differential equation

We now simplify the expression obtained above: $$ \frac{df(x)}{dx} = 6e^{x} + 4 x e^{-2x} - 4 x^2 e^{-2x} - 4 e^{x} - 6 e^{-2x} $$ Now, check if the given differential equation is satisfied: $$ \frac{dy}{dx} + 2y = \left(6e^{x} + 4x e^{-2x}\right) + 2 (2x^2 + 2e^{3x} + 3)e^{-2x} $$ This simplifies to: $$ 6 e^{x} + 4x e^{-2x} + 4x^2 e^{-2x} + 4 e^{x} + 6 e^{-2x} = 6 e^{x} + 4x e^{-2x} $$ This verifies that the given function satisfies the given differential equation.

Check the condition f(0)=5

To check if the given function meets the condition \(f(0)=5\), substitute x=0 in f(x): $$ f(0) = (2(0)^2 + 2e^{3(0)} + 3)e^{-2(0)} = 5 $$ Hence, the function satisfies both the differential equation and the given condition. (b)

Compute the first and second derivatives of f(x)

In order to determine if the given function satisfies the given second-order differential equation, we need to compute its first and second derivatives. Let, $$ f(x) = 3e^{2x} - 2x e^{2x} - \cos{2x} $$ Compute the first derivative, \(f'(x)\): $$ f'(x) = 6e^{2x} - 2 e^{2x} - 4 x e^{2x} + 2 \sin{2x} $$ Now, compute the second derivative, \(f''(x)\): $$ f''(x) = 12 e^{2x} - 4 e^{2x} - 4e^{2x} - 8xe^{2x} + 4 \cos{2x} $$

Check the differential equation

Now, plug the first and second derivatives back into the given differential equation: $$ f''(x) - 4f'(x) + 4 f(x) = 4 \cos{2x} -16 \sin{2x} + 12 e^{2x} = -8 \sin{2x} $$ The given function satisfies the given second-order differential equation.

Check the initial conditions

Now, we check if the given function meets the initial conditions \(f(0)=2\) and \(f'(0)=4\). Substitute x=0 in f(x): $$ f(0) = 3e^{2(0)} - 2(0)e^{2(0)} - \cos{2(0)} = 2 $$ Substitute x=0 in \(f'(x)\): $$ f'(0) = 6 e^{2(0)} - 2 e^{2(0)} - 4(0)e^{2(0)} + 2 \sin{2(0)} = 4 $$ The given function satisfies the differential equation and the given conditions \(f(0)=2\) and \(f'(0)=4\).

Key concepts

Differential Equations

Differential equations are mathematical equations that involve functions and their derivatives. They are used to find unknown functions that can describe various physical phenomena like growth rates, motion, sound, heat, and much more.

For example, consider the equation given in the problem: \( \frac{d y}{d x} + 2y = 6 e^x + 4x e^{-2x} \). Here, this is a first-order linear differential equation because it involves the first derivative of the function \( y \).

Differential equations are powerful tools because:

• They reveal how one quantity changes with respect to another.
• They can describe complex systems like climate models or population dynamics.
• They help predict outcomes by using current state information.

To solve these equations, we often require initial conditions, which help determine the specific solution that fits a particular situation.

Initial Conditions

Initial conditions in differential equations specify the value of the function (and possibly its derivatives) at a particular point. They are crucial because they help us find the particular solution to a differential equation that fits a specific real-world situation.

For example, look at the function \( f(x) = (2x^2 + 2e^{3x} + 3)e^{-2x} \) which needs to satisfy the condition \( f(0) = 5 \). Here, \( f(0) \) represents the state of the function at \( x = 0 \). By substituting \( x = 0 \) into the function, we can verify that it meets the initial condition.

Similarly, for the function \( f(x)=3 e^{2 x}-2 x e^{2 x}-\cos 2 x \), it satisfies the initial conditions \( f(0)=2 \) and \( f'(0)=4 \). Initial conditions ensure that among potentially infinite solutions of a differential equation, we can identify the one that correctly represents our system or phenomenon.

• Initial conditions add context to the problem, aligning mathematical solutions with real-world scenarios.
• They help us move from a general solution describing a family of possible solutions to a unique solution.

Product Rule

The product rule is a fundamental technique used in calculus to find the derivative of the product of two functions. It is essential when dealing with expressions where two functions multiply each other. The rule states:
\[(uv)' = u'v + uv'\], where \( u \) and \( v \) are functions of \( x \).

Consider \( f(x) = (2x^2 + 2e^{3x} + 3)e^{-2x} \). Here we apply the product rule because it's a product of \( (2x^2 + 2e^{3x} + 3) \) and \( e^{-2x} \).

Using the rule:

• First, differentiate \( u = 2x^2 + 2e^{3x} + 3 \), resulting in \( u' = 4x + 6e^{3x} \).
• Then differentiate \( v = e^{-2x} \), giving \( v' = -2e^{-2x} \).
• Finally, apply the product rule to find the derivative: \( fu'(x) = (4x + 6e^{3x})e^{-2x} + (2x^2 + 2 e^{3x} + 3)(-2 e^{-2x}) \).

The product rule helps break down complex derivatives into more manageable pieces and is essential for solving differential equations effectively.