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Chapter 4: Problem 16

Find all solutions of \(y^{\prime}+y \cot x=2 \cos x\) on the interval \((0, \pi)\). Prove that exactly one of these is also a solution on \((-\infty, \infty)\)

Short Answer

Expert verified
The general solution to the ODE is given by \(y(x) = \frac{\sin^2(x) + C_1}{\sin(x)}\). There is exactly one solution that is valid on the entire real line, as the solution is undefined at the multiples of \(\pi\) except for \(x = \frac{\pi}{2}\).

Step by step solution

01

Identify the Integrating Factor

The integrating factor (I.F.) is given by the formula: \(I.F. = e^{\int P(x) dx}\). In our case, \(P(x) = \cot(x)\). So, we have: \(I.F. = e^{\int \cot(x) dx}\). Now, we'll find the integrating factor by computing the integral.
02

Compute the Integral

Integrate \(\cot(x)\) with respect to x: \(e^{\int \cot(x) dx} = e^{\int \frac{\cos(x)}{\sin(x)} dx} = e^{\ln(\sin(x))}\). Now, we can simplify the expression: \(I.F. = e^{\ln(\sin(x))} = \sin(x)\).
03

Multiply the Integrating Factor

Now, we'll multiply the integrating factor to both sides of the ODE: \(y^{\prime}\sin(x) + y\cot(x)\sin(x) = 2\cos(x)\sin(x)\). This simplifies to: \(y^{\prime}\sin(x) + y\cos(x) = 2\cos(x)\sin(x)\).
04

Integrate Both Sides

Integrate both sides with respect to x: \(\int\left(y^{\prime}\sin(x) + y\cos(x)\right) dx = \int 2\cos(x)\sin(x) dx\). Observe that the left side is an exact derivative. We can rewrite it as \(\int(d(y\sin(x))) = \int 2\cos(x)\sin(x) dx\). So, we have: \(y\sin(x) = \int 2\cos(x)\sin(x) dx + C\).
05

Solve for y(x)

Now, we need to solve for y(x). First, we'll compute the integral: \(\int 2\cos(x)\sin(x) dx = \sin^2(x) + C_1\). Next, rewrite the equation for y(x): \(y(x) = \frac{\sin^2(x) + C_1}{\sin(x)}\).
06

Check for Solution Validity

Let's analyze the solution for the whole real line: If \(\sin(x) = 0\), the solution is undefined. Since we're interested in solutions on \((-\infty, \infty)\), we should check for values at the multiples of \(\pi\). For those values, the left side of the ODE is zero (because \(\sin(x) = 0\)), and the right side is \(\cos(x)\). For the right side to be zero as well, \(\cos(x)\) needs to be 0 for that particular \(x\), otherwise, the solution won't be valid for the whole real line. Only one such value of \(x\) exists at \(x = \frac{\pi}{2}\). Thus, the general solution to the ODE is: \(y(x) = \frac{\sin^2(x) + C_1}{\sin(x)}\), with exactly one solution being also valid on \((-\infty, \infty)\).

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Most popular questions from this chapter

Solve the differential equation \(\frac{d y}{d x}=\frac{1}{x \cos y+\sin 2 y}\).

If \(y_{\llcorner}\)is a solution of \(\frac{d^{2} y}{d x^{2}}+a \frac{d y}{d x}+b y=0\) and \(y_{p}\) is a solution of \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{a} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{by}=\mathrm{h}(\mathrm{x})\), show that \(\mathrm{y}_{\mathrm{h}}+\mathrm{y}_{p}\) is also a solution of \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}+\mathrm{a} \frac{\mathrm{dy}}{\mathrm{d} \mathrm{x}}+\mathrm{by}=\mathrm{h}(\mathrm{x})\)

Show that a linear equation remains linear whatever replacements of the independent variable \(\mathrm{x}=\varphi(\mathrm{t})\), where \(\varphi(\mathrm{t})\) is a differentiable function, are made.

A function of \(\mathrm{x}\) is a solution of a differential equation if it and its derivatives make the equation true. For what value (or values) of \(\mathrm{m}\) is \(\mathrm{y}=\mathrm{e}^{\mathrm{mx}}\) a solution of \(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}-3 \frac{\mathrm{dy}}{\mathrm{dx}}+2 \mathrm{y}=0\) ?

\(x y^{\prime 2}-y y^{\prime}-y^{\prime}+1=0\)
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