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Chapter 13: Problem 1167

The solution of the differential equation \((\mathrm{dy} / \mathrm{dx})=[(\mathrm{x}+\mathrm{y}) / \mathrm{x}]\) satisfying the condition \(\mathrm{y}(1)=1\) is: (A) \(\mathrm{y}=\mathrm{x} \log \mathrm{x}+\mathrm{x}\) (B) \(\mathrm{y}=\log \mathrm{x}+\mathrm{x}\) (C) \(\mathrm{y}=\mathrm{x} \log \mathrm{x}+\mathrm{x}^{2}\) (D) \(y=x \cdot e^{x-1}\)

Short Answer

Expert verified
The short answer to the given differential equation problem is: (A) \(y = x\log x + x\).

Step by step solution

01

Identify the type of differential equation

We recognize that the given differential equation is a first-order linear differential equation of the form \( \frac{dy}{dx} - \frac{y}{x} = 1 \).
02

Find the integrating factor

To solve the linear differential equation, we need to multiply the equation by the integrating factor. The integrating factor is given by \(e^{\int{-\frac{1}{x}dx}}\), which simplifies to: \(e^{\int{-\frac{1}{x}dx}}\) = \(e^{-\ln{|x|}}\) = \(x^{-1}\).
03

Multiply the equation by the integrating factor

Multiply both sides of the differential equation by the integrating factor, i.e., \(x^{-1}\), to get: Amount of \(x^{-1}\left(\frac{dy}{dx} - \frac{y}{x}\right)\) = \(x^{-1}(1)\). Which simplifies to: \(\frac{1}{x}\frac{dy}{dx} - \frac{y}{x^2} = 1\).
04

Rewrite the equation into a product rule form

After multiplying both sides by the integrating factor, the left part of the equation should be a product rule in the form of \(\frac{d}{dx}(uv)\) where u is y and v is the integrating factor. So, \(\frac{d}{dx}\left(x^{-1}y\right) = 1\).
05

Integrate both sides

Now, integrate both sides: \(\int\frac{d}{dx}(x^{-1}y)dx = \int 1 dx\) Which leads to: \(x^{-1}y = x + C\).
06

Apply the initial condition

We are given the initial condition y(1) = 1. Apply this condition to our equation: \((1)^{-1}(1) = (1) + C\Rightarrow C = 0\).
07

Solve for y

We can now write our equation as: \(x^{-1}y = x\), Multiplying both sides by x, we get the final solution: \(y = x\log x + x\). Comparing this solution to the given options, we conclude that the correct answer is: (A) \(y = x\log x + x\).

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Most popular questions from this chapter

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) \(\mathrm{y} \cdot \Phi(\mathrm{y} / \mathrm{x})=\mathrm{k}\)

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

The differential equation of all circles passing through the origin and having their centers on the y-axis is: OR The differential equation for the family of curves \(\mathrm{x}^{2}+\mathrm{y}^{2}-2 \mathrm{ay}=0\), where a is an arbitrary constant is: (A) \(\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=2 \mathrm{xy}\) (B) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (C) \(2\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) \mathrm{y}^{1}=\mathrm{xy}\) (D) \(\left(x^{2}+y^{2}\right) y^{1}=2 x y\)

The Integrating factor of the differential equation \(\left(1-\mathrm{y}^{2}\right)(\mathrm{dy} / \mathrm{dx})-\mathrm{yx}=1\) is: (A) \(\left[1 / \sqrt{ \left.\left(1-\mathrm{y}^{2}\right)\right]}\right.\) (B) \(\sqrt{\left(1-\mathrm{y}^{2}\right)}\) (C) \(\left[1 /\left(1-\mathrm{y}^{2}\right)\right]\) (D) \(1-\mathrm{y}^{2}\)

The differential equation which represents the family of curves \(\mathrm{y}=\mathrm{c}_{1} \mathrm{e}^{(\mathrm{c}) 2 \mathrm{x}}\), where \(\mathrm{c}_{1}\) and \(\mathrm{c}_{2}\) are arbitarary constants, is: (A) \(y^{\prime}=y^{2}\) (B) \(\mathrm{y}^{\prime \prime}=\mathrm{y}^{\prime} \mathrm{y}\) (C) \(\mathrm{yy}^{\prime \prime}=\left(\mathrm{y}^{1}\right)^{2}\) (D) \(\mathrm{yy}^{\prime \prime}=\mathrm{y}^{\prime}\)
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