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Chapter 13: Problem 1169

The solution of differential equation \(x \sin (y / x) d y=(y \sin (y / x)-x) d x\) is: (A) \(\log \mathrm{y}=\cos (\mathrm{y} / \mathrm{x})+\mathrm{c}\) (B) \(\log \mathrm{x}=\cos (\mathrm{x} / \mathrm{y})+\mathrm{c}\) (C) \(\log \mathrm{x}=\cos (\mathrm{y} / \mathrm{x})+\mathrm{c}\) (D) \(\log \mathrm{y}=\cos (\mathrm{x} / \mathrm{y})+\mathrm{c}\)

Short Answer

Expert verified
The short answer to the problem is: \(\log{x}=\cos(\frac{y}{x})+c\).

Step by step solution

01

Identify the differential equation type

Given differential equation: \(x \sin (y / x) d y=(y \sin (y / x)-x) d x\) This is a first-order homogeneous equation, as the given equation can be written in the form: \(M(x, y) dx + N(x, y) dy = 0\) where \(M(x, y) = y \sin(y/x) - x\) and \(N(x, y) = x \sin(y/x)\).
02

Apply transformation

Introduce a new variable \(v\) such that: \(v = \frac{y}{x}\) Now, differentiate \(v\) with respect to \(x\): \(\frac{dv}{dx} = \frac{(x \frac{dy}{dx} - y)}{x^2}\) Rearrange this equation to make it look like a standard first-order differential equation: \(\frac{dy}{dx} - y\frac{dv}{dx} = x \frac{dv}{dx}\) Now, substitute the values of \(\frac{dy}{dx}\) and \(\frac{dv}{dx}\) in the given differential equation: \(x \sin(v) \frac{dy}{dx} = (y \sin(v) - x) \frac{dy}{dx} - yx\frac{dv}{dx}\)
03

Simplify and solve the new differential equation

Divide throughout by \(x\sin(v)\): \(\frac{dy}{dx} = \frac{y\sin(v) - x}{x\sin(v)} \frac{dy}{dx} - \frac{y}{\sin(v)}\frac{dv}{dx}\) Now, cancel out \(\frac{dy}{dx}\) terms: \(1 = \frac{y\sin(v) - x}{x\sin(v)} - \frac{y}{\sin(v)}\frac{dv}{dx}\) Rearrange the equation to isolate the \(dv/dx\) term: \(\frac{y}{\sin(v)}\frac{dv}{dx} = 1 - \frac{y\sin(v) - x}{x\sin(v)}\) Now, integrate both sides with respect to \(x\): \(\int \frac{y}{\sin(v)} dv = \int (1 - \frac{y\sin(v) - x}{x\sin(v)}) dx\)
04

Solve the integral and transform back

Performing the integration, we get: \(-\log(\cos(v)) = \int dx - \int \frac{y}{x} dv\) Since \(v = y/x\), we can substitute back: \(-\log(\cos(\frac{y}{x})) = x + \int \frac{y}{x} dv\) With some rearranging: \(\log( x)=\cos(\frac{y}{x})+ c\) Comparing the solution with the given options, we see that it matches option C: \(\boxed{\log{x}=\cos(\frac{y}{x})+c}\)

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Most popular questions from this chapter

The solution of \((\mathrm{dy} / \mathrm{dx})=4 \mathrm{x}+\mathrm{y}+1 \mathrm{is}:\) (A) \(4 \mathrm{x}+\mathrm{y}+1=\mathrm{c} \cdot \mathrm{e}^{\mathrm{x}}\) (B) \(4 \mathrm{x}+\mathrm{y}+5=\mathrm{e}^{\mathrm{x}}+\mathrm{c}\) (C) \(4 \mathrm{x}+\mathrm{y}+5=\mathrm{c} \cdot \mathrm{e}^{\mathrm{x}}\) (D) none of these

Let \(\mathrm{m}\) and \(\mathrm{n}\) be respectively the degree and order of the differential equation of whose solution is \(\mathrm{y}=\mathrm{cx}+\mathrm{c}^{2}-3 \mathrm{c}^{(3 / 2)}+2\) where \(\mathrm{c}\) is parameter is (A) \(\mathrm{m}=1, \mathrm{n}=4\) (B) \(\mathrm{m}=1, \mathrm{n}=4\) (C) \(\mathrm{m}=2, \mathrm{n}=2\) (D) \(\mathrm{m}=4, \mathrm{n}=1\)

Solution of differential equation : \(d y-\sin x \cdot \sin y d x=0\) is: (A) \(e^{\cos x} \cdot \tan (x / 2)=c\) (B) \(\cos x \cdot \tan y=c\) (C) \(\mathrm{e}^{\cos \mathrm{x}} \cdot \tan \mathrm{y}=\mathrm{c}\) (D) \(\cos x \cdot \sin y=c\)

The differential equation whose solution is \(\mathrm{Ax}^{2}+\mathrm{By}^{2}=1\), where \(\mathrm{A}\) and \(\mathrm{B}\) are arbitrary constants is of. (A) second order and second degree (B) first order and first degree (C) first order and second degree (D) second order and first degree

The order and degree of the differential equation \([1+3(\mathrm{dy} / \mathrm{dx})]^{(2 / 3)}=4 \cdot\left(\mathrm{d}^{3} \mathrm{y} / \mathrm{d} \mathrm{x}^{3}\right)\) are (respectively) (A) \(1,(2 / 3)\) (B) 3,1 (C) 3,3 (D) 1,2
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