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TS EAMCET 2015 Solved Paper

Section: Mathematics

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Question : 1 of 160

Marks: +1, -0

Match the differential equation in List 1 to their Integrating factors in List-II

List-I	List-II
Differential Equation	Integrating factor
(i) $(x^3+1)\frac{dy}{dx}+x^2y=3x^2$	(a) $x^3$
(ii) $x^2\frac{dy}{dx}+3xy=x^6$	(b) $(x^3+1)^2$
(iii) $(x^3+1)^2\frac{dy}{dx}+6x^2(x^3+1)y=x^2$	(c) $(x^2+1)^2$
(iv) $(x^2+1)\frac{dy}{dx}+4xy=\ln x$	(d) $x^2+1$
	(e) $(x^3+1)^{1/3}$
	(f) $(x^3+1)^{1/2}$

	i	ii	iii	iv
(1)	(d)	(a)	(b)	(c)
(2)	(e)	(a)	(b)	(c)
(3)	(e)	(b)	(c)	(f)
(4)	(e)	(a)	(c)	(d)

1
3
2
4

Solution:

(i)

\frac{dy}{dx}+\frac{x^2}{1+x^3}y=\frac{3x^2}{1+x^3}

\int Pdx = \frac{1}{3}\int \frac{3x^2}{1+x^3}dx = \log(1+x^3)^{1/3}

I.F = (1+x^3)^{1/3}

––––––––>key (e)

(ii)

\frac{dy}{dx}+\frac{3x}{x^2}y=x^4

\int Pdx = 3\int \frac{1}{x}dx = \log x^3

I.F

= x^3

––––––––>key (a)

(iii)

\frac{dy}{dx}+\left(\frac{6x^2}{x^3+1}\right)y=\frac{x^2}{(x^3+1)^2}

\int Pdx = 2\int \frac{3x^2}{1+x^3}dx = \log(1+x^3)^2

I.F

= (1+x^3)^2

––––––––>key (b)

(iv)

\frac{dy}{dx}+\frac{4x}{x^2+1}y=\frac{\log x}{x^2+1}

\int Pdx = 2\int \frac{2x}{x^2+1}dx = \log(x^2+1)^2

I.F

= (x^2+1)^2

––––––––>key (c)

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