QBManager

Differential Equation

87229 The solution of differential equation
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$ is

1

\log (1 + y) = \log x + 2 \tan x +

constant

2

\log (1 + y^{2}) = 3 \log (\frac{1}{x}) + 6 \tan^{- 1} x +

constant

3

2 \log (1 + y^{2}) = 3 \log x + 6 \tan^{- 1} x +

constant

4 None of the above

Explanation:

(C) : We have differential equation,
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$
Now separating the variable, we get-
$\frac{4 y d y}{(1 + y^{2})} = \frac{3 (1 + x)^{2}}{x (1 + x^{2})} \Rightarrow \frac{4 y d y}{1 + y^{2}} = (\frac{3}{x} + \frac{6}{1 + x^{2}})$
Now integrating both side, we get-
$\int \frac{4 y d y}{1 + y^{2}} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) d x$
Let, $1 + y^{2} = t$
$2 ydy = dt$
$2 \int \frac{dt}{t} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) dx$
$2 \log t = 3 \log | x | + 6 \tan^{- 1} x + c$
$2 \log (1 + y^{2}) = 3 \log | x | + 6 \tan^{- 1} x + c$

UPSEE-2011

Differential Equation

87230 If $y^{2} = P (x)$ be a cubic polynomial, then
$2 \frac{d}{d x} (y^{3} \frac{d^{2} y}{d x^{2}})$ is equal to

1

P^{'''} (x) + P^{'} (x)

2

P^{''} (x) P^{'''} (x)

3

P (x) P^{'''} (x)

4 constant

Explanation:

(C) : Given,
$y^{2} = P (x)$
Differentiating w.r.t $x$ , we get-
$2 y \frac{d y}{d x} = P^{'} (x)$
$\frac{d y}{d x} = \frac{P^{'} (x)}{2 y}$
Again differentiating w.r.t $x$ , we get-
$2 [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = P^{''} (x)$
$2 y \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} = P^{''} (x)$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 (d y / d x)^{2}}{2 y}$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 {(\frac{P^{'} (x)}{2 y})}^{2}}{2 y}$
Multiplying by $y^{3}$ in both side we get-
$y^{3} (\frac{d^{2} y}{d^{2}}) = \frac{y^{3} P^{''} (x)}{2 y} - \frac{2 y^{3}}{2 y} {(\frac{P^{'} (x)}{2 y})}^{2}$
$y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{y^{2} P^{''} (x)}{2} - y^{2} {(\frac{P^{'} (x)}{2 y})}^{2}$
Again differentiating with respect to $x$ we get-
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{2}{2} \frac{d}{d x} {y^{2} P^{''} (x)} - \frac{2}{4} \frac{d}{d x} {P^{'} (x)}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{d}{d x} [P (x) P^{''} (x)] - \frac{1}{2} \frac{d}{d x} {[P^{'} (x)]}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = [P (x) P^{''} (x) + P^{''} (x) P^{'} (x)]$
$- \frac{1}{2} 2 P^{'} (x) P^{''} (x)$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = P (x) P^{'''} (x)$

UPSEE-2010

Differential Equation

87231 The solution of differential equation $x \cos^{2} y dx$ $= y \cos^{2} x d y$ is

1

x \tan x - y \tan y - \log (\sec x / \sec y) = c

2

y \tan x - x \tan x - \log (\sec x \cdot \sec y) = c

3

x \tan x - y \tan y + \log (\sec x \cdot \sec y) = c

4 None of the above

Explanation:

(A) : Thę differential ęquation-
$x \cos^{2} y d x = y \cos^{2} x d y$
Separating the variable, we get-
$\frac{x}{\cos^{2} x} d x = \frac{y}{\cos^{2} y} d y$
Integrating both side, we get-
$\int x \sec^{2} x d x = \int y \sec^{2} y d y$
$x \int \sec^{2} x d x - \int \frac{d}{d x} (x) \cdot \int \sec^{2} x d x = y \int \sec^{2} y d y$
$x \tan x - \log | \sec x | = y \tan y - \log | \sec y | + c$
$- \int \frac{d}{d x} (y) \cdot \int \sec^{2} y d y$
$x \tan x - y \tan y = \log (\sec x / \sec y) + c$
$x \tan x - y \tan y - \log (\sec x / \sec y) = c$

