Question

\(y^{\prime}=(1+y) \sin (t)\), with \(y(\pi)=0\), on \([0,2 \pi]\).

Short answer

The solution is \( y(t) = e^{-\cos(t)} - 1 \).

Step-by-step solution

Analyze the Differential Equation

The given differential equation is \( y' = (1+y) \sin(t) \). This is a first-order non-linear ordinary differential equation because the term \( (1+y) \) makes it non-linear.

Separate Variables

To solve the equation, we need to separate variables. We can rewrite the equation as \( \frac{dy}{1+y} = \sin(t) \, dt \). This allows us to integrate both sides independently to find \( y(t) \).

Integrate Both Sides

Integrate the left side with respect to \( y \) and the right side with respect to \( t \):\[ \int \frac{dy}{1+y} = \int \sin(t) \, dt \]The left side integrates to \( \ln|1+y| \) and the right side to \( -\cos(t) + C \), where \( C \) is the integration constant.

Solve the Integral Equation

After integration, we have:\[ \ln|1+y| = -\cos(t) + C \]Exponentiate both sides to solve for \( 1+y \):\[ |1+y| = e^{-\cos(t) + C} \]

Use Initial Condition

Apply the initial condition \( y(\pi) = 0 \) to determine the constant \( C \). At \( t = \pi \), \( -\cos(\pi) = 1 \), so:\[ \ln|1+0| = 1 + C \]Solving yields \( C = 0 \). Thus, \(|1+y| = e^{-\cos(t)} \).

Resolve Absolute Value

Since \( y(\pi) = 0 \), it indicates that \( 1+y \) is positive at \( t=\pi \). Therefore, we can omit the absolute value, giving:\[ 1+y = e^{-\cos(t)} \]

Solve for \( y \)

Rearrange \( 1+y = e^{-\cos(t)} \) to solve for \( y \):\[ y = e^{-\cos(t)} - 1 \]

Verify and Conclude

Verify that \( y(t) = e^{-\cos(t)} - 1 \) satisfies both the differential equation and the initial condition. Substitute this form back into the original equation and confirm it holds, finally ensuring the initial condition is met at \( t=\pi \).

Key concepts

First-Order Non-Linear Equation

In the realm of differential equations, first-order equations involve derivatives with respect to one independent variable and are characterized by the highest derivative being the first. The equation given in the exercise, \( y' = (1+y) \sin(t) \), represents a non-linear form due to the presence of the term \((1+y)\). Such terms prevent the equation from being a straightforward linear equation.
Non-linear equations can depict complex real-world phenomena much better than linear equations, allowing the representation of more intricate relationships.
Understanding the nature of non-linear equations is critical as they often require unique solution strategies, different from their linear counterparts.

Separation of Variables

Separation of variables is a powerful method used to solve differential equations by rearranging them so that each variable appears on one side of the equation. In our exercise, the equation \( y' = (1+y) \sin(t) \) was rewritten as \( \frac{dy}{1+y} = \sin(t) \, dt \). This allows us to treat \( y \) and \( t \) independently.
This technique is particularly useful for equations where variables are multiplicatively separable, enabling integration on both sides independently. The goal is to eventually express the solution in terms of a simple equation, yielding functions of one variable on each side.

• It simplifies the problem, allowing for straightforward integration.
• Makes it possible to solve differential equations that might otherwise seem challenging.

Initial Condition

An initial condition allows us to find a particular solution of a differential equation from a family of potential solutions. It is essential because it provides a specific value that satisfies the equation at a certain point.
In the exercise, the initial condition \( y(\pi) = 0 \) was used to determine the integration constant \( C \). After solving the integration part, the initial condition helps fine-tune the general solution to fit exactly the conditions specified by the problem.
Applying initial conditions is a critical step in the process of solving differential equations as it transforms a general solution into a specific one that meets the problem's criteria.

Exponential Function Integration

The integration process in our exercise leads us to exponential functions when solving the differential equation. Once we have separated variables and integrated both sides, we reach a natural logarithm form: \[ \ln|1+y| = -\cos(t) + C \]Exponentiating both sides helps remove the logarithm, giving \[ |1+y| = e^{-\cos(t) + C} \]
This step is key because exponential functions often appear as solutions for differential equations involving separation of variables, particularly when a logarithm is involved.

• Exponential integration helps encapsulate the behavior of the solution over a given interval or domain.
• Provides insight into the growth and decay behaviors captured by the solution.

Understanding exponential function integration is crucial for solving many complex equations and accurately interpreting their solutions.