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Chapter 13: Problem 1187

The solution of the differential equation \((\mathrm{dy} / \mathrm{dx})=(\mathrm{y} / \mathrm{x})+\left[\\{\Phi(\mathrm{y} / \mathrm{x})\\} /\left\\{\Phi^{1}(\mathrm{y} / \mathrm{x})\right\\}\right]\) is: (A) \(\phi(\mathrm{y} / \mathrm{x})=\mathrm{kx}\) (B) \(\Phi(\mathrm{y} / \mathrm{x})=\mathrm{ky}\) (C) \(\mathrm{x} \cdot \Phi(\mathrm{y} / \mathrm{x})=\mathrm{k}\) (D) \(\mathrm{y} \cdot \Phi(\mathrm{y} / \mathrm{x})=\mathrm{k}\)

Short Answer

Expert verified
The short answer is: \(\boxed{\phi(\mathrm{y} / \mathrm{x})=\mathrm{kx}}\).

Step by step solution

01

Rewrite the differential equation

We are given the differential equation \[\frac{dy}{dx} = \frac{y}{x} + \frac{\Phi(\frac{y}{x})}{\Phi'(\frac{y}{x})},\] where \(\Phi'(\frac{y}{x})\) represents the derivative of \(\Phi(\frac{y}{x})\) with respect to \(\frac{y}{x}\).
02

Separate variables

In order to separate the variables, we can rewrite the given equation as follows: \[\frac{dy}{dx} - \frac{y}{x} = \frac{\Phi(\frac{y}{x})}{\Phi'(\frac{y}{x})}.\] Now, let \(v = \frac{y}{x}\), meaning that \(y = xv\). Taking the derivative of \(y\) with respect to \(x\), we get \[\frac{dy}{dx} = x \frac{dv}{dx} + v.\] Substituting for \(\frac{dy}{dx}\) in the separated variables equation, we obtain \[x \frac{dv}{dx} + v - \frac{xv}{x} = \frac{\Phi(v)}{\Phi'(v)}.\] Simplifying the equation, we have \[x \frac{dv}{dx} = \frac{\Phi(v)}{\Phi'(v)}.\]
03

Integrate both sides

To solve for the relationship between \(y\) and \(x\), integrate both sides with respect to \(x\), noting that we are integrating for \(v\) on the right-hand side: \[\int x \frac{dv}{dx} dx = \int \frac{\Phi(v)}{\Phi'(v)} dv.\] Evaluating the integral, we get \[xv = \int \frac{\Phi(v)}{\Phi'(v)} dv + C,\] where \(C\) is the constant of integration.
04

Rewrite in terms of y and x and compare to the given choices

Recall that \(v = \frac{y}{x}\) and \(y = xv\). Thus, our found solution can be rewritten as \[y = x \int \frac{\Phi(\frac{y}{x})}{\Phi'(\frac{y}{x})} d\frac{y}{x} + Cx.\] Comparing this solution to the given choices, we find that choice (A) is in the form of our found solution: \[\boxed{\phi(\mathrm{y} / \mathrm{x})=\mathrm{kx}}.\]

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Most popular questions from this chapter

The solution of the differential equation \(\mathrm{ydx}+\left(\mathrm{x}+\mathrm{x}^{2} \mathrm{y}\right) \mathrm{dy}=0\) is: (A) \((1 / \mathrm{xy})+\log \mathrm{y}=\mathrm{c}\) (B) \(-(1 / x y)+\log y=c\) (C) \(-(1 / \mathrm{xy})=\mathrm{c}\) (D) \(\log \mathrm{y}=\mathrm{cx}\)

The general solution of \([\mathrm{x}(\mathrm{dy} / \mathrm{dx})-\mathrm{y}] \mathrm{e}^{(\mathrm{y} / \mathrm{x})}=\mathrm{x}^{2} \cos \mathrm{x}\) is: (A) \(\mathrm{e}^{(\mathrm{x} / \mathrm{y})}=\cos \mathrm{x}+\mathrm{c}\) (B) \(\mathrm{e}^{(\mathrm{x} / \mathrm{y})}=\sin \mathrm{x}+\mathrm{c}\) (C) \(e^{(y / x)}=\sin x+c\) (D) \(e^{(y / x)}=\cos x+c\)

The solution of \(\mathrm{y}^{5} \mathrm{x}+\mathrm{y}-\mathrm{x}(\mathrm{dy} / \mathrm{dx})=0\) is: (A) \((\mathrm{x} / \mathrm{y})^{5}+\left(\mathrm{x}^{4} / 4\right)=\mathrm{c}\) (B) \((\mathrm{xy})^{4}+\left(\mathrm{x}^{5} / 5\right)=\mathrm{c}\) (C) \(\left(x^{5} / 5\right)+(1 / 4)(x / y)^{4}=c\) (D) \(\left(x^{4} / y\right)+(1 / 5)(x / y)^{5}=c\)

The general solution of the equation \((d y / d x)=\left(x^{2} / y^{2}\right)\) is: (A) \(x^{3}+y^{3}=c\) (B) \(x^{3}-y^{3}=c\) (C) \(x^{2}+y^{2}=c\) (D) \(x^{2}-y^{2}=c\)

If \(\mathrm{m}\) and \(\mathrm{n}\) are order and degree of the equation \(\left(d^{2} y / d x^{2}\right)^{5}+4\left[\left(d^{2} y / d x^{2}\right)^{3} /\left(d^{3} y / d x^{3}\right)\right]+\left(d^{3} y / d x^{3}\right)=x^{2}-1\) then: (A) \(\mathrm{m}=3, \mathrm{n}=2\) (B) \(\mathrm{m}=3, \mathrm{n}=3\) (C) \(\mathrm{m}=3, \mathrm{n}=5\) (D) \(\mathrm{m}=3, \mathrm{n}=1\)
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