UPSEE-2010]**#

Differential Equation

87232 The solution of the differential equation
$\frac{d y}{d x} = y \tan x - 2 \sin x is$

1

y \sin x = c + \sin 2 x

2

y \cos x = c + \frac{1}{2} \sin 2 x

3

y \cos x = c - \sin 2 x

4

y \cos x = c + \frac{1}{2} \cos 2 x

Explanation:

(D) : We have differential equation-
$\frac{d y}{d x} = y \tan x - 2 \sin x \Rightarrow \frac{d y}{d x} - y \tan x = - 2 \sin x$
Which is form of $\frac{d y}{d x} + P y = Q$ $P = - \tan x$ and $Q = - 2 \sin x$
$I . F = e^{\int P d x} = e^{\int - \tan x d x} = e^{- \log \sec x} = \cos x$
Now particular solution, we get-
$y \cos x = \int \cos x (- 2 \sin x) d x + c$
$y \cos x = - \int \sin 2 x + c$
$y \cos x = \frac{1}{2} \cos 2 x + c$

UPSEE-2007

Differential Equation

87229 The solution of differential equation
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$ is

1

\log (1 + y) = \log x + 2 \tan x +

constant

2

\log (1 + y^{2}) = 3 \log (\frac{1}{x}) + 6 \tan^{- 1} x +

constant

3

2 \log (1 + y^{2}) = 3 \log x + 6 \tan^{- 1} x +

constant

4 None of the above

Explanation:

(C) : We have differential equation,
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$
Now separating the variable, we get-
$\frac{4 y d y}{(1 + y^{2})} = \frac{3 (1 + x)^{2}}{x (1 + x^{2})} \Rightarrow \frac{4 y d y}{1 + y^{2}} = (\frac{3}{x} + \frac{6}{1 + x^{2}})$
Now integrating both side, we get-
$\int \frac{4 y d y}{1 + y^{2}} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) d x$
Let, $1 + y^{2} = t$
$2 ydy = dt$
$2 \int \frac{dt}{t} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) dx$
$2 \log t = 3 \log | x | + 6 \tan^{- 1} x + c$
$2 \log (1 + y^{2}) = 3 \log | x | + 6 \tan^{- 1} x + c$

UPSEE-2011

Differential Equation

87230 If $y^{2} = P (x)$ be a cubic polynomial, then
$2 \frac{d}{d x} (y^{3} \frac{d^{2} y}{d x^{2}})$ is equal to

1

P^{'''} (x) + P^{'} (x)

2

P^{''} (x) P^{'''} (x)

3

P (x) P^{'''} (x)

4 constant

Explanation:

(C) : Given,
$y^{2} = P (x)$
Differentiating w.r.t $x$ , we get-
$2 y \frac{d y}{d x} = P^{'} (x)$
$\frac{d y}{d x} = \frac{P^{'} (x)}{2 y}$
Again differentiating w.r.t $x$ , we get-
$2 [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = P^{''} (x)$
$2 y \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} = P^{''} (x)$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 (d y / d x)^{2}}{2 y}$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 {(\frac{P^{'} (x)}{2 y})}^{2}}{2 y}$
Multiplying by $y^{3}$ in both side we get-
$y^{3} (\frac{d^{2} y}{d^{2}}) = \frac{y^{3} P^{''} (x)}{2 y} - \frac{2 y^{3}}{2 y} {(\frac{P^{'} (x)}{2 y})}^{2}$
$y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{y^{2} P^{''} (x)}{2} - y^{2} {(\frac{P^{'} (x)}{2 y})}^{2}$
Again differentiating with respect to $x$ we get-
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{2}{2} \frac{d}{d x} {y^{2} P^{''} (x)} - \frac{2}{4} \frac{d}{d x} {P^{'} (x)}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{d}{d x} [P (x) P^{''} (x)] - \frac{1}{2} \frac{d}{d x} {[P^{'} (x)]}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = [P (x) P^{''} (x) + P^{''} (x) P^{'} (x)]$
$- \frac{1}{2} 2 P^{'} (x) P^{''} (x)$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = P (x) P^{'''} (x)$

UPSEE-2010

Differential Equation

87231 The solution of differential equation $x \cos^{2} y dx$ $= y \cos^{2} x d y$ is

1

x \tan x - y \tan y - \log (\sec x / \sec y) = c

2

y \tan x - x \tan x - \log (\sec x \cdot \sec y) = c

3

x \tan x - y \tan y + \log (\sec x \cdot \sec y) = c

4 None of the above

Explanation:

(A) : Thę differential ęquation-
$x \cos^{2} y d x = y \cos^{2} x d y$
Separating the variable, we get-
$\frac{x}{\cos^{2} x} d x = \frac{y}{\cos^{2} y} d y$
Integrating both side, we get-
$\int x \sec^{2} x d x = \int y \sec^{2} y d y$
$x \int \sec^{2} x d x - \int \frac{d}{d x} (x) \cdot \int \sec^{2} x d x = y \int \sec^{2} y d y$
$x \tan x - \log | \sec x | = y \tan y - \log | \sec y | + c$
$- \int \frac{d}{d x} (y) \cdot \int \sec^{2} y d y$
$x \tan x - y \tan y = \log (\sec x / \sec y) + c$
$x \tan x - y \tan y - \log (\sec x / \sec y) = c$

UPSEE-2010]**#

Differential Equation

87232 The solution of the differential equation
$\frac{d y}{d x} = y \tan x - 2 \sin x is$

1

y \sin x = c + \sin 2 x

2

y \cos x = c + \frac{1}{2} \sin 2 x

3

y \cos x = c - \sin 2 x

4

y \cos x = c + \frac{1}{2} \cos 2 x

Explanation:

(D) : We have differential equation-
$\frac{d y}{d x} = y \tan x - 2 \sin x \Rightarrow \frac{d y}{d x} - y \tan x = - 2 \sin x$
Which is form of $\frac{d y}{d x} + P y = Q$ $P = - \tan x$ and $Q = - 2 \sin x$
$I . F = e^{\int P d x} = e^{\int - \tan x d x} = e^{- \log \sec x} = \cos x$
Now particular solution, we get-
$y \cos x = \int \cos x (- 2 \sin x) d x + c$
$y \cos x = - \int \sin 2 x + c$
$y \cos x = \frac{1}{2} \cos 2 x + c$

UPSEE-2007

Differential Equation

87229 The solution of differential equation
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$ is

1

\log (1 + y) = \log x + 2 \tan x +

constant

2

\log (1 + y^{2}) = 3 \log (\frac{1}{x}) + 6 \tan^{- 1} x +

constant

3

2 \log (1 + y^{2}) = 3 \log x + 6 \tan^{- 1} x +

constant

4 None of the above

Explanation:

(C) : We have differential equation,
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$
Now separating the variable, we get-
$\frac{4 y d y}{(1 + y^{2})} = \frac{3 (1 + x)^{2}}{x (1 + x^{2})} \Rightarrow \frac{4 y d y}{1 + y^{2}} = (\frac{3}{x} + \frac{6}{1 + x^{2}})$
Now integrating both side, we get-
$\int \frac{4 y d y}{1 + y^{2}} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) d x$
Let, $1 + y^{2} = t$
$2 ydy = dt$
$2 \int \frac{dt}{t} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) dx$
$2 \log t = 3 \log | x | + 6 \tan^{- 1} x + c$
$2 \log (1 + y^{2}) = 3 \log | x | + 6 \tan^{- 1} x + c$

UPSEE-2011

Differential Equation

87230 If $y^{2} = P (x)$ be a cubic polynomial, then
$2 \frac{d}{d x} (y^{3} \frac{d^{2} y}{d x^{2}})$ is equal to

1

P^{'''} (x) + P^{'} (x)

2

P^{''} (x) P^{'''} (x)

3

P (x) P^{'''} (x)

4 constant

Explanation:

(C) : Given,
$y^{2} = P (x)$
Differentiating w.r.t $x$ , we get-
$2 y \frac{d y}{d x} = P^{'} (x)$
$\frac{d y}{d x} = \frac{P^{'} (x)}{2 y}$
Again differentiating w.r.t $x$ , we get-
$2 [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = P^{''} (x)$
$2 y \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} = P^{''} (x)$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 (d y / d x)^{2}}{2 y}$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 {(\frac{P^{'} (x)}{2 y})}^{2}}{2 y}$
Multiplying by $y^{3}$ in both side we get-
$y^{3} (\frac{d^{2} y}{d^{2}}) = \frac{y^{3} P^{''} (x)}{2 y} - \frac{2 y^{3}}{2 y} {(\frac{P^{'} (x)}{2 y})}^{2}$
$y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{y^{2} P^{''} (x)}{2} - y^{2} {(\frac{P^{'} (x)}{2 y})}^{2}$
Again differentiating with respect to $x$ we get-
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{2}{2} \frac{d}{d x} {y^{2} P^{''} (x)} - \frac{2}{4} \frac{d}{d x} {P^{'} (x)}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{d}{d x} [P (x) P^{''} (x)] - \frac{1}{2} \frac{d}{d x} {[P^{'} (x)]}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = [P (x) P^{''} (x) + P^{''} (x) P^{'} (x)]$
$- \frac{1}{2} 2 P^{'} (x) P^{''} (x)$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = P (x) P^{'''} (x)$

