QBManager

Differential Equation

87545 Which one of the following is correct solution $(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y) ?$

1

\sec x = C - 2 \sec y

2

\sec y = C + 2 \cos y

3

\sec y = C - 2 \cos x

4

\sec x = C - 2 \cos y

Explanation:

(C) : We have
$(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y)$
$\frac{dy}{dx} (\tan y) = 2 \sin x \cos y$
$\frac{\sin y d y}{\cos^{2} y} = 2 \sin x d x$
On integrating both side we get
$\int \frac{\sin y}{\cos^{2} y} d y = \int 2 \sin x d x$
Put $\cos y = t$
$- \sin y d y = d t$
$- \int \frac{d t}{t^{2}} = 2 \int \sin x d x + C$
$- (\frac{1}{- t}) = - 2 \cos x + C$
$\frac{1}{t} = - 2 \cos x + C$
$\frac{1}{\cos y} = C - 2 \cos x$
$\sec y = C - 2 \cos x$

AMU-2014

Differential Equation

87546 If $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$ , then $| f (x y) |$ is equal to

1

{ke}^{x^{2} / 2}

2

{ke}^{y^{2} / 2}

3

{ke}^{x^{2}}

4

{ke}^{y^{2}}

Explanation:

(A): Given, $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$
$\frac{d}{d x} (x y) = x \frac{f (x y)}{f^{'} (x y)}$
$\frac{f^{'} (x y)}{f (x y)} d (x y) = x d x$
By integrating
$\int \frac{f^{'} (x y)}{f (x y)} d (x y) = \int x d x$
$\log [f (x y)] = \frac{x^{2}}{2} + C$
$f (x y) = e^{\frac{x^{2}}{2} + C}$
$f (x y) = e^{\frac{x^{2}}{2}} e^{C} \Rightarrow f (x y) = k e^{\frac{x^{2}}{2}}$

WB JEE-2021

Differential Equation

87547 Find the solution of the following differential equation : $\sin^{- 1} (\frac{d y}{d x}) = x + y$

1

x = \tan (x + y) + \sec (x + y) + c

2

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

3

x = \tan (x + y) + \sec^{2} (x + y) + c

4

x = \tan (x + y) - \sec^{2} (x + y) + c

Explanation:

(B) : Given differential equation-
$\sin^{- 1} (\frac{d y}{d x}) = x + y$
$\frac{d y}{d x} = \sin (x + y)$
Put $x + y = t$
$1 + \frac{dy}{dx} = \frac{dt}{dx}$
Or $\frac{d t}{d x} - 1 = \frac{d y}{d x}$
Put the above value in equation (i) we get-
$\frac{dt}{dx} - 1 = \sin t or \frac{dt}{1 + \sin t} = dx$
$\frac{1 - \sin t}{1 - \sin^{2} t} dt = dx or \frac{1 - \sin t}{\cos^{2} t}$
$Or \sec^{2} tdt - \sec t \tan t dt = dx$
$Now, integrating we get-$
$\tan t - \sec t = x + C$
$Put, t = x + y$
$\tan (x + y) - \sec (x + y) = x + C$
$x = \tan (x + y) - \sec (x + y) + C$

AP EAMCET-05.10.2021

Differential Equation

87549 In a triangle $P Q R, A, B$ and $C$ are the angles opposite the corresponding sides of lengths $a, b$ and $c$ respectively. If the side are $a = 5, b = 13$ and $c = 12$ then $\sin \frac{B}{2} + \cos \frac{B}{2} =$

1

\frac{1}{2}

2

\frac{1}{\sqrt{2}}

3

\sqrt{2}

4 1

Explanation:

(C) : Given,
$a_{2} = 5, b = 13, c = 12$
$a^{2} + c^{2} = b^{2},$
$(5)^{2} + (12)^{2} = (13)^{2}$
$169 = 169$
$∴ △ PQR$ is right angled triangle at $Q$
$∠ B = 90^{\circ} \Rightarrow ∠ \frac{B}{2} = 45^{\circ}$
So, $\sin \frac{B}{2} + \cos \frac{B}{2}$
$= \sin 45^{\circ} + \cos 45^{\circ}$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$

J and K CET-2017

Differential Equation

87550 Let $y = y (x)$ be the solution of the differential equation. $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) =$ $\log_{e} 4 - 1$ , then $y (e)$ is equal to

