/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 25 Solve the given differential equ... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the given differential equation by using an appropriate substitution. $$ \frac{d y}{d x}=\tan ^{2}(x+y) $$

Short Answer

Expert verified
The solution to \( \frac{dy}{dx} = \tan^2(x+y) \) is implicit: \( x + y = \tan(x + y) + C \).

Step by step solution

01

Recognize the Need for Appropriate Substitution

The given differential equation is \( \frac{dy}{dx} = \tan^2(x+y) \). Notice that the expression on the right-hand side is \( \tan^2(x+y) \), a composite angle, suggesting a substitution where \( x+y \) is a single variable.
02

Introduce the Substitution

Let's set \( u = x + y \). This substitution helps simplify the differential equation by unraveling the composite argument into a single variable form.
03

Differentiate the Substitution

By differentiating \( u = x + y \) with respect to \( x \), we get \( \frac{du}{dx} = 1 + \frac{dy}{dx} \). This allows us to express \( \frac{dy}{dx} \) in terms of \( \frac{du}{dx} \).
04

Substitute into the Differential Equation

Substitute \( u = x + y \) and its derivative into the original differential equation. The equation \( \frac{dy}{dx} = \tan^2(u) \) turns into \( \frac{du}{dx} - 1 = \tan^2(u) \). Simplifying gives \( \frac{du}{dx} = 1 + \tan^2(u) \).
05

Recognize a Trigonometric Identity

Recall the trigonometric identity: \( 1 + \tan^2(u) = \sec^2(u) \). Thus, the differential equation becomes \( \frac{du}{dx} = \sec^2(u) \).
06

Solve the Simplified Differential Equation

This differential equation \( \frac{du}{dx} = \sec^2(u) \) is separable. It can be rewritten as \( du = \sec^2(u) \, dx \). Integrate both sides, which gives \( \int du = \int \sec^2(u) \, dx \).
07

Perform the Integration

Integrate \( \int du \) to get \( u \), and \( \int \sec^2(u) \, dx \) to get \( \tan(u) + C \). So the equation becomes \( u = \tan(u) + C \). However, since \( u = x + y \), substitute back to get \( x + y = \tan(x + y) + C \), leading to the relation for \( y \).
08

Solve for the Function y(x)

Since the equation \( x + y = \tan(x + y) + C \) is implicit, it describes a relationship between \( x \), \( y \), and a constant \( C \), defining \( y(x) \) implicitly.

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Key Concepts

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

Substitution Method
The substitution method is a powerful tool used in solving differential equations, making it simpler by transforming complex equations into more manageable forms. In this particular problem, the need for substitution arises when we notice the composite argument \(x+y\) within the tangent squared term, \(\tan^2(x+y)\).

By introducing a substitution such as \(u = x + y\), we convert the variable expression into a single variable form. This substitution makes the equation easier to manipulate and solve. Differentiating \(u = x + y\) with respect to \(x\) gives us \(\frac{du}{dx} = 1 + \frac{dy}{dx}\), providing an equation that relates \(\frac{dy}{dx}\) to \(\frac{du}{dx}\).

After substitution, the original equation \(\frac{dy}{dx} = \tan^2(u)\) gets modified to \(\frac{du}{dx} - 1 = \tan^2(u)\), which simplifies further into \(\frac{du}{dx} = 1 + \tan^2(u)\). This demonstrates how a clever substitution can transform complex expressions, revealing a simpler path to the solution.
Trigonometric Identities
Trigonometric identities are fundamental relationships involving trigonometric functions that allow for the simplification of expressions. In this exercise, the identity \(1 + \tan^2(u) = \sec^2(u)\) plays a crucial role.

This identity is derived from the Pythagorean identities and enables us to rewrite expressions involving \(\tan^2(u)\) in a simplified form. It essentially tells us that the sum of 1 and the square of the tangent is equivalent to the square of the secant function.

By applying this identity, the equation \(\frac{du}{dx} = 1 + \tan^2(u)\) is transformed into \(\frac{du}{dx} = \sec^2(u)\). Trigonometric identities like this one are key to breaking down and solving complex trigonometric equations. Mastering them provides an invaluable shortcut in a variety of mathematical problems, especially those involving trigonometric expressions.
Separable Differential Equations
Separable differential equations are a type of differential equations that can be written in a form where all terms involving one variable can be separated from all terms involving another variable. This makes them relatively straightforward to integrate.

In the problem at hand, once the substitution and trigonometric identity have been applied, the equation becomes \( \frac{du}{dx} = \sec^2(u)\). This equation is naturally separable. We can re-arrange it to get \( du = \sec^2(u) \, dx\).

The separable form allows us to integrate both sides independently. Integrating the left side \( \int du \) gives us \( u \), and the integral of \( \int \sec^2(u) \, dx \) results in \( \tan(u) + C \). The result is a relationship \( u = \tan(u) + C \) which, through substitution back, describes a relationship between the original variables \(x\) and \(y\).

This method highlights the power and simplicity of separable differential equations, offering a structured approach to solving what might initially seem like complex differential problems.

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Most popular questions from this chapter

(a) Find an implicit solution of the IVP $$ (2 y+2) d y-\left(4 x^{3}+6 x\right) d x=0, \quad y(0)=-3 $$ (b) Use part (a) to find an explicit solution \(y=\phi(x)\) of the IVP. (c) Consider your answer to part (b) as a function only. Use a graphing utility or a CAS to graph this function, and then use the graph to estimate its domain. (d) With the aid of a root-finding application of a CAS, determine the approximate largest interval \(I\) of definition of the solution \(y=\phi(x)\) in part (b). Use a graphing utility or a CAS to graph the solution curve for the IVP on this interval.

(a) The solution of the differential equation $$ \frac{2 x y}{\left(x^{2}+y^{2}\right)^{2}} d x+\left[1+\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}\right] d y=0 $$ is a family of curves that can be interpreted as streamlines of a fluid flow around a circular object whose boundary is described by the equation \(x^{2}+y^{2}=1\). Solve this \(\mathrm{DE}\) and note the solution \(f(x, y)=c\) for \(c=0\). (b) Use a CAS to plot the streamlines for \(c=0, \pm 0.2, \pm 0.4\) \(\pm 0.6\), and \(\pm 0.8\) in three different ways. First, use the contourplot of a CAS. Second, solve for \(x\) in terms of the variable \(y .\) Plot the resulting two functions of \(y\) for the given values of \(c\), and then combine the graphs. Third, use the CAS to solve a cubic equation for \(y\) in terms of \(x\).

A dead body was found within a closed room of a bouse where the temperature was a constant \(70^{\circ} \mathrm{F}\). At the time of discovery, the core temperature of the body was determined to be \(85^{\circ} \mathrm{F}\). One hour later a second measurement showed that the core temperature of the body was \(80^{\circ} \mathrm{F}\). Assume that the time of death corresponds to \(t=0\) and that the core temperature at that time was \(98.6^{\circ} \mathrm{F}\). Determine how many hours elapsed before the body was found.

A 30-volt electromotive force is applied to an \(L R\) -series circuit in which the inductance is \(0.1\) henry and the resistance is 50 ohms. Find the current \(i(t)\) if \(i(0)=0\). Determine the current as \(t \rightarrow \infty\).

Use a numerical solver and Euler'smethodto approximate \(y(1.0)\) where \(y(x)\) is the solution to \(y^{\prime}=2 x y^{2}, y(0)=1\). First use \(h=0.1\) and then \(h=0.05\). Repeat using the \(\mathrm{RK} 4\) method. Discuss what might cause the approximations of \(y(1.0)\) to differ so greatly.

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