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Chapter 8: Problem 5

Find the general solution of each of the following differential equations. \(y^{\prime} \cos x+y=\cos ^{2} x\)

Short Answer

Expert verified
The general solution is \( y = 1 + C e^{ -\sin x} \).

Step by step solution

01

Rewrite the equation

Start by rewriting the given differential equation. The equation is given as: \( y' \cos x + y = \cos^2 x \).
02

Identify the integrating factor

To solve this linear first-order differential equation, identify the integrating factor. The standard form for a first-order linear differential equation is \( y' + P(x)y = Q(x) \). For our equation, \( P(x) = \cos x \). The integrating factor, \( \mu(x) \), is given by:\[ \mu(x) = e^{\int P(x) \, dx } = e^{\int \cos x \, dx} \]
03

Compute the integrating factor

Integrate \( \cos x \) with respect to \( x \):\[ \int \cos x \, dx = \sin x \]Thus, the integrating factor is:\[ \mu(x) = e^{\sin x} \]
04

Multiply through by the integrating factor

Multiply both sides of the original differential equation by the integrating factor \( e^{\sin x} \):\( e^{\sin x} y' \cos x + e^{\sin x} y = e^{\sin x} \cos^2 x \)
05

Simplify the left-hand side

Recognize that the left-hand side is the derivative of \( y e^{\sin x} \):\[ (y e^{\sin x})' = e^{\sin x} \cos^2 x \]
06

Integrate both sides

Integrate both sides with respect to \( x \):\[ \int (y e^{\sin x})' \, dx = \int e^{\sin x} \cos^2 x \, dx \]The left-hand side integrates to:\[ y e^{\sin x} \]The integral on the right-hand side can be simplified using the substitution \( u = \sin x \), \( du = \cos x \, dx \):\[ \int e^u \cos x \, dx = \int e^u \, du = e^u + C = e^{\sin x} + C \]
07

Solve for y

Finally, solve for \( y \):\[ y e^{\sin x} = e^{\sin x} + C \]Divide by \( e^{\sin x} \) to isolate \( y \):\[ y = 1 + C e^{ -\sin x} \]

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Key Concepts

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

first-order linear differential equation
A differential equation relates a function to its derivatives. A first-order linear differential equation is a specific type, where the highest order derivative is the first derivative. The general form is:
\( y' + P(x)y = Q(x) \).
Here, \(y'\) is the first derivative and \(P(x)\) and \(Q(x)\) are functions of \(x\).
In our problem, the given equation is: \( y' \, \cos x + y = \cos^2 x \). By dividing the entire equation by \( \cos x \), we can rewrite it in the standard form:
\( y' + \frac{1}{ \cos x } y = \cos x \). Notice how it fits the general form: \( y' + P(x) y = Q(x) \) with \( P(x) = \cos x \) and \( Q(x) = \cos^2 x \).
integrating factor
To solve first-order linear differential equations, we use the method of integrating factors. This technique makes it easier to integrate the equation and find a solution. The integrating factor, \( \mu(x) \), is calculated using:
\( \mu(x) = e^{ \int P(x) \, dx } \). For our problem, \( P(x) = \cos x \). Let's find the integrating factor step by step:
1. Compute the integral of \( P(x) \):
\( \int \cos x \, dx = \sin x \). 2. Exponentiate the result to find the integrating factor:
\( \mu(x) = e^{ \sin x } \). Now, we multiply the original differential equation by this integrating factor to simplify it and make the left-hand side a perfect derivative. This allows us to integrate both sides of the equation more easily.
solving differential equations
After finding the integrating factor \( \mu(x) = e^{ \sin x } \), we proceed to solve the differential equation. Multiplying both sides by the integrating factor gives us:
\( e^{ \sin x } y' \, \cos x + e^{ \sin x } y = e^{ \sin x } \cos^2 x \). We recognize that the left-hand side is the derivative of \( y e^{ \sin x } \):
\[ ( y e^{ \sin x })' = e^{ \sin x } \cos^2 x \]. To solve, integrate both sides with respect to \( x \):
\[ \int ( y e^{ \sin x })' \, dx = \int e^{ \sin x } \cos^2 x \, dx \]. Using the substitution \( u = \sin x \), \( du = \cos x \, dx \), the right-hand side integrates to:
\[ \int e^u \, du = e^u + C = e^{ \sin x } + C \]. Thus, we get:
\[ y e^{ \sin x } = e^{ \sin x } + C \]. Finally, solve for \( y \) by dividing both sides by \( e^{ \sin x } \):
\[ y = 1 + C e^{- \sin x } \]. This is the general solution of the differential equation.