1

- \frac{e}{2}

2

- \frac{e^{2}}{2}

3

\frac{e}{4}

4

\frac{e^{2}}{4}

Explanation:

(C) : Given differential equation is-
$\frac{x d y}{d x} + y = x \log_{e} x (x > 1)$
$\frac{d y}{d x} + \frac{1}{x} y = \log_{e} x$
So, IF $= e^{\int \frac{1}{x} dx} = e^{\log_{c} x} = x$
Now, solution of differential equation (i), is-
$y \times x = \int (\log_{e} x) x d x + C$
$y x = \frac{x^{2}}{2} \log_{e} x - \int \frac{x^{2}}{2} \times \frac{1}{x} d x + C$
(Using integration by parts)
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4} + C$
Given that $2 y (2) = \log_{e} 4 - 1$
On substituting, $x = 2$ in equation (ii) we get-
$2 y (2) = \frac{4}{2} \log_{e} 2 - \frac{4}{4} + C$
[Where, $y (2)$ represents value of $x$ at $x = 2$ ]
$2 y (2) = \log_{e} 4 - 1 + C$
$(∵ m \log a = \log a^{m})$
From equation (iii) and (iv), we get-
$C = 0$
So, required solution is,
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4}$
Now, at $x = e, e y (e) = \frac{e^{2}}{2} \log_{e} e - \frac{e^{2}}{4}$
(Where, $y (e)$ represents value of $y$ at $x = e)$
$y (e) = \frac{e}{4} (∵ \log_{e} e = 1)$

JEE Main 12.01.2019. Shift-I]**#

Differential Equation

87545 Which one of the following is correct solution $(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y) ?$

1

\sec x = C - 2 \sec y

2

\sec y = C + 2 \cos y

3

\sec y = C - 2 \cos x

4

\sec x = C - 2 \cos y

Explanation:

(C) : We have
$(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y)$
$\frac{dy}{dx} (\tan y) = 2 \sin x \cos y$
$\frac{\sin y d y}{\cos^{2} y} = 2 \sin x d x$
On integrating both side we get
$\int \frac{\sin y}{\cos^{2} y} d y = \int 2 \sin x d x$
Put $\cos y = t$
$- \sin y d y = d t$
$- \int \frac{d t}{t^{2}} = 2 \int \sin x d x + C$
$- (\frac{1}{- t}) = - 2 \cos x + C$
$\frac{1}{t} = - 2 \cos x + C$
$\frac{1}{\cos y} = C - 2 \cos x$
$\sec y = C - 2 \cos x$

AMU-2014

Differential Equation

87546 If $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$ , then $| f (x y) |$ is equal to

1

{ke}^{x^{2} / 2}

2

{ke}^{y^{2} / 2}

3

{ke}^{x^{2}}

4

{ke}^{y^{2}}

Explanation:

(A): Given, $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$
$\frac{d}{d x} (x y) = x \frac{f (x y)}{f^{'} (x y)}$
$\frac{f^{'} (x y)}{f (x y)} d (x y) = x d x$
By integrating
$\int \frac{f^{'} (x y)}{f (x y)} d (x y) = \int x d x$
$\log [f (x y)] = \frac{x^{2}}{2} + C$
$f (x y) = e^{\frac{x^{2}}{2} + C}$
$f (x y) = e^{\frac{x^{2}}{2}} e^{C} \Rightarrow f (x y) = k e^{\frac{x^{2}}{2}}$

WB JEE-2021

Differential Equation

87547 Find the solution of the following differential equation : $\sin^{- 1} (\frac{d y}{d x}) = x + y$

1

x = \tan (x + y) + \sec (x + y) + c

2

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

3

x = \tan (x + y) + \sec^{2} (x + y) + c

4

x = \tan (x + y) - \sec^{2} (x + y) + c

Explanation:

(B) : Given differential equation-
$\sin^{- 1} (\frac{d y}{d x}) = x + y$
$\frac{d y}{d x} = \sin (x + y)$
Put $x + y = t$
$1 + \frac{dy}{dx} = \frac{dt}{dx}$
Or $\frac{d t}{d x} - 1 = \frac{d y}{d x}$
Put the above value in equation (i) we get-
$\frac{dt}{dx} - 1 = \sin t or \frac{dt}{1 + \sin t} = dx$
$\frac{1 - \sin t}{1 - \sin^{2} t} dt = dx or \frac{1 - \sin t}{\cos^{2} t}$
$Or \sec^{2} tdt - \sec t \tan t dt = dx$
$Now, integrating we get-$
$\tan t - \sec t = x + C$
$Put, t = x + y$
$\tan (x + y) - \sec (x + y) = x + C$
$x = \tan (x + y) - \sec (x + y) + C$

AP EAMCET-05.10.2021

Differential Equation

87549 In a triangle $P Q R, A, B$ and $C$ are the angles opposite the corresponding sides of lengths $a, b$ and $c$ respectively. If the side are $a = 5, b = 13$ and $c = 12$ then $\sin \frac{B}{2} + \cos \frac{B}{2} =$

1

\frac{1}{2}

2

\frac{1}{\sqrt{2}}

3

\sqrt{2}

4 1

Explanation:

(C) : Given,
$a_{2} = 5, b = 13, c = 12$
$a^{2} + c^{2} = b^{2},$
$(5)^{2} + (12)^{2} = (13)^{2}$
$169 = 169$
$∴ △ PQR$ is right angled triangle at $Q$
$∠ B = 90^{\circ} \Rightarrow ∠ \frac{B}{2} = 45^{\circ}$
So, $\sin \frac{B}{2} + \cos \frac{B}{2}$
$= \sin 45^{\circ} + \cos 45^{\circ}$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$

J and K CET-2017

Differential Equation

87550 Let $y = y (x)$ be the solution of the differential equation. $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) =$ $\log_{e} 4 - 1$ , then $y (e)$ is equal to

1

- \frac{e}{2}

2

- \frac{e^{2}}{2}

3

\frac{e}{4}

4

\frac{e^{2}}{4}

Explanation:

(C) : Given differential equation is-
$\frac{x d y}{d x} + y = x \log_{e} x (x > 1)$
$\frac{d y}{d x} + \frac{1}{x} y = \log_{e} x$
So, IF $= e^{\int \frac{1}{x} dx} = e^{\log_{c} x} = x$
Now, solution of differential equation (i), is-
$y \times x = \int (\log_{e} x) x d x + C$
$y x = \frac{x^{2}}{2} \log_{e} x - \int \frac{x^{2}}{2} \times \frac{1}{x} d x + C$
(Using integration by parts)
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4} + C$
Given that $2 y (2) = \log_{e} 4 - 1$
On substituting, $x = 2$ in equation (ii) we get-
$2 y (2) = \frac{4}{2} \log_{e} 2 - \frac{4}{4} + C$
[Where, $y (2)$ represents value of $x$ at $x = 2$ ]
$2 y (2) = \log_{e} 4 - 1 + C$
$(∵ m \log a = \log a^{m})$
From equation (iii) and (iv), we get-
$C = 0$
So, required solution is,
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4}$
Now, at $x = e, e y (e) = \frac{e^{2}}{2} \log_{e} e - \frac{e^{2}}{4}$
(Where, $y (e)$ represents value of $y$ at $x = e)$
$y (e) = \frac{e}{4} (∵ \log_{e} e = 1)$

JEE Main 12.01.2019. Shift-I]**#

Differential Equation

87545 Which one of the following is correct solution $(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y) ?$

1

\sec x = C - 2 \sec y

2

\sec y = C + 2 \cos y

3

\sec y = C - 2 \cos x

4

\sec x = C - 2 \cos y

Explanation:

(C) : We have
$(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y)$
$\frac{dy}{dx} (\tan y) = 2 \sin x \cos y$
$\frac{\sin y d y}{\cos^{2} y} = 2 \sin x d x$
On integrating both side we get
$\int \frac{\sin y}{\cos^{2} y} d y = \int 2 \sin x d x$
Put $\cos y = t$
$- \sin y d y = d t$
$- \int \frac{d t}{t^{2}} = 2 \int \sin x d x + C$
$- (\frac{1}{- t}) = - 2 \cos x + C$
$\frac{1}{t} = - 2 \cos x + C$
$\frac{1}{\cos y} = C - 2 \cos x$
$\sec y = C - 2 \cos x$

