Problem 28 - Calculus Early Transcendentals

Question

Solve the following problems. $$\frac{d u}{d x}=2 u+6, u(1)=6$$

Step-by-Step Solution

Answer

Question: Find the function \(u(x)\) that satisfies the given first-order ordinary differential equation (ODE) and the initial condition: $$\frac{d u}{d x}=2 u+6, \quad u(1)=6$$ Answer: The function \(u(x)\) that satisfies the given ODE and initial condition is: $$u(x)=-3+\frac{9}{e^2}e^{2x}$$

Step 1: Write the given ODE and initial condition

The given ODE is: $$\frac{d u}{d x}=2 u+6$$ And the initial condition is: $$u(1)=6$$

Step 2: Solve the ODE using the Integrating Factor method

First, we notice that this is a first-order linear ODE, which can be written in the standard form: $$\frac{d u}{d x}-2u=6$$ To solve this ODE, we can use the Integrating Factor (IF) method. We identify the function multiplying the dependent variable, which is \(-2\), and calculate the integrating factor, IF: $$\text{IF}=e^{\int{-2 dx}}=e^{-2x}$$ Next, multiply the entire ODE by the Integrating Factor: $$e^{-2x}\frac{d u}{d x} - 2ue^{-2x} = 6e^{-2x}$$ Now, notice that the left side of the equation is the derivative of \(u(x)e^{-2x}\): $$\frac{d}{d x}\left(u(x)e^{-2x}\right) = 6e^{-2x}$$ Next, we integrate both sides with respect to \(x\): $$u(x)e^{-2x} = -3e^{-2x} + C_1$$

Step 3: Solve for \(u(x)\)

Now, isolate \(u(x)\) by multiplying both sides by \(e^{2x}\): $$u(x) = -3e^{2x}e^{2x} + C_1e^{2x}$$ Which simplifies to: $$u(x)=-3+C_1e^{2x}$$

Step 4: Apply the initial condition and find the constant \(C_1\)

To find the constant \(C_1\), insert the initial condition \(u(1)=6\) into the equation of \(u(x)\): $$6=-3+C_1e^{2\cdot 1}$$ $$9=C_1e^2$$ Divide both sides by \(e^2\) to solve for \(C_1\): $$C_1=\frac{9}{e^2}$$

Step 5: Write the final solution for \(u(x)\)

Finally, plug the found value for \(C_1\) back into the equation for \(u(x)\): $$u(x)=-3+\frac{9}{e^2}e^{2x}$$ So the solution of the given ODE with the specified initial condition is: $$u(x)=-3+\frac{9}{e^2}e^{2x}$$

Key Concepts

First-order Linear ODE

A first-order linear ordinary differential equation (ODE) is a differential equation that involves a function and its first derivative. It can be expressed in the form: \[-rac{d y}{d x} + P(x) y = Q(x)\]where:

• \(\frac{d y}{d x}\) is the derivative of \(y\) with respect to \(x\).
• \(P(x)\) and \(Q(x)\) are functions of \(x\).
• \(y\) is the dependent variable.

This structure allows us to identify the linear nature of the ODE, meaning that \(y\) and its derivative are both raised to the power of one and are not multiplied together or with other non-linear terms.
In our example, the given equation \(\frac{d u}{d x}=2u+6\) can be rearranged to match the standard form by writing it as \(\frac{d u}{d x} - 2u = 6\).
Notice how the equation is linear with \(u\) and its derivative appearing only to the first power.

Integrating Factor Method

The Integrating Factor Method is a common technique used to solve first-order linear ODEs. The key idea is to multiply the entire differential equation by a specially chosen function, called the integrating factor, which simplifies the equation so that it can be easily integrated.
To apply this method, follow these steps:

• Rewrite the ODE in the standard form \(\frac{d y}{d x} + P(x) y = Q(x)\).
• The integrating factor (IF) is given by \(e^{\int P(x) \, dx}\).
• Multiply the entire ODE by this integrating factor.
• Notice that the left-hand side becomes the derivative of a product, \(\frac{d}{d x} [y(x) \cdot \, \text{IF}]\).
• Integrate both sides with respect to \(x\).

In the provided example, after finding the integrating factor \(e^{-2x}\), multiplying through simplifies the equation to a format that can be directly integrated. The process allows us to solve for the dependent variable \(u(x)\) efficiently.

Initial Value Problem

An initial value problem (IVP) is a type of differential equation problem where the solution is required to satisfy certain given values at a specified point, known as initial conditions.
Typically, an IVP takes the form of a differential equation along with a condition like \(y(x_0) = y_0\), where:

• \(x_0\) is the specific value of \(x\) where the initial condition is applied.
• \(y_0\) is the given value of \(y\) at \(x_0\).

In solving such problems, these initial conditions help determine the constants of integration that arise when solving the differential equation.
For our problem, the ODE \(\frac{d u}{d x}=2u+6\) comes with the initial condition \(u(1)=6\). This condition is crucial as it allows us to solve for the constant \(C_1\) in the solution \(u(x) = -3 + C_1 e^{2x}\), leading to a final specific solution that satisfies both the differential equation and the initial state of the system.