UPSEE-2010

Differential Equation

87231 The solution of differential equation $x \cos^{2} y dx$ $= y \cos^{2} x d y$ is

1

x \tan x - y \tan y - \log (\sec x / \sec y) = c

2

y \tan x - x \tan x - \log (\sec x \cdot \sec y) = c

3

x \tan x - y \tan y + \log (\sec x \cdot \sec y) = c

4 None of the above

Explanation:

(A) : Thę differential ęquation-
$x \cos^{2} y d x = y \cos^{2} x d y$
Separating the variable, we get-
$\frac{x}{\cos^{2} x} d x = \frac{y}{\cos^{2} y} d y$
Integrating both side, we get-
$\int x \sec^{2} x d x = \int y \sec^{2} y d y$
$x \int \sec^{2} x d x - \int \frac{d}{d x} (x) \cdot \int \sec^{2} x d x = y \int \sec^{2} y d y$
$x \tan x - \log | \sec x | = y \tan y - \log | \sec y | + c$
$- \int \frac{d}{d x} (y) \cdot \int \sec^{2} y d y$
$x \tan x - y \tan y = \log (\sec x / \sec y) + c$
$x \tan x - y \tan y - \log (\sec x / \sec y) = c$

UPSEE-2010]**#

Differential Equation

87232 The solution of the differential equation
$\frac{d y}{d x} = y \tan x - 2 \sin x is$

1

y \sin x = c + \sin 2 x

2

y \cos x = c + \frac{1}{2} \sin 2 x

3

y \cos x = c - \sin 2 x

4

y \cos x = c + \frac{1}{2} \cos 2 x

Explanation:

(D) : We have differential equation-
$\frac{d y}{d x} = y \tan x - 2 \sin x \Rightarrow \frac{d y}{d x} - y \tan x = - 2 \sin x$
Which is form of $\frac{d y}{d x} + P y = Q$ $P = - \tan x$ and $Q = - 2 \sin x$
$I . F = e^{\int P d x} = e^{\int - \tan x d x} = e^{- \log \sec x} = \cos x$
Now particular solution, we get-
$y \cos x = \int \cos x (- 2 \sin x) d x + c$
$y \cos x = - \int \sin 2 x + c$
$y \cos x = \frac{1}{2} \cos 2 x + c$

UPSEE-2007

Differential Equation

87229 The solution of differential equation
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$ is

1

\log (1 + y) = \log x + 2 \tan x +

constant

2

\log (1 + y^{2}) = 3 \log (\frac{1}{x}) + 6 \tan^{- 1} x +

constant

3

2 \log (1 + y^{2}) = 3 \log x + 6 \tan^{- 1} x +

constant

4 None of the above

Explanation:

(C) : We have differential equation,
$4 x y \frac{d y}{d x} = \frac{3 (1 + x)^{2} (1 + y^{2})}{(1 + x^{2})}$
Now separating the variable, we get-
$\frac{4 y d y}{(1 + y^{2})} = \frac{3 (1 + x)^{2}}{x (1 + x^{2})} \Rightarrow \frac{4 y d y}{1 + y^{2}} = (\frac{3}{x} + \frac{6}{1 + x^{2}})$
Now integrating both side, we get-
$\int \frac{4 y d y}{1 + y^{2}} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) d x$
Let, $1 + y^{2} = t$
$2 ydy = dt$
$2 \int \frac{dt}{t} = \int (\frac{3}{x} + \frac{6}{1 + x^{2}}) dx$
$2 \log t = 3 \log | x | + 6 \tan^{- 1} x + c$
$2 \log (1 + y^{2}) = 3 \log | x | + 6 \tan^{- 1} x + c$