AMU-2014

Differential Equation

87546 If $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$ , then $| f (x y) |$ is equal to

1

{ke}^{x^{2} / 2}

2

{ke}^{y^{2} / 2}

3

{ke}^{x^{2}}

4

{ke}^{y^{2}}

Explanation:

(A): Given, $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$
$\frac{d}{d x} (x y) = x \frac{f (x y)}{f^{'} (x y)}$
$\frac{f^{'} (x y)}{f (x y)} d (x y) = x d x$
By integrating
$\int \frac{f^{'} (x y)}{f (x y)} d (x y) = \int x d x$
$\log [f (x y)] = \frac{x^{2}}{2} + C$
$f (x y) = e^{\frac{x^{2}}{2} + C}$
$f (x y) = e^{\frac{x^{2}}{2}} e^{C} \Rightarrow f (x y) = k e^{\frac{x^{2}}{2}}$

WB JEE-2021

Differential Equation

87547 Find the solution of the following differential equation : $\sin^{- 1} (\frac{d y}{d x}) = x + y$

1

x = \tan (x + y) + \sec (x + y) + c

2

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

3

x = \tan (x + y) + \sec^{2} (x + y) + c

4

x = \tan (x + y) - \sec^{2} (x + y) + c

Explanation:

(B) : Given differential equation-
$\sin^{- 1} (\frac{d y}{d x}) = x + y$
$\frac{d y}{d x} = \sin (x + y)$
Put $x + y = t$
$1 + \frac{dy}{dx} = \frac{dt}{dx}$
Or $\frac{d t}{d x} - 1 = \frac{d y}{d x}$
Put the above value in equation (i) we get-
$\frac{dt}{dx} - 1 = \sin t or \frac{dt}{1 + \sin t} = dx$
$\frac{1 - \sin t}{1 - \sin^{2} t} dt = dx or \frac{1 - \sin t}{\cos^{2} t}$
$Or \sec^{2} tdt - \sec t \tan t dt = dx$
$Now, integrating we get-$
$\tan t - \sec t = x + C$
$Put, t = x + y$
$\tan (x + y) - \sec (x + y) = x + C$
$x = \tan (x + y) - \sec (x + y) + C$

AP EAMCET-05.10.2021

Differential Equation

87549 In a triangle $P Q R, A, B$ and $C$ are the angles opposite the corresponding sides of lengths $a, b$ and $c$ respectively. If the side are $a = 5, b = 13$ and $c = 12$ then $\sin \frac{B}{2} + \cos \frac{B}{2} =$

1

\frac{1}{2}

2

\frac{1}{\sqrt{2}}

3

\sqrt{2}

4 1

Explanation:

(C) : Given,
$a_{2} = 5, b = 13, c = 12$
$a^{2} + c^{2} = b^{2},$
$(5)^{2} + (12)^{2} = (13)^{2}$
$169 = 169$
$∴ △ PQR$ is right angled triangle at $Q$
$∠ B = 90^{\circ} \Rightarrow ∠ \frac{B}{2} = 45^{\circ}$
So, $\sin \frac{B}{2} + \cos \frac{B}{2}$
$= \sin 45^{\circ} + \cos 45^{\circ}$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$

J and K CET-2017

Differential Equation

87550 Let $y = y (x)$ be the solution of the differential equation. $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) =$ $\log_{e} 4 - 1$ , then $y (e)$ is equal to

1

- \frac{e}{2}

2

- \frac{e^{2}}{2}

3

\frac{e}{4}

4

\frac{e^{2}}{4}

Explanation:

(C) : Given differential equation is-
$\frac{x d y}{d x} + y = x \log_{e} x (x > 1)$
$\frac{d y}{d x} + \frac{1}{x} y = \log_{e} x$
So, IF $= e^{\int \frac{1}{x} dx} = e^{\log_{c} x} = x$
Now, solution of differential equation (i), is-
$y \times x = \int (\log_{e} x) x d x + C$
$y x = \frac{x^{2}}{2} \log_{e} x - \int \frac{x^{2}}{2} \times \frac{1}{x} d x + C$
(Using integration by parts)
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4} + C$
Given that $2 y (2) = \log_{e} 4 - 1$
On substituting, $x = 2$ in equation (ii) we get-
$2 y (2) = \frac{4}{2} \log_{e} 2 - \frac{4}{4} + C$
[Where, $y (2)$ represents value of $x$ at $x = 2$ ]
$2 y (2) = \log_{e} 4 - 1 + C$
$(∵ m \log a = \log a^{m})$
From equation (iii) and (iv), we get-
$C = 0$
So, required solution is,
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4}$
Now, at $x = e, e y (e) = \frac{e^{2}}{2} \log_{e} e - \frac{e^{2}}{4}$
(Where, $y (e)$ represents value of $y$ at $x = e)$
$y (e) = \frac{e}{4} (∵ \log_{e} e = 1)$

