Problem 18
Question
Solve the differential equation and then use a graphing utility to generate five integral curves for the equation. $$ (\cos y) y^{\prime}=\cos x $$
Step-by-Step Solution
Verified Answer
The solution is \(\sin y = \sin x + C\), with integral curves for various \(C\) values.
1Step 1: Identify the Type of Differential Equation
The given differential equation is \((\cos y) y' = \cos x\). This is a first-order, separable differential equation because it can be rewritten in the form \(f(y)y' = g(x)\).
2Step 2: Separate Variables
Rearrange the equation to separate the variables. Divide both sides by \(\cos y\) to get: \[ y' = \frac{\cos x}{\cos y} \] This can be written as: \[ \cos y \, dy = \cos x \, dx \] The variables are now separated.
3Step 3: Integrate Both Sides
Integrate both sides of the equation to solve for \(y\). The left side becomes: \[ \int \cos y \, dy = \sin y + C_1 \] The right side becomes: \[ \int \cos x \, dx = \sin x + C_2 \] Combine the integration constants into one constant \(C\): \[ \sin y = \sin x + C \]
4Step 4: Solve for y
Since the equation \(\sin y = \sin x + C\) doesn’t lend itself easily to solving explicitly for \(y\), the solution will be implicit in terms of the relationship between \(x\) and \(y\).
5Step 5: Generate Integral Curves
Use a graphing utility to plot the curves represented by\[ \sin y = \sin x + C \] Select different values for \(C\) (e.g., \(C = -1, -0.5, 0, 0.5, 1\)) to plot five integral curves. Each curve will represent a different solution based on the initial condition.
Key Concepts
Separable Differential EquationsIntegral CurvesFirst-Order Differential EquationsVariable Separation
Separable Differential Equations
Separable differential equations are a special type of differential equations where the variables can be separated onto different sides of the equation. This makes these equations particularly straightforward to solve.
In a separable differential equation, the general form can be represented as:
This structured separation and integration facilitate many real-world calculations very efficiently.
In a separable differential equation, the general form can be represented as:
- \( f(y) y' = g(x) \)
- \( f(y) dy = g(x) dx \)
This structured separation and integration facilitate many real-world calculations very efficiently.
Integral Curves
Integral curves are graphical representations of solutions to differential equations. They can offer visual insights into the behavior of the differential solutions. Each integral curve corresponds to a particular solution that satisfies the given differential equation under specific initial conditions.
In the context of the equation \(\sin y = \sin x + C\), these curves are generated for various constant values \( C \). Each curve illustrates how as \( C \) varies, the solution to the equation behaves differently. You can think of each integral curve as a path that a point following the differential equation might trace out over time.
By using a graphing utility to visualize these curves, it becomes easier to understand complex relationships and dynamics described by differential equations. It's particularly useful because many differential equations can't be easily solved or interpreted without the aid of such visual tools.
In the context of the equation \(\sin y = \sin x + C\), these curves are generated for various constant values \( C \). Each curve illustrates how as \( C \) varies, the solution to the equation behaves differently. You can think of each integral curve as a path that a point following the differential equation might trace out over time.
By using a graphing utility to visualize these curves, it becomes easier to understand complex relationships and dynamics described by differential equations. It's particularly useful because many differential equations can't be easily solved or interpreted without the aid of such visual tools.
First-Order Differential Equations
First-order differential equations involve derivatives of the first degree. They have the general form \( \frac{dy}{dx} = f(x, y) \). These equations are fundamental and often appear in natural and physical sciences because they describe how changes in one variable relate directly to changes in another.
There are different methods to tackle first-order differential equations, but when they are separable (like in our exercise), it allows us to use variable separation to find the solution. First-order differential equations can take in a wide variety of forms, but if you can convert them into a separable form, it simplifies the solution process significantly, allowing you to focus on the pure integration part which you typically already know how to handle.
There are different methods to tackle first-order differential equations, but when they are separable (like in our exercise), it allows us to use variable separation to find the solution. First-order differential equations can take in a wide variety of forms, but if you can convert them into a separable form, it simplifies the solution process significantly, allowing you to focus on the pure integration part which you typically already know how to handle.
Variable Separation
Variable separation is a technique used to solve certain types of differential equations, particularly separable ones. This method involves rearranging the equation in such a manner that all terms involving one variable are on one side and all terms involving the other variable are on the opposite side.
For example, starting with this equation:
For example, starting with this equation:
- \( (\cos y) y' = \cos x \)
- \( \cos y \, dy = \cos x \, dx \)
Other exercises in this chapter
Problem 16
(a) Show that the solution of the initial-value problem \(y^{\prime}=e^{-x^{2}}, y(0)=0\) is $$ y(x)=\int_{0}^{x} e^{-t^{2}} d t $$ (b) Use Euler's Method with
View solution Problem 17
Solve the differential equation and then use a graphing utility to generate five integral curves for the equation. $$ \left(x^{2}+4\right) \frac{d y}{d x}+x y=0
View solution Problem 19
Solve the differential equation. If you have a CAS with implicit plotting capability, use the CAS to generate five integral curves for the equation. $$ y^{\prim
View solution Problem 19
(a) Use Euler's Method with a step size of \(\Delta x=0.2\) to approximate the solution of the initial-value problem $$ y^{\prime}=x+y, \quad y(0)=1 $$ over the
View solution