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Most popular questions from this chapter

(a) Show that $$ \begin{aligned} D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\ D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y \end{aligned} $$ and so on, that is, for any positive integral \(n_{y}\) $$ D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y $$ Thus show that if \(L(D)\) is any polynomial in the operator \(D\), then $$L(D)\left(e^{a x} y\right)=e^{\alpha x} L(D+a) y$$ This is called the exponential shift.

Consider an equation for damped forced vibrations (mechanical or electrical) in which the right-hand side is a sum of several forces or emfs of different frequencies. For example, in (6.32) let the right-hand side be $$ F_{1} e^{i \omega_{1}^{\prime} t}+F_{2} e^{\operatorname{ior}_{2} l}+F_{3} e^{i \operatorname{cog} T} $$ Write the solution by the principle of superposition. Suppose, for given \(\omega_{1}^{\prime}, \omega_{2}^{\prime}, \omega_{3}^{\prime}\), that we adjust the system so that \(\omega=\omega_{1}^{\prime}\); show that the principal term in the solution is then the first one. Thus the system acts as a "filter" to select vibrations of one frequency from a given set (for example, a radio tuned to one station selects principally the vibrations of the frequency of that station).

(a) Consider a light beam traveling downward into the ocean. As the beam progresses, it is partially absorbed and its intensity decreases. The rate at whieh the intensity is decreasing with depth at any point is proportional to the intensity at that depth. The proportionality constant \(\mu\) is called the linear absorption coefficient. Show that if the intensity at the surface is \(I_{0}\), the intensity at a distance \(s\) below the surface is \(l=\) \(I_{0} e^{-\mu s} .\) The linear absorption coefficient for water is of the order of \(10^{-2} \mathrm{ft}^{-1}\) (the exact value depending on the wavelength of the light and the impurities in the water). For this value of \(\mu\), find the intensity as a fraction of the surface intensity at a depth of \(1 \mathrm{ft}, 50 \mathrm{ft}, 500 \mathrm{ft}, 1\) mile. When the intensity of a light beam has been reduced to half. its surface intensity \(\left(I=\frac{1}{2} I_{0}\right)\), the distance the light has penetrated into the absorbing substance is called the half-value thickness of the substance. Find the half-value thickness in terms of \(\mu\). Find the half-value thickness for water for the value of \(\mu\) given above. (b) Note that the differential equation and its solution in this problem are mathematically the same as those in Example I, although the physical problem and the terminology are different. In discussing radioactive decay, we call \(\lambda\) the decay constant, and we define the half-life \(T\) of a radioactive substance as the time when \(N=\frac{1}{2} N_{0}\) (compare half-value thickness). Find the relation between \(\lambda\) and \(T\).

Solve the following differential equations. \(y^{\prime \prime}+9 y=0\)

By separation of variables, solve the differential equation \(d y / d x=\sqrt{1-y^{2}}\) to obtain solution containing one arbitrary constant. Although this solution may be referred to as the " general solution," show that \(y=1\) is a solution of the differential equation not obtainablc from the "general solution" by any choice of the arbitrary constant. The solution \(y=1\) is called a singular selution; \(y=-1\) is another singular solution. Sketch a number of graphs of the "general solution" for different values of the arbitrary constant and observe that \(y=1\) is tangent to all of them. This is characteristic of a singular solution -its graph is tangent at each point to one of the graphs of the "general solution." Note that the given differential equation is not linear; for linear equations, all solutions are contained in the general solution, but nonlinear equations may have singular solutions which cannot be obtained from the "general solution" by specializing the arbitrary constant (or constants). Thus a nonlinear first-order equation in \(x\) and \(y\) may have two (or more) solutions passing through a given point in the \((x, y)\) plane, whereas a linear first-order equation always has just one such solution. Show that any continuous curve made up of pieces of \(y=1, y=-1\), and the sinc curves of the "general solution," gives a solution of the above differential equation. Sketch such a solution curve on your graphs.
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