JEE Main 12.01.2019. Shift-I]**#

Differential Equation

87545 Which one of the following is correct solution $(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y) ?$

1

\sec x = C - 2 \sec y

2

\sec y = C + 2 \cos y

3

\sec y = C - 2 \cos x

4

\sec x = C - 2 \cos y

Explanation:

(C) : We have
$(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y)$
$\frac{dy}{dx} (\tan y) = 2 \sin x \cos y$
$\frac{\sin y d y}{\cos^{2} y} = 2 \sin x d x$
On integrating both side we get
$\int \frac{\sin y}{\cos^{2} y} d y = \int 2 \sin x d x$
Put $\cos y = t$
$- \sin y d y = d t$
$- \int \frac{d t}{t^{2}} = 2 \int \sin x d x + C$
$- (\frac{1}{- t}) = - 2 \cos x + C$
$\frac{1}{t} = - 2 \cos x + C$
$\frac{1}{\cos y} = C - 2 \cos x$
$\sec y = C - 2 \cos x$

AMU-2014

Differential Equation

87546 If $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$ , then $| f (x y) |$ is equal to

1

{ke}^{x^{2} / 2}

2

{ke}^{y^{2} / 2}

3

{ke}^{x^{2}}

4

{ke}^{y^{2}}

Explanation:

(A): Given, $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$
$\frac{d}{d x} (x y) = x \frac{f (x y)}{f^{'} (x y)}$
$\frac{f^{'} (x y)}{f (x y)} d (x y) = x d x$
By integrating
$\int \frac{f^{'} (x y)}{f (x y)} d (x y) = \int x d x$
$\log [f (x y)] = \frac{x^{2}}{2} + C$
$f (x y) = e^{\frac{x^{2}}{2} + C}$
$f (x y) = e^{\frac{x^{2}}{2}} e^{C} \Rightarrow f (x y) = k e^{\frac{x^{2}}{2}}$

WB JEE-2021

Differential Equation

87547 Find the solution of the following differential equation : $\sin^{- 1} (\frac{d y}{d x}) = x + y$

1

x = \tan (x + y) + \sec (x + y) + c

2

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

3

x = \tan (x + y) + \sec^{2} (x + y) + c

4

x = \tan (x + y) - \sec^{2} (x + y) + c

Explanation:

(B) : Given differential equation-
$\sin^{- 1} (\frac{d y}{d x}) = x + y$
$\frac{d y}{d x} = \sin (x + y)$
Put $x + y = t$
$1 + \frac{dy}{dx} = \frac{dt}{dx}$
Or $\frac{d t}{d x} - 1 = \frac{d y}{d x}$
Put the above value in equation (i) we get-
$\frac{dt}{dx} - 1 = \sin t or \frac{dt}{1 + \sin t} = dx$
$\frac{1 - \sin t}{1 - \sin^{2} t} dt = dx or \frac{1 - \sin t}{\cos^{2} t}$
$Or \sec^{2} tdt - \sec t \tan t dt = dx$
$Now, integrating we get-$
$\tan t - \sec t = x + C$
$Put, t = x + y$
$\tan (x + y) - \sec (x + y) = x + C$
$x = \tan (x + y) - \sec (x + y) + C$

AP EAMCET-05.10.2021

Differential Equation

87549 In a triangle $P Q R, A, B$ and $C$ are the angles opposite the corresponding sides of lengths $a, b$ and $c$ respectively. If the side are $a = 5, b = 13$ and $c = 12$ then $\sin \frac{B}{2} + \cos \frac{B}{2} =$

1

\frac{1}{2}

2

\frac{1}{\sqrt{2}}

3

\sqrt{2}

4 1

Explanation:

