Question

Solve the initial-value problems. $$ \frac{d x}{d t}-x=\sin 2 t, \quad x(0)=0 $$

Short answer

The short answer is: \[ x(t) = -\sin 2t + 2\cos 2t - 2e^{t} \]

Step-by-step solution

Identify the problem type

The given differential equation is: \[ \frac{d x}{d t} - x = \sin 2t \] This equation can be rewritten as: \[ \frac{d x}{d t} - x - \sin 2t = 0 \] It is a first-order linear differential equation.

Find the integrating factor

We will use the integrating factor method to solve this differential equation. The integrating factor is given by: \[ \mu(t) = e^{\int P(t) dt}, \] where \(P(t)\) is the coefficient of the variable x(t) in the differential equation. In this case, \(P(t) = -1\). So, the integrating factor is: \[ \mu(t) = e^{\int -1 dt} = e^{-t}. \]

Multiply the entire equation by the integrating factor

Now, multiply the entire differential equation by the integrating factor: \[ e^{-t}\frac{d x}{dt} - e^{-t}x = e^{-t}\sin 2t. \] Next, observe that the left side of the equation is the derivative of the product of x(t) and the integrating factor: \[ \frac{d}{dt}(x e^{-t}) = e^{-t}\sin 2t. \]

Integrate both sides of the equation

Now, integrate both sides of the equation with respect to t: \[ \int \frac{d}{dt}(x e^{-t}) dt = \int e^{-t}\sin 2t dt. \] Integrate by parts on the right side of the equation: Let \(u = \sin 2t\) and \(dv = e^{-t} dt\). Then, \(du = 2 \cos 2t dt\) and \(v = -e^{-t}\). Applying integration by parts, we have: \[ \int u dv = uv - \int v du \] \[ xe^{-t} = [-e^{-t}\sin 2t] - \int (-e^{-t}) 2\cos 2t dt. \] Next, integrate by parts again for the remaining integral: Let \(U = \cos 2t\) and \(dV = e^{-t}dt\). Then, \(dU = -2\sin 2t dt\) and \(V = -e^{-t}\). Applying integration by parts again, we have: \[ -2\int U dV = -2[UV - \int V dU] \] \[ -2\int \cos 2t(-e^{-t}) dt = -2[-e^{-t}\cos 2t - \int (-e^{-t})(-2\sin 2t) dt] \] Thus, we have: \[ xe^{-t} = -e^{-t}\sin 2t + 2e^{-t}\cos 2t + 4 \int e^{-t}\sin 2t dt. \] Finally, we have: \[ xe^{-t} = -e^{-t}\sin 2t + 2e^{-t}\cos 2t + C, \] where C is a constant.

Apply the initial condition

Now, we'll use the initial condition, \(x(0) = 0\), to find the value of C. \[ x(0)e^{-0} = -e^{-0}\sin 0 + 2e^{-0}\cos 0 + C \] So, we have: \[ 0 = 0 + 2(1) + C \implies C = -2. \]

Solve for the dependent variable

Now that we have the value of C, we can solve for x(t): \[ xe^{-t} = -e^{-t}\sin 2t + 2e^{-t}\cos 2t - 2 \] Multiplying both sides by \(e^t\), we have: \[ x(t) = -\sin 2t + 2\cos 2t - 2e^{t}. \] Therefore, the solution to the initial-value problem is: \[ x(t) = -\sin 2t + 2\cos 2t - 2e^{t} \]

Key concepts

Integrating Factor Method

The integrating factor method is a powerful technique used to solve first-order linear differential equations of the form \( \frac{dy}{dt} + P(t)y = Q(t) \). It simplifies the equation into something easier to solve by making the left-hand side a derivative of a product. The key to the method is to multiply the entire equation by an integrating factor \( \mu(t) \), which is determined by the formula \( \mu(t) = e^{\int P(t) \, dt} \).

Here's how it works step-by-step:

• Calculate the integrating factor \( \mu(t) \) using the function \( P(t) \).
• Multiply every term of the differential equation by \( \mu(t) \).
• The left side of the equation becomes a single derivative, typically \( \frac{d}{dt}[\mu(t)y] \).
• Integrate both sides to solve for the function \( y(t) \).

In our original problem, this method transformed the difficult-to-solve equation into \( e^{-t}\frac{d x}{dt} - e^{-t}x = e^{-t}\sin 2t \), allowing for a straightforward integration and eventual solution.

Initial-Value Problems

Initial-value problems are a fundamental part of differential equations where in addition to the differential equation, an initial condition is given. This initial condition specifies the value of the function (often at time \( t=0 \)) and enables finding a unique solution that fits both the differential equation and the specified value.

For the problem \( \frac{d x}{d t} - x = \sin 2 t \) with \( x(0)=0 \), it was crucial to apply this initial condition after solving the differential equation. By substituting \( t = 0 \) into the general solution and ensuring it equals zero, the constant of integration, \( C \), was determined. This lead directly to the specific solution \( x(t) \) by eliminating other possibilities and ensuring it fits perfectly from the beginning point described by the initial condition.

Integration by Parts

Integration by parts is a technique in calculus used for integrating the product of two functions. This method comes in handy when integrating complex products where simpler methods do not suffice. It is based on the product rule for differentiation and is essential for solving certain integrals that appear in differential equations.

The formula is:
\[ \int u \cdot dv = uv - \int v \cdot du \]
In our solution, integration by parts was applied twice to handle the terms involving trigonometric and exponential functions.

• First, \( u = \sin 2t \) and \( dv = e^{-t} dt \) were chosen for the initial integration.
• Then, some subsequent terms required another round of integration by parts, using \( U = \cos 2t \) and \( dV = e^{-t}dt \).

This helped efficiently manage the integration process, turning intricate integrals into solvable chunks.

Differential Equations Solutions

Solving differential equations involves finding a function or set of functions that satisfy the given equation. In our case, the solution process transforms the expression into something that reveals how variables change over time or space as constrained by the equation.

For our specific problem, by following the solution steps:

• The integrating factor method consolidated the differential equation into a manageable form.
• Initial conditions helped specify the solution precisely, eliminating other possible functions fitting the general solution form.
• Integration by parts tackled complex integrals within the equation.

Thus, combining these techniques yielded the solution \( x(t) = -\sin 2t + 2\cos 2t - 2e^{t} \) which satisfies both the equation and initial conditions, illustrating how differential equations model dynamic systems in a wide array of fields.