UPSEE-2011

Differential Equation

87230 If $y^{2} = P (x)$ be a cubic polynomial, then
$2 \frac{d}{d x} (y^{3} \frac{d^{2} y}{d x^{2}})$ is equal to

1

P^{'''} (x) + P^{'} (x)

2

P^{''} (x) P^{'''} (x)

3

P (x) P^{'''} (x)

4 constant

Explanation:

(C) : Given,
$y^{2} = P (x)$
Differentiating w.r.t $x$ , we get-
$2 y \frac{d y}{d x} = P^{'} (x)$
$\frac{d y}{d x} = \frac{P^{'} (x)}{2 y}$
Again differentiating w.r.t $x$ , we get-
$2 [y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}] = P^{''} (x)$
$2 y \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{2} = P^{''} (x)$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 (d y / d x)^{2}}{2 y}$
$\frac{d^{2} y}{d x^{2}} = \frac{P^{''} (x) - 2 {(\frac{P^{'} (x)}{2 y})}^{2}}{2 y}$
Multiplying by $y^{3}$ in both side we get-
$y^{3} (\frac{d^{2} y}{d^{2}}) = \frac{y^{3} P^{''} (x)}{2 y} - \frac{2 y^{3}}{2 y} {(\frac{P^{'} (x)}{2 y})}^{2}$
$y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{y^{2} P^{''} (x)}{2} - y^{2} {(\frac{P^{'} (x)}{2 y})}^{2}$
Again differentiating with respect to $x$ we get-
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{2}{2} \frac{d}{d x} {y^{2} P^{''} (x)} - \frac{2}{4} \frac{d}{d x} {P^{'} (x)}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = \frac{d}{d x} [P (x) P^{''} (x)] - \frac{1}{2} \frac{d}{d x} {[P^{'} (x)]}^{2}$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = [P (x) P^{''} (x) + P^{''} (x) P^{'} (x)]$
$- \frac{1}{2} 2 P^{'} (x) P^{''} (x)$
$2 \frac{d}{d x} y^{3} (\frac{d^{2} y}{d x^{2}}) = P (x) P^{'''} (x)$

UPSEE-2010

Differential Equation

87231 The solution of differential equation $x \cos^{2} y dx$ $= y \cos^{2} x d y$ is

1

x \tan x - y \tan y - \log (\sec x / \sec y) = c

2

y \tan x - x \tan x - \log (\sec x \cdot \sec y) = c

3

x \tan x - y \tan y + \log (\sec x \cdot \sec y) = c

4 None of the above

Explanation:

(A) : Thę differential ęquation-
$x \cos^{2} y d x = y \cos^{2} x d y$
Separating the variable, we get-
$\frac{x}{\cos^{2} x} d x = \frac{y}{\cos^{2} y} d y$
Integrating both side, we get-
$\int x \sec^{2} x d x = \int y \sec^{2} y d y$
$x \int \sec^{2} x d x - \int \frac{d}{d x} (x) \cdot \int \sec^{2} x d x = y \int \sec^{2} y d y$
$x \tan x - \log | \sec x | = y \tan y - \log | \sec y | + c$
$- \int \frac{d}{d x} (y) \cdot \int \sec^{2} y d y$
$x \tan x - y \tan y = \log (\sec x / \sec y) + c$
$x \tan x - y \tan y - \log (\sec x / \sec y) = c$

UPSEE-2010]**#

Differential Equation

87232 The solution of the differential equation
$\frac{d y}{d x} = y \tan x - 2 \sin x is$

1

y \sin x = c + \sin 2 x

2

y \cos x = c + \frac{1}{2} \sin 2 x

3

y \cos x = c - \sin 2 x

4

y \cos x = c + \frac{1}{2} \cos 2 x

Explanation:

(D) : We have differential equation-
$\frac{d y}{d x} = y \tan x - 2 \sin x \Rightarrow \frac{d y}{d x} - y \tan x = - 2 \sin x$
Which is form of $\frac{d y}{d x} + P y = Q$ $P = - \tan x$ and $Q = - 2 \sin x$
$I . F = e^{\int P d x} = e^{\int - \tan x d x} = e^{- \log \sec x} = \cos x$
Now particular solution, we get-
$y \cos x = \int \cos x (- 2 \sin x) d x + c$
$y \cos x = - \int \sin 2 x + c$
$y \cos x = \frac{1}{2} \cos 2 x + c$

UPSEE-2007

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Question Bank Series

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