(C) : Given,
$a_{2} = 5, b = 13, c = 12$
$a^{2} + c^{2} = b^{2},$
$(5)^{2} + (12)^{2} = (13)^{2}$
$169 = 169$
$∴ △ PQR$ is right angled triangle at $Q$
$∠ B = 90^{\circ} \Rightarrow ∠ \frac{B}{2} = 45^{\circ}$
So, $\sin \frac{B}{2} + \cos \frac{B}{2}$
$= \sin 45^{\circ} + \cos 45^{\circ}$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$

J and K CET-2017

Differential Equation

87550 Let $y = y (x)$ be the solution of the differential equation. $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) =$ $\log_{e} 4 - 1$ , then $y (e)$ is equal to

1

- \frac{e}{2}

2

- \frac{e^{2}}{2}

3

\frac{e}{4}

4

\frac{e^{2}}{4}

Explanation:

(C) : Given differential equation is-
$\frac{x d y}{d x} + y = x \log_{e} x (x > 1)$
$\frac{d y}{d x} + \frac{1}{x} y = \log_{e} x$
So, IF $= e^{\int \frac{1}{x} dx} = e^{\log_{c} x} = x$
Now, solution of differential equation (i), is-
$y \times x = \int (\log_{e} x) x d x + C$
$y x = \frac{x^{2}}{2} \log_{e} x - \int \frac{x^{2}}{2} \times \frac{1}{x} d x + C$
(Using integration by parts)
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4} + C$
Given that $2 y (2) = \log_{e} 4 - 1$
On substituting, $x = 2$ in equation (ii) we get-
$2 y (2) = \frac{4}{2} \log_{e} 2 - \frac{4}{4} + C$
[Where, $y (2)$ represents value of $x$ at $x = 2$ ]
$2 y (2) = \log_{e} 4 - 1 + C$
$(∵ m \log a = \log a^{m})$
From equation (iii) and (iv), we get-
$C = 0$
So, required solution is,
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4}$
Now, at $x = e, e y (e) = \frac{e^{2}}{2} \log_{e} e - \frac{e^{2}}{4}$
(Where, $y (e)$ represents value of $y$ at $x = e)$
$y (e) = \frac{e}{4} (∵ \log_{e} e = 1)$

JEE Main 12.01.2019. Shift-I]**#

Differential Equation

87545 Which one of the following is correct solution $(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y) ?$

1

\sec x = C - 2 \sec y

2

\sec y = C + 2 \cos y

3

\sec y = C - 2 \cos x

4

\sec x = C - 2 \cos y

Explanation:

(C) : We have
$(\frac{d y}{d x}) \tan y = \sin (x + y) + \sin (x - y)$
$\frac{dy}{dx} (\tan y) = 2 \sin x \cos y$
$\frac{\sin y d y}{\cos^{2} y} = 2 \sin x d x$
On integrating both side we get
$\int \frac{\sin y}{\cos^{2} y} d y = \int 2 \sin x d x$
Put $\cos y = t$
$- \sin y d y = d t$
$- \int \frac{d t}{t^{2}} = 2 \int \sin x d x + C$
$- (\frac{1}{- t}) = - 2 \cos x + C$
$\frac{1}{t} = - 2 \cos x + C$
$\frac{1}{\cos y} = C - 2 \cos x$
$\sec y = C - 2 \cos x$

AMU-2014

Differential Equation

87546 If $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$ , then $| f (x y) |$ is equal to

1

{ke}^{x^{2} / 2}

2

{ke}^{y^{2} / 2}

3

{ke}^{x^{2}}

4

{ke}^{y^{2}}

Explanation:

(A): Given, $x \frac{d y}{d x} + y = \frac{x f (x y)}{f^{'} (x y)}$
$\frac{d}{d x} (x y) = x \frac{f (x y)}{f^{'} (x y)}$
$\frac{f^{'} (x y)}{f (x y)} d (x y) = x d x$
By integrating
$\int \frac{f^{'} (x y)}{f (x y)} d (x y) = \int x d x$
$\log [f (x y)] = \frac{x^{2}}{2} + C$
$f (x y) = e^{\frac{x^{2}}{2} + C}$
$f (x y) = e^{\frac{x^{2}}{2}} e^{C} \Rightarrow f (x y) = k e^{\frac{x^{2}}{2}}$

