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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

If the slope of tangent at \((\mathrm{x}, \mathrm{y})\) to the curve passing through \((2,1)\) is \(\left[\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right) /(2 \mathrm{xy})\right]\). The equation of the curve is: (A) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)=6 \mathrm{y}\) (B) \(2\left(\mathrm{x}^{2}-\mathrm{y}^{2}\right)=3 \mathrm{x}\) (C) \(x\left(x^{2}+y^{2}\right)=10\) (D) \(x\left(x^{2}-y^{2}\right)=6\)

The solution of \(\mathrm{xdy}-\mathrm{ydx}=0\) represents: (A) parabola having vertex at \((0,0)\) (B) circle having centre at \((0,0)\) (C) a straight line passing through \((0,0)\) (D) a rectangular hyperbola

The differential equation of family of curves \(\mathrm{y}=\mathrm{Ax}+(\mathrm{B} / \mathrm{x})\) is: (A) \(\left.\mathrm{y}\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{dx}^{2}\right)+\mathrm{x}^{2} \mathrm{dy} / \mathrm{dx}\right)-\mathrm{y}=0\) (B) \(y\left(d^{2} y / d x^{2}\right)+x^{2}(d y / d x)+y=0\) (C) \(x^{2}\left(d^{2} y / d x^{2}\right)+x(d y / d x)-y=0\) (D) \(x^{2}\left(d^{2} y / d x^{2}\right)+x(d y / d x)+y=0\)

The solution of \(\mathrm{x}^{2} \mathrm{y}-\mathrm{x}^{3}(\mathrm{dy} / \mathrm{dx})=\mathrm{y}^{4} \cos \mathrm{x} ; \mathrm{y}(0)=1\) is: (A) \(x^{3}=y^{3} \sin x\) (B) \(x^{3}=3 y^{3} \sin x\) (C) \(y^{3}=3 x^{3} \sin x\) (D) none the these

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