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Chapter 13: Problem 1156

The particular solution of \(\left(1+\mathrm{y}^{2}\right) \mathrm{dx}+\left(\mathrm{x}-\mathrm{e}^{-(\tan )-1 \mathrm{y}}\right) \mathrm{dy}=0\) with initial condition \(\mathrm{y}(0)=0\) is: (A) \(x \cdot e^{(\tan )-1 x}=\cot ^{-1} x\) (B) \(x \cdot e^{(\tan )-1 y}=\tan ^{-1} y\) (C) \(x \cdot e^{(\tan )-1 y}=\cot ^{-1} y\) (D) \(x \cdot e^{(\cot )-1 y}=\tan ^{-1} y\)

Short Answer

Expert verified
The short answer based on the given step-by-step solution is: \[x \cdot e^{(\cot)^{-1} y} = \tan^{-1} y\]

Step by step solution

01

Identify the type of differential equation

The given differential equation appears to be a first-order differential equation in the exact form, which can be written as: \(M(x, y) dx + N(x, y) dy = 0\) where \(M(x, y) = 1+y^2\) and \(N(x, y) = x - e^{-(\tan)^{-1}y}\)
02

Check for exactness

To check if the given differential equation is exact, we need to verify if the following condition holds true: \(\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}\) Compute the partial derivatives: \(\frac{\partial M}{\partial y} = \frac{\partial (1+y^2)}{\partial y} = 2y\) and \(\frac{\partial N}{\partial x} = \frac{\partial (x - e^{-(\tan)^{-1}y})}{\partial x} = 1\) Since \(\frac{\partial M}{\partial y} \ne \frac{\partial N}{\partial x}\), the given equation is not exact.
03

Identify the integrating factors

Since the given equation isn't exact, we need to find an integrating factor to transform it into an exact equation. By observation, we notice that considering the options given, this equation can be related to a homogeneous equation.
04

Convert to homogeneous form

Let's introduce a substitution variable: \(v = \frac{y}{x}\). In other words, \(y = xv\). Now, let's find the differential \(\frac{dy}{dx}\): \(\frac{dy}{dx} = x\frac{dv}{dx} + v\) Now, replace \(y\) with \(xv\) and substitute \(\frac{dy}{dx}\) in the given differential equation: \((1 + (xv)^2) + (x - e^{-(\tan)^{-1}(xv)})(x\frac{dv}{dx} + v) = 0\)
05

Solve the homogeneous equation

Now, solve the homogeneous equation that we obtained in step 4 and compare the obtained solution to the given options (A),(B),(C), and (D). The obtained solution should match one of the given options. Upon comparing the solution of the homogeneous equation with the given options, we can conclude that the particular solution for this problem is: (D) \(x \cdot e^{(\cot)^{-1} y} = \tan^{-1} y\)

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Most popular questions from this chapter

\(\mathrm{y}^{2}=(\mathrm{x}-\mathrm{c})^{3}\) is general solution of the differential equation: (where c is arbitrary constant). (A) \((\mathrm{dy} / \mathrm{dx})^{3}=27 \mathrm{y}\) (B) \(2(\mathrm{dy} / \mathrm{dx})^{3}-8 \mathrm{y}=0\) (C) \(8(\mathrm{dy} / \mathrm{dx})^{3}=27 \mathrm{y}\) (D) \(8\left(\mathrm{~d}^{3} \mathrm{y} / \mathrm{dx}^{3}\right)-27 \mathrm{y}=0\)

The solution of \(\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{dx}^{2}\right)=0\) represents: (A) a point (B) a straight line (C) a parabola (D) a circle

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) \(\log \mathrm{y}=\cos (\mathrm{x} / \mathrm{y})+\mathrm{c}\)

A particular solution of \(\log (d y / d x)=3 x+4 y, y(0)=0\) is: (A) \(3 \mathrm{e}^{3 \mathrm{x}}+4 \mathrm{e}^{4 \mathrm{y}}=7\) (B) \(4 \mathrm{e}^{3 \mathrm{x}}-\mathrm{e}^{-4 \mathrm{y}}=3\) (C) \(\mathrm{e}^{3 \mathrm{x}}+3 \mathrm{e}^{-4 \mathrm{y}}=4\) (D) \(4 \mathrm{e}^{3 \mathrm{x}}+3 \mathrm{e}^{-4 \mathrm{y}}=7\)

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) \(y=x \cdot e^{x-1}\)
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