Problem 12

Question

From $2 y d y=-\frac{\sin 3 x}{\cos ^{3} 3 x} d x$ or $2 y d y=-\tan 3 x \sec ^{2} 3 x d x$ we obtain $y^{2}=-\frac{1}{6} \sec ^{2} 3 x+c$.

Step-by-Step Solution

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Answer

The solution is $y^2 = -\frac{1}{6}\sec^2 3x + C$.

1Step 1: Identify the Type of Differential Equation

The given differential equation is separable, meaning we can separate the variables on each side of the equation to solve for the general solution.

2Step 2: Separate the Variables

We begin by rewriting the equation to separate variables. The equation is given as $2ydy = -\tan 3x \sec^2 3x \, dx$. We can then write it as $2y \, dy = -\frac{\sin 3x}{\cos^3 3x} \, dx$.

3Step 3: Integrate Both Sides

Integrate both sides separately: on the left, $\int 2y \, dy$ and on the right, $-\int \frac{\sin 3x}{\cos^3 3x} \, dx$.

4Step 4: Perform the Integrations

For part $\int 2y \, dy$, it results in $ y^2$.For $-\int \frac{\sin 3x}{\cos^3 3x} \, dx$, perform a substitution $u = \cos 3x$, so that $du = -3 \sin 3x \, dx$. The integral can be written as $-\frac{1}{3} \int u^{-3} \, du$, which is solved as $-\frac{1}{6} u^{-2} = -\frac{1}{6} \sec^2 3x$.

5Step 5: Combine the Results

Combine the integration results to get the general solution: $y^2 = -\frac{1}{6}\sec^2 3x + C$, where $C$ is the constant of integration.

Key Concepts

Integration TechniquesVariable SeparationTrigonometric SubstitutionConstant of Integration

Integration Techniques

Working with differential equations often requires a solid understanding of various integration techniques. When confronted with an equation like $ 2y \, dy = -\tan 3x \sec^2 3x \, dx $, integrating both sides to find a solution is key. Here, we use different integration approaches for each side:

**Left Side**: $ \int 2y \, dy $ is a straightforward integral. Recognizing it's of the form $ \int a\,f'(y) \, dy $, we use the basic power rule $ \int y^n \, dy = \frac{y^{n+1}}{n+1} $. This results in $ y^2 $.
**Right Side**: For $ -\int \frac{\sin 3x}{\cos^3 3x} \, dx $, things get trickier. This requires substitution and manipulation of trigonometric identities, which we'll discuss in detail in a later section.

These techniques reflect essential skills when solving separable differential equations. Integrating each side carefully lets us obtain a general solution that includes a constant of integration.

Variable Separation

A separable differential equation is a common type that allows you to isolate the variables on opposite sides of the equation. The goal is to split the equation into two separate integrals. Let's break down the steps for separating variables in our example:

Start with the given differential equation. Initially, we have: $ 2y \, dy = -\tan 3x \sec^2 3x \, dx $.
Notice that the terms involving $ y $ and $ x $ can be separated onto different sides. This results in $ 2y \, dy = -\frac{\sin 3x}{\cos^3 3x} \, dx $.

By having each side depend on only one variable, you can then integrate each side individually. Variable separation is powerful because it allows a complex differential equation to become a more straightforward integration problem.

Trigonometric Substitution

In solving the integral $ -\int \frac{\sin 3x}{\cos^3 3x} \, dx $, trigonometric substitution is an advantageous technique. Let's see what steps are involved:

Set $ u = \cos 3x $. This naturally simplifies the expression since $ du = -3\sin 3x \, dx $.
The integral changes to $ -\frac{1}{3} \int u^{-3} \, du $. Simplifying trigonometric expressions via substitution often turns complex fractions into polynomial-like integrals.
This integral is straightforward now: apply the power rule for integration, resulting in $ -\frac{1}{6} u^{-2} $.
Replace $ u = \cos 3x $ back into the expression to achieve $ -\frac{1}{6} \sec^2 3x $.

Trigonometric substitution is a handy tool for transforming tricky integrals into manageable forms by using substitutions based on trigonometric identities.

Constant of Integration

When integrating both sides of a differential equation, a constant of integration arises from indefinite integrals. These constants ensure the most general solution possible:

After integrating, you get expressions that might be identical yet differ by a constant. This constant represents all possible vertical shifts of the graph of an antiderivative.
In our exercise, after finding $ y^2 = -\frac{1}{6} \sec^2 3x + C $, the $ C $ captures any initial conditions or boundaries that define a particular solution to the equation.
This constant ensures that all potential solutions of the differential equation are considered.

In solving differential equations, including this constant is critical until specific conditions allow you to solve for its exact value, thus offering a complete picture of all solution families.

Problem 11

Problem 12

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Problem 11

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Problem 12

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Problem 12

Let $M=3 x^{2} y+e^{y}$ and $N=x^{3}+x e^{y}-2 y$ so that $M_{y}=3 x^{2}+e^{y}=N_{x} .$ From $f_{x}=3 x^{2} y+e^{y}$ we obtain \(f=x^{3} y+x e^{y}+h(y),

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