WB JEE-2021

Differential Equation

87547 Find the solution of the following differential equation : $\sin^{- 1} (\frac{d y}{d x}) = x + y$

1

x = \tan (x + y) + \sec (x + y) + c

2

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

3

x = \tan (x + y) + \sec^{2} (x + y) + c

4

x = \tan (x + y) - \sec^{2} (x + y) + c

Explanation:

(B) : Given differential equation-
$\sin^{- 1} (\frac{d y}{d x}) = x + y$
$\frac{d y}{d x} = \sin (x + y)$
Put $x + y = t$
$1 + \frac{dy}{dx} = \frac{dt}{dx}$
Or $\frac{d t}{d x} - 1 = \frac{d y}{d x}$
Put the above value in equation (i) we get-
$\frac{dt}{dx} - 1 = \sin t or \frac{dt}{1 + \sin t} = dx$
$\frac{1 - \sin t}{1 - \sin^{2} t} dt = dx or \frac{1 - \sin t}{\cos^{2} t}$
$Or \sec^{2} tdt - \sec t \tan t dt = dx$
$Now, integrating we get-$
$\tan t - \sec t = x + C$
$Put, t = x + y$
$\tan (x + y) - \sec (x + y) = x + C$
$x = \tan (x + y) - \sec (x + y) + C$

AP EAMCET-05.10.2021

Differential Equation

87549 In a triangle $P Q R, A, B$ and $C$ are the angles opposite the corresponding sides of lengths $a, b$ and $c$ respectively. If the side are $a = 5, b = 13$ and $c = 12$ then $\sin \frac{B}{2} + \cos \frac{B}{2} =$

1

\frac{1}{2}

2

\frac{1}{\sqrt{2}}

3

\sqrt{2}

4 1

Explanation:

(C) : Given,
$a_{2} = 5, b = 13, c = 12$
$a^{2} + c^{2} = b^{2},$
$(5)^{2} + (12)^{2} = (13)^{2}$
$169 = 169$
$∴ △ PQR$ is right angled triangle at $Q$
$∠ B = 90^{\circ} \Rightarrow ∠ \frac{B}{2} = 45^{\circ}$
So, $\sin \frac{B}{2} + \cos \frac{B}{2}$
$= \sin 45^{\circ} + \cos 45^{\circ}$
$= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \sqrt{2}$

J and K CET-2017

Differential Equation

87550 Let $y = y (x)$ be the solution of the differential equation. $x \frac{d y}{d x} + y = x \log_{e} x, (x > 1)$ . If $2 y (2) =$ $\log_{e} 4 - 1$ , then $y (e)$ is equal to

1

- \frac{e}{2}

2

- \frac{e^{2}}{2}

3

\frac{e}{4}

4

\frac{e^{2}}{4}

Explanation:

(C) : Given differential equation is-
$\frac{x d y}{d x} + y = x \log_{e} x (x > 1)$
$\frac{d y}{d x} + \frac{1}{x} y = \log_{e} x$
So, IF $= e^{\int \frac{1}{x} dx} = e^{\log_{c} x} = x$
Now, solution of differential equation (i), is-
$y \times x = \int (\log_{e} x) x d x + C$
$y x = \frac{x^{2}}{2} \log_{e} x - \int \frac{x^{2}}{2} \times \frac{1}{x} d x + C$
(Using integration by parts)
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4} + C$
Given that $2 y (2) = \log_{e} 4 - 1$
On substituting, $x = 2$ in equation (ii) we get-
$2 y (2) = \frac{4}{2} \log_{e} 2 - \frac{4}{4} + C$
[Where, $y (2)$ represents value of $x$ at $x = 2$ ]
$2 y (2) = \log_{e} 4 - 1 + C$
$(∵ m \log a = \log a^{m})$
From equation (iii) and (iv), we get-
$C = 0$
So, required solution is,
$y x = \frac{x^{2}}{2} \log_{e} x - \frac{x^{2}}{4}$
Now, at $x = e, e y (e) = \frac{e^{2}}{2} \log_{e} e - \frac{e^{2}}{4}$
(Where, $y (e)$ represents value of $y$ at $x = e)$
$y (e) = \frac{e}{4} (∵ \log_{e} e = 1)$

JEE Main 12.01.2019. Shift-I]**#

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