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Chapter 13: Problem 1111

If \(\sin x\) is an Integrating factor of \((d y / d x)+p \cdot y=Q\) then \(p\) is: (A) \(\sin x\) (B) \(\log \sin x\) (C) \(\cot x\) (D) \(\log \cos x\)

Short Answer

Expert verified
\(P = \cot x\)

Step by step solution

01

Write down the given integrating factor and given equation

We are given that the integrating factor is \(\sin x\). The differential equation is given in the form: \[ \frac{dy}{dx} + Py = Q \]
02

Identify the form of the integrating factor

We know that the general formula for the integrating factor is \(e^{\int P dx}\). In this case, the integrating factor is given as \(\sin x\). So, we can rewrite the given integrating factor as: \[ \text{IF} = e^{\int P dx} = \sin x \]
03

Find P using the integrating factor formula

We want to find P such that the integrating factor is \(\sin x\). So we have the equation: \[ e^{\int P dx} = \sin x \] To solve for P, first, take the natural logarithm on both sides: \[ \int P dx = \ln(\sin x) + C \] Now, differentiate both sides with respect to x: \[ P = \frac{d}{dx}(\ln(\sin x)) \] Use the chain rule on the right side: \[ P = \frac{1}{\sin x} \cdot (\frac{d}{dx}(\sin x)) = \cot x \] So, based on our analysis, the correct value for P is \(\cot x\), which corresponds to answer choice (C).

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Most popular questions from this chapter

The order of the differential equation whose general solution is given by \(\mathrm{y}=\mathrm{C}_{1} \mathrm{e}^{(\mathrm{x}+\mathrm{C}) 2}+\left(\mathrm{C}_{3}+\mathrm{C}_{4}\right) \cdot \sin \left(\mathrm{x}+\mathrm{C}_{5}\right)\), where \(\mathrm{C}_{1}\), \(\mathrm{C}_{2}, \mathrm{C}_{3}, \mathrm{C}_{4}, \mathrm{C}_{5}\) are arbitrary Constant is (A) 5 (B) 4 (C) 3 (D) 2

Let \(\mathrm{m}\) and \(\mathrm{n}\) be respectively the degree and order of the differential equation of whose solution is \(\mathrm{y}=\mathrm{cx}+\mathrm{c}^{2}-3 \mathrm{c}^{(3 / 2)}+2\) where \(\mathrm{c}\) is parameter is (A) \(\mathrm{m}=1, \mathrm{n}=4\) (B) \(\mathrm{m}=1, \mathrm{n}=4\) (C) \(\mathrm{m}=2, \mathrm{n}=2\) (D) \(\mathrm{m}=4, \mathrm{n}=1\)

If \(\mathrm{f}(\mathrm{x})\) and \(\mathrm{g}(\mathrm{x})\) are two solutions of the differential equation \(\mathrm{q}\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{d} \mathrm{x}^{2}\right)+\mathrm{x}^{2}(\mathrm{dy} / \mathrm{d} \mathrm{x})+\mathrm{y}=\mathrm{e}^{\mathrm{x}}\), then \(\mathrm{f}(\mathrm{x})-\mathrm{g}(\mathrm{x})\) is the solution of: (A) \(\mathrm{q}\left(\mathrm{d}^{2} \mathrm{y} / \mathrm{d} \mathrm{x}^{2}\right)+\mathrm{y}=\mathrm{e}^{\mathrm{x}}\) (B) \(q^{2}\left(d^{2} y / d x^{2}\right)+(d y / d x)+y=e^{x}\) (C) \(q^{2}\left(d^{2} y / d x^{2}\right)+y=e^{x}\) (D) \(q\left(d^{2} y / d x^{2}\right)+x^{2}(d y / d x)+y=0\)

The solution of the equation \(\left(d^{2} y / d x^{2}\right)=e^{-2 x}\) is: \(y=\) (A) \((1 / 4) \mathrm{e}^{-2 \mathrm{x}}+\mathrm{cx}+\mathrm{d}\) (B) \((1 / 4) e^{-2 \mathrm{x}}\) (C) \((1 / 4) e^{-2 x}+c x^{2}+d\) (D) \((1 / 4) e^{-2 x}+c x+d\)

The differential equation whose solution is \(\mathrm{Ax}^{2}+\mathrm{By}^{2}=1\), where \(\mathrm{A}\) and \(\mathrm{B}\) are arbitrary constants is of. (A) second order and second degree (B) first order and first degree (C) first order and second degree (D) second order and first degree
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