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Chapter 13: Problem 1112

Integrating factor of differential equation \((1+x)(d y / d x)-x \cdot y=1-x\) is: (A) \(1+\mathrm{x}\) (B) \(\log (1+x)\) (C) \(e^{-x}(1+x)\) (D) \(x \cdot e^{x}\)

Short Answer

Expert verified
The integrating factor for the given differential equation is (A) \(1+x\).

Step by step solution

01

Write the equation in standard form

In the standard form of a first-order linear differential equation, we should have the equation written as: \(\frac{dy}{dx} + P(x)y = Q(x)\) Rearrange the given equation to fit this format: \(\frac{dy}{dx} + \frac{x}{1+x}y = \frac{1-x}{1+x}\) Now, the differential equation is in the standard form with \(P(x) = \frac{x}{1+x}\) and \(Q(x) = \frac{1-x}{1+x}\).
02

Compute the integrating factor

To find the integrating factor, integrate \(P(x)\) with respect to \(x\). So, we have: \(I(x)= e^{\int P(x) dx}\) Now, substitute \(P(x) = \frac{x}{1+x}\) to find the integrating factor, \(I(x)\): \(I(x)= e^{\int \frac{x}{1+x} dx}\)
03

Solve the integration

You can use substitution to integrate the function. Let \(u = 1+x\), which gives \(du = dx\): \(I(x) = e^{\int \frac{u-1}{u} du}\) This can be separated into two separate integrations: \(I(x) = e^{\int 1 - \frac{1}{u} du}\) Now, integrate both parts: \(I(x) = e^{\int 1 du - \int \frac{1}{u} du}\) \(I(x) = e^{u - \log(u)}\) Now, substitute back \(u = 1+x\) to find the integrating factor, leaving: \(I(x) = e^{(1+x) - \log(1+x)}\) Since \(e^{\log(a)} = a\) and the properties of exponentials: \(I(x) = (1+x)e^0\) \(I(x) = (1+x)\) Thus, the integrating factor for the given differential equation is (A) \(1+x\).

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Most popular questions from this chapter

The degree and order of the differential equation of the family of all parabolas whose axis is \(\mathrm{x}\) -axis, are respectively. (A) 1,2 (B) 3,2 (C) 2,3 (D) 2,1

Solution of \((\mathrm{y} / \mathrm{x}) \cos (\mathrm{y} / \mathrm{x})[(\mathrm{dy} / \mathrm{dx})-(\mathrm{y} / \mathrm{x})]\) \(+\sin (\mathrm{y} / \mathrm{x})[(\mathrm{dy} / \mathrm{dx})+(\mathrm{y} / \mathrm{x})]=0 ; \mathrm{y}(1)=(\pi / 2)\) is: (A) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / 2 \mathrm{x})\) (B) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / \mathrm{x})\) (C) \(\mathrm{y} \sin (\mathrm{y} / \mathrm{x})=(\pi / 3 \mathrm{x})\) (D) none of these

The equation of a curve passing through \([2,(7 / 2)]\) and having gradient \(1-\left(1 / \mathrm{x}^{2}\right)\) at \((\mathrm{x}, \mathrm{y})\) is: (A) \(x y=x+1\) (B) \(x^{2}+x+1\) (C) \(x y=x^{2}+x+1\) (D) none of these

If \(\sin (x+y)(d y / d x)=5\) then (A) \(5 \int[(\mathrm{dt}) /(5+\sin \mathrm{t})]=\mathrm{t}+\mathrm{x}(\) where \(\mathrm{t}=\mathrm{x}+\mathrm{y})\) (B) \(5 \int[(\mathrm{dt}) /(5+\sin \mathrm{t})]=\mathrm{t}-\mathrm{x}(\) where \(\mathrm{t}=\mathrm{x}+\mathrm{y})\) (C) \(\int[(\mathrm{dt}) /(5+\operatorname{cosec} \mathrm{t})]=\mathrm{d} \mathrm{x}(\) where \(\mathrm{t}=\mathrm{x}+\mathrm{y})\) (D) \(\int[(\mathrm{dt}) /(5 \sin t+1)]=\mathrm{dt}(\) where \(t=x+y)\)

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) \(1-\mathrm{y}^{2}\)
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