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Chapter 10: Problem 24

Solve the following differential equations with the given initial conditions. $$y^{\prime}=-y^{2} \sin t, y\left(\frac{\pi}{2}\right)=1$$

Short Answer

Expert verified
The solution is \(y = \frac{1}{1 - \cos t}\).

Step by step solution

01

Rewrite the differential equation

Rewrite the differential equation in the form \ \( \frac{dy}{dt} = -y^2 \sin t \). This shows that the independent variable is \(t\) and the dependent variable is \(y\).
02

Separate the variables

Separate the variables so that all terms involving \(y\) and \(dy\) are on one side and all terms involving \(t\) and \(dt\) are on the other side. \ \(\frac{dy}{y^2} = -\sin t \, dt\).
03

Integrate both sides

Integrate both sides of the equation. \ \( \int \frac{1}{y^2} \, dy = \int -\sin t \, dt \). \ The left integral becomes \ \[-\frac{1}{y}\], and the right integral becomes \ \[\cos t\]. Thus, \ \(-\frac{1}{y} = \cos t + C\), where \(C\) is the constant of integration.
04

Solve for the constant \(C\)

Use the initial condition \( y\left(\frac{\pi}{2}\right) = 1 \) to solve for \(C\). \ Substitute \( t = \frac{\pi}{2} \) and \( y = 1 \) into the equation \ \(-\frac{1}{1} = \cos \left( \frac{\pi}{2} \right) + C\). \ This simplifies to \ \(-1 = 0 + C \). Thus, \( C = -1 \).
05

Write the particular solution

Substitute \( C = -1 \) back into the equation to find the particular solution. \ \(-\frac{1}{y} = \cos t - 1 \). \ Solve for \( y \), giving the final solution \ y = \frac{1}{1 - \cos t}.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Variable Separation
Variable separation is a fundamental method for solving simple differential equations. The goal is to isolate the variables on different sides of the equation to make it easier to integrate. Start by identifying the dependent and independent variables. In our case, the dependent variable is \( y \) and the independent variable is \( t \).
To separate variables, rewrite the differential equation so terms involving \( y \) and \( dy \) are on one side, while terms involving \( t \) and \( dt \) are on the other side.
From \( \frac{dy}{dt} = -y^2 \sin t \), we rearrange to \( \frac{dy}{y^2} = -\sin t \, dt \). This separation sets the stage for integration.
Integration
Integration is a key step in solving differential equations after separating variables. By integrating both sides, we find the integral of each side which leads us to a function. In our example, integrating \( \frac{dy}{y^2} \) and \( -\sin t \, dt \) transforms the equation.
For the left side, \( \frac{dy}{y^2} \) integrates to \(-\frac{1}{y} \). For the right side, the integral of \( -\sin t \) is \( -\cos t \). Thus, the equation after integration is \(-\frac{1}{y} = \cos t + C \), where \( C \) is the constant of integration.
Remember, \(C\) represents an unknown constant derived from indefinite integration.
Initial Conditions
Initial conditions are specific values that allow us to solve for the constant of integration \( C \). They provide a real-world context to the differential equation, making the solution unique.
In our problem, the initial condition is given by \( y(\frac{\pi}{2}) = 1 \). To use this condition, plug \( t = \frac{\pi}{2} \) and \( y = 1 \) into the integrated equation. We get:
\(-\frac{1}{1} = \cos(\frac{\pi}{2})+C \). Since \( \cos(\frac{\pi}{2}) = 0 \), the equation simplifies to \( -1 = 0 + C \). Thus, \( C = -1 \). Initial conditions ensure that the differential equation aligns with specific values at specific points.
Constant of Integration
The constant of integration, \( C \), appears whenever we perform indefinite integration. It represents an arbitrary constant that accounts for the family of solutions to the differential equation.
To find a particular solution, we use initial conditions to solve for \( C \). As we discovered with \( C = -1 \) in our problem, this constant personalizes the general solution.
Substitute \( C \) back into the equation to get the specific solution. For us, it is:
\(-\frac{1}{y} = \cos t - 1\).
Finally, solve for \( y \) to get \( y = \frac{1}{1 - \cos t} \). The constant of integration ties together the particular and general forms of the solution.

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Most popular questions from this chapter

Solve the following differential equations with the given initial conditions. $$y^{\prime}=2 t e^{-2 y}-e^{-2 y}, y(0)=3$$

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