Question

Solve the following initial value problems. $$u^{\prime}(x)=2 u+6, u(1)=6$$

Short answer

Question: Determine the explicit solution for the function \(u(x)\) that satisfies the given initial value problem: \(u^{\prime}(x)=2 u(x) + 6,\) with the initial condition \(u(1)=6\). Answer: The explicit solution for \(u(x)\) is \(u(x) = 3 + \frac{3(e^2-1)}{e^{2x}}.\)

Step-by-step solution

Identify the differential equation and initial condition

The given differential equation is: $$u^{\prime}(x)=2 u(x) + 6,$$ and the initial condition is: $$u(1)=6.$$

Use the integration factor method to solve the ODE

First, we determine the integrating factor: $$\mu(x) = e^{\int 2 \mathrm{d}x} = e^{2x}.$$ Multiply both sides of the ODE by the integrating factor: $$e^{2x}u^{\prime}(x) = 2e^{2x}u(x) + 6e^{2x}.$$ Now, we notice that the left side of the above equation is the derivative of the product \(e^{2x}u(x)\). So, rewrite the equation as: $$(e^{2x}u(x))^{\prime} = 6e^{2x}.$$

Integrate both sides of the equation

Integrate both sides of the equation with respect to x: $$\int(e^{2x}u(x))^{\prime} \mathrm{d}x = \int 6e^{2x} \mathrm{d}x.$$ This results in: $$e^{2x}u(x) = 3e^{2x} + C,$$ where C is the constant of integration.

Solve for u(x)

Divide both sides of the equation by \(e^{2x}\) to find \(u(x)\): $$u(x) = 3 + \frac{C}{e^{2x}}.$$

Use the initial condition to find the constant of integration

Apply the initial condition \(u(1) = 6\) to determine C: $$6 = 3 + \frac{C}{e^{2(1)}}.$$ Solve for C: $$C = 3(e^2 - 1).$$

Write the final solution for u(x)

Substitute the value of C back into the expression for \(u(x)\): $$u(x) = 3 + \frac{3(e^2-1)}{e^{2x}}.$$ This is the explicit solution for the function \(u(x)\) that satisfies the given ODE and initial condition.