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Chapter 3: Problem 13

Identifying \(P(x)=-1 / x\) we have $$y_{2}=x \sin (\ln x) \int \frac{e^{-\int-d x / x}}{x^{2} \sin ^{2}(\ln x)} d x=x \sin (\ln x) \int \frac{x}{x^{2} \sin ^{2}(\ln x)} d x$$ $$=x \sin (\ln x) \int \frac{\csc ^{2}(\ln x)}{x} d x=[x \sin (\ln x)][-\cot (\ln x)]=-x \cos (\ln x)$$ A second solution is \(y_{2}=x \cos (\ln x)\)

Short Answer

Expert verified
The second solution is \(y_2 = x \cos(\ln x)\).

Step by step solution

01

Simplify the Problem

We start by understanding the problem, which involves finding a function solution that satisfies the differential equation using the integrating factor method. The given function to identify is \(P(x)=-\frac{1}{x}\). We need to work with the solution \(y_2 = x \sin (\ln x)\).
02

Compute the Integrating Factor

The integrating factor \(I(x)\) is obtained by exponentiating the integral of \(P(x)\). This means we have:\[e^{\int (-1/x) \, dx} = \frac{1}{x}\].
03

Use the Integrating Factor in the Integral

Using the integrating factor, the expression simplifies within the integral:\[\int \frac{e^{-\int (-1/x) \,dx}}{x^2 \sin^2(\ln x)} \, dx = \int \frac{x}{x^2 \sin^2(\ln x)} \, dx\].
04

Simplify the Integral

Simplify the expression: \[\int \frac{x}{x^2 \sin^2(\ln x)} \, dx = \int \frac{1}{x \sin^2(\ln x)} \, dx\]which simplifies to \(\int \csc^2(\ln x) \, dx\).
05

Solve the Simplified Integral

The antiderivative of \(\csc^2(\ln x)\) involves a substitution for trigonometric identities giving us \[-\cot (\ln x)\].
06

Combine the Results

Multiply the solution obtained from the integral by the outside function:\[x \sin(\ln x) [-\cot(\ln x)] = -x \cos(\ln x)\].
07

Verify the Solution

The second solution is confirmed as \(y_2 = x \cos(\ln x)\) through verification of the computation steps.

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Key Concepts

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

Integrating Factor
The concept of an integrating factor is a widely used technique in solving linear differential equations. It primarily helps to transform a non-exact differential equation into an exact one, making it easier to solve.

An integrating factor is derived by recognizing a specific function within the differential equation. In the context of the original exercise, we identify that function as \(P(x) = -\frac{1}{x}\). This function is used to obtain the integrating factor by taking its exponential of the integral:
  • The integral of \(P(x)\) becomes \( \int -\frac{1}{x} \, dx \), which simplifies to \(- \ln|x|\).
  • Exponentiating it gives the integrating factor \(I(x) = e^{-\ln x} = \frac{1}{x}\).
This transformation allows us to rearrange and simplify the given differential equation, making it possible to solve it systematically. By multiplying the entire equation by this integrating factor, we transform it into a form that can be integrated directly, as explored in the next steps.
Trigonometric Substitution
Trigonometric substitution is a technique often used to solve integrals that may seem complex at first glance. In solving differential equations, such substitutions can simplify expressions containing trigonometric functions by using known identities.

In the provided exercise, trigonometric substitution plays a key role in simplifying the integral\[\int \csc^2(\ln x) \, dx.\]The function \(\csc^2(\text{something})\) tends to hint at the derivative of a trigonometric identity. The antiderivative of \(\csc^2(u)\) is \(-\cot(u)\).
  • This relation allows us to transform the integrand into an expression easier to integrate.
  • We simplify our work by directly using the identity \(\int \csc^2(u) \, du = -\cot(u) + C\), where \(C\) is the integration constant.
By applying this substitution, we achieve the desired solution for the integral, which greatly simplifies the expression and helps move forward in solving the differential equation.
Second Solution Verification
Verification is a critical step in confirming that a proposed solution of a differential equation is indeed correct. After deriving potential solutions, it鈥檚 essential to check them against the original equation to ensure they hold true.

In this particular step, the focus is on ensuring that the derived function, \( y_2 = x \cos(\ln x) \), satisfies the conditions set by the differential equation. This involves:
  • Substituting the solution \(y_2\) back into the modified differential equation to check if it reduces the equation to a true identity.
  • Evaluating whether all transformations and integrations made during the solution process have remained valid.
Through verification, it is confirmed that \(y_2 = x \cos(\ln x)\) is indeed a correct solution, aligning with the original formulation of the differential equation. This step ensures accuracy and provides confidence in the solution's authenticity.

