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Chapter 13: Problem 1181

If integrating factor of \(: x\left(1-x^{2}\right) d y+\left(2 x^{2} y-y-a x^{3}\right) d x=0\) is \(\mathrm{e}^{\int \mathrm{p} \cdot \mathrm{d} \mathrm{x}}\), then \(\mathrm{p}\) is equal to: (A) \(2 \mathrm{x}^{2}-1\) (B) \(\left[\left\\{2 \mathrm{x}^{2}-1\right\\} /\left\\{\mathrm{x}\left(1-\mathrm{x}^{2}\right)\right\\}\right]\) (C) \(\left[\left\\{2 \mathrm{x}^{2}-\mathrm{ax}^{3}\right\\} /\left\\{\mathrm{x}\left(1-\mathrm{x}^{2}\right)\right]\right.\) (D) \(\mathrm{ax}^{3}\)

Short Answer

Expert verified
The value of p(x) is: \(\boxed{(B) \frac{2x^2-1}{x(1-x^2)}}\).

Step by step solution

01

Rewrite the given differential equation in the standard form (dy/dx)

Given the differential equation: \(x(1-x^2)~dy+(2x^2y-y-ax^3)~dx=0\). We can rewrite it in the standard form (dy/dx): \(dy/dx =-\frac{2x^2y-y-ax^3}{x(1-x^2)}\)
02

Identify the P(x) term

In the standard form, the P(x) term is the term next to the y variable: \(\frac{dy}{dx} + P(x)y = Q(x)\) Here, we have: \(\frac{dy}{dx} -\frac{2x^2-1}{x(1-x^2)}y =\frac{-ax^3}{x(1-x^2)}\) So, the P(x) term is: \(P(x) = -\frac{2x^2-1}{x(1-x^2)}\)
03

Use the integrating factor formula and match with the given exponential term with integration

The given integrating factor is: \(e^{\int p~dx}\) Replacing 'p' with P(x) we have: Integrating factor: \(e^{\int -\frac{2x^2-1}{x(1-x^2)}~dx}\)
04

Extract the value of p(x) from that integrating factor

Now, we have to match this integrating factor with the given options to find the correct value of p(x): (A) \(2x^2-1\) (B) \(\frac{2x^2-1}{x(1-x^2)}\) (C) \(\frac{2x^2-ax^3}{x(1-x^2)}\) (D) \(ax^3\) Comparing the integrating factor expression, we can conclude that the correct option is: \(p(x) = \boxed{(B) \frac{2x^2-1}{x(1-x^2)}}\)

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Most popular questions from this chapter

Solution of differential equation \((\mathrm{dy} / \mathrm{dx})+\mathrm{ay}=\mathrm{e}^{\mathrm{mx}}\) is: (A) \(y=e^{m x}+c \cdot e^{-a x}\) (B) \((\mathrm{a}+\mathrm{m}) \mathrm{y}=\mathrm{e}^{\mathrm{mx}}+\mathrm{c}\) (C) \((a+m) y=e^{m x}+c \cdot e^{-a x}\) (D) \(\mathrm{y} \cdot \mathrm{e}^{\mathrm{ax}}=\mathrm{m} \cdot \mathrm{e}^{\mathrm{mx}}+\mathrm{c}\)

The equation of the curve passing through \([1,(\pi / 4)]\) and having the slope \([(\sin 2 \mathrm{y}) /(\mathrm{x}+\tan \mathrm{y})]\) at \((\mathrm{x}, \mathrm{y})\) is: (A) \(x=\tan y\) (B) \(\mathrm{y}=2 \tan \mathrm{x}\) (C) \(y=\tan x\) (D) \(x=2 \tan y\)

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) \(x^{2}\left(d^{2} y / d x^{2}\right)+x(d y / d x)+y=0\)

A solution of the differential equation \((\mathrm{dy} / \mathrm{dx})^{2}-\mathrm{x}(\mathrm{dy} / \mathrm{dx})\) \(+y=0\) is: (A) \(y=2\) (B) \(y=2 x^{2}-4\) (C) \(\mathrm{y}=2 \mathrm{x}\) (D) \(\mathrm{y}=2 \mathrm{x}-4\)

Order and degree of differential equation of all tangent lines to the parabola \(\mathrm{y}^{2}=4 \mathrm{ax}\) is (A) 2,2 (B) 3,1 (C) 1,2 (D) 4,1
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