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Most popular questions from this chapter

Write the equation in the form $$y^{\prime \prime}+\frac{1}{x} y^{\prime}+\frac{1}{x^{2}} y=\frac{\sec (\ln x)}{x^{2}}$$ and identify \(f(x)=\sec (\ln x) / x^{2} .\) From \(y_{1}=\cos (\ln x)\) and \(y_{2}=\sin (\ln x)\) we compute $$W=\left|\begin{array}{cc}\cos (\ln x) & \sin (\ln x) \\ -\frac{\sin (\ln x)}{x} & \frac{\cos (\ln x)}{x}\end{array}\right|=\frac{1}{x}$$ Now $$u_{1}^{\prime}=-\frac{\tan (\ln x)}{x} \quad \text { so } \quad u_{1}=\ln |\cos (\ln x)|,$$ and $$u_{2}^{\prime}=\frac{1}{x} \quad \text { so } \quad u_{2}=\ln x$$ Thus, a particular solution is $$y_{p}=\cos (\ln x) \ln |\cos (\ln x)|+(\ln x) \sin (\ln x),$$ and the general solution is $$y=c_{1} \cos (\ln x)+c_{2} \sin (\ln x)+\cos (\ln x) \ln |\cos (\ln x)|+(\ln x) \sin (\ln x).$$

Using the double-angle formula for the cosine, we have $$\sin x \cos 2 x=\sin x\left(\cos ^{2} x-\sin ^{2} x\right)=\sin x\left(1-2 \sin ^{2} x\right)=\sin x-2 \sin ^{3} x$$ since \(\sin x\) is a solution of the related homogeneous differential equation we look for a particular solution of the form \(y_{p}=A x \sin x+B x \cos x+C \sin ^{3} x .\) Substituting into the differential equation we obtain $$2 A \cos x+(6 C-2 B) \sin x-8 C \sin ^{3} x=\sin x-2 \sin ^{3} x$$ Equating coefficients we find \(A=0, C=\frac{1}{4},\) and \(B=\frac{1}{4} .\) Thus, a particular solution is $$y_{p}=\frac{1}{4} x \cos x+\frac{1}{4} \sin ^{3} x$$

Using a CAS to solve the auxiliary equation \(3.15 m^{4}-5.34 m^{2}+6.33 m-2.03=0\) we find \(m_{1}=-1.74806\) \(m_{2}=0.501219, m_{3}=0.62342+0.588965 i,\) and \(m_{4}=0.62342-0.588965 i .\) The general solution is $$y=c_{1} e^{-1.74806 x}+c_{2} e^{0.501219 x}+e^{0.62342 x}\left(c_{3} \cos 0.588965 x+c_{4} \sin 0.588965 x\right)$$

(a) The general solution of the differential equation is \(y=c_{1} \cos 4 x+c_{2} \sin 4 x .\) From \(y_{0}=y(0)=c_{1}\) we see that \(y=y_{0} \cos 4 x+c_{2} \sin 4 x .\) From \(y_{1}=y(\pi / 2)=y_{0}\) we see that any solution must satisfy \(y_{0}=y_{1} .\) We also see that when \(y_{0}=y_{1}, y=y_{0} \cos 4 x+c_{2} \sin 4 x\) is a solution of the boundary-value problem for any choice of \(c_{2}\). Thus, the boundary-value problem does not have a unique solution for any choice of \(y_{0}\) and \(y_{1}\). (b) Whenever \(y_{0}=y_{1}\) there are infinitely many solutions. (c) When \(y_{0} \neq y_{1}\) there will be no solutions. (d) The boundary-value problem will have the trivial solution when \(y_{0}=y_{1}=0 .\) This solution will not be unique.

Solving \(\frac{1}{2} q^{\prime \prime}+10 q^{\prime}+100 q=150\) we obtain \(q(t)=e^{-10 t}\left(c_{1} \cos 10 t+c_{2} \sin 10 t\right)+3 / 2 .\) The initial conditions \(q(0)=1\) and \(q^{\prime}(0)=0\) imply \(c_{1}=c_{2}=-1 / 2 .\) Thus \\[q(t)=-\frac{1}{2} e^{-10 t}(\cos 10 t+\sin 10 t)+\frac{3}{2}\\] As \(t \rightarrow \infty, q(t) \rightarrow 3 / 2\)
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