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Differential Equation

87180 The differential equation of all lines making intercept 3 on the $X$ -axis is

1

x \frac{d y}{d x} = 3 y

2

(x - 3) \frac{d y}{d x} = y

3

\frac{d y}{d x} = x - 3

4

\frac{d y}{d x} = x + 3

Explanation:

(B) : Given, differential equation of straight line. $y = m x + c$
Making intercept 3 on the $x$ axis then $y = 0$ $0 = 3 m + c$ $\begin{array}{r} ∴ c = - 3 m \\ ∴ y = mx - 3 m \end{array}$
Differentiating both side w.r.t. $x$ , we get-
$\frac{dy}{dx} = m$
Putting the value of $c$ and $m$ in equation (i), we get -
$y = m x + (- 3 m) \Rightarrow y = m (x - 3)$
$y = \frac{d y}{d x} (x - 3)$

MHT CET-2019

Differential Equation

87181 The particular solution of the differential equation $\log (\frac{d y}{d x}) = x$ , when $x = 0, y = 1$ , is

1

y = - e^{x} + 2

2

y = e^{x} + 2

3

y = e^{x}

4

y = - e^{x}

Explanation:

(C) : Given, differential equation,
$\log (\frac{d y}{d x}) = x$
Taking antilog on both sides, we get-
$\frac{d y}{d x} = e^{x} \Rightarrow d y = e^{x} d x$
Integrating on both sides, we get-
$\int d y = \int e^{x} d x \Rightarrow y = e^{x} + c$
When, $x = 0$ and $y = 1$
$1 = e^{0} + c \Rightarrow 1 = 1 + c$
$c = 0$
Putting the value of $c$ in equation (i)
Then, $y = e^{x}$

MHT CET-2019

Differential Equation

87192 The equation of the curve through the point
$(1, 0)$ and whose slope is $\frac{y - 1}{x^{2} + x}$ is

1

(y - 1) (x + 1) + 2 x = 0

2

2 x (y - 1) + x + 1 = 0

3

x (y - 1) (x + 1) + 2 = 0

4

y (x + 1) - x = 0

Explanation:

(A) : Given,
slope $= (\frac{d y}{d x}) = \frac{y - 1}{x^{2} + x} \Rightarrow \frac{d y}{y - 1} = \frac{d x}{x^{2} + x}$
Integrating on both sides we get,
$\int \frac{d y}{y - 1} = \int \frac{d x}{x^{2} + x} \Rightarrow \int \frac{d y}{y - 1} = \int \frac{d x}{x (x + 1)}$
$\int \frac{d y}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) d x$
$\log (y - 1) = \log x - \log (x + 1) + \log c$
$\log (y - 1) = \log (\frac{c x}{x + 1})$
$y - 1 = \frac{c x}{x + 1}$
Since, curve passes through point $(1, 0)$
$\begin{array}{r} 0 - 1 = \frac{1 . c}{1 + 1} \\ c = - 2 \end{array}$
Put the above value of 'c' in equation (i) we get -
$y - 1 = \frac{- 2 x}{x + 1}$
$(y - 1) (x + 1) = - 2 x$
$(y - 1) (x + 1) + 2 x = 0$

SRM JEEE-2012

Differential Equation

87193 The solution of the equation $\frac{d^{2} y}{d^{2}} = e^{- 2 x}$ is

1

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

2

\frac{e^{- 2 x}}{4} + c x + d

3

\frac{1}{4} e^{- 2 x} + {cx}^{2} + d

4

\frac{e^{- 4 x}}{4} + c x + d

Explanation:

(B) :Given, differential equation,
$\frac{d^{2} y}{d x^{2}} = e^{- 2 x}$
Integrating both sides we get,
$\frac{dy}{dx} = \frac{e^{- 2 x}}{- 2} + C$
Again integrating both sides we get,
$y = \frac{e^{- 2 x}}{(- 2) (- 2)} + c x + d$
$y = \frac{e^{- 2 x}}{4} + c x + d$

Differential Equation

87194 The differential equation of the family of curves $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

1

2 (x^{2} - y^{2}) y^{'} = x y

2

2 (x^{2} + y^{2}) y^{'} = x y

3

(x^{2} - y^{2}) y^{'} = 2 x y

4

(x^{2} + y^{2}) y^{'} = 2 x y

Explanation:

(C): $\begin{array}{r} Given, \\ x^{2} + y^{2} - 2 a y = 0 \end{array}$
Differentiating w.r.t.x we get,
$2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0 \Rightarrow x + (y - a) \frac{d y}{d x} = 0$
$x + {y - \frac{x^{2} + y^{2}}{2 y}} \frac{dy}{dx} = 0 [∵ a = \frac{x^{2} + y^{2}}{2 y}]$
$x + {\frac{y^{2} - x^{2}}{2 y}} \frac{d y}{d x} = 0 \Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} = 2 x y$
$∵ \frac{dy}{dx} = y^{'}$
$∴ (x^{2} - y^{2}) y^{'} = 2 xy$

SRM JEEE-2015

Differential Equation

87180 The differential equation of all lines making intercept 3 on the $X$ -axis is

1

x \frac{d y}{d x} = 3 y

2

(x - 3) \frac{d y}{d x} = y

3

\frac{d y}{d x} = x - 3

4

\frac{d y}{d x} = x + 3

Explanation:

(B) : Given, differential equation of straight line. $y = m x + c$
Making intercept 3 on the $x$ axis then $y = 0$ $0 = 3 m + c$ $\begin{array}{r} ∴ c = - 3 m \\ ∴ y = mx - 3 m \end{array}$
Differentiating both side w.r.t. $x$ , we get-
$\frac{dy}{dx} = m$
Putting the value of $c$ and $m$ in equation (i), we get -
$y = m x + (- 3 m) \Rightarrow y = m (x - 3)$
$y = \frac{d y}{d x} (x - 3)$

MHT CET-2019

Differential Equation

87181 The particular solution of the differential equation $\log (\frac{d y}{d x}) = x$ , when $x = 0, y = 1$ , is

1

y = - e^{x} + 2

2

y = e^{x} + 2

3

y = e^{x}

4

y = - e^{x}

Explanation:

(C) : Given, differential equation,
$\log (\frac{d y}{d x}) = x$
Taking antilog on both sides, we get-
$\frac{d y}{d x} = e^{x} \Rightarrow d y = e^{x} d x$
Integrating on both sides, we get-
$\int d y = \int e^{x} d x \Rightarrow y = e^{x} + c$
When, $x = 0$ and $y = 1$
$1 = e^{0} + c \Rightarrow 1 = 1 + c$
$c = 0$
Putting the value of $c$ in equation (i)
Then, $y = e^{x}$

MHT CET-2019

Differential Equation

87192 The equation of the curve through the point
$(1, 0)$ and whose slope is $\frac{y - 1}{x^{2} + x}$ is

1

(y - 1) (x + 1) + 2 x = 0

2

2 x (y - 1) + x + 1 = 0

3

x (y - 1) (x + 1) + 2 = 0

4

y (x + 1) - x = 0

Explanation:

(A) : Given,
slope $= (\frac{d y}{d x}) = \frac{y - 1}{x^{2} + x} \Rightarrow \frac{d y}{y - 1} = \frac{d x}{x^{2} + x}$
Integrating on both sides we get,
$\int \frac{d y}{y - 1} = \int \frac{d x}{x^{2} + x} \Rightarrow \int \frac{d y}{y - 1} = \int \frac{d x}{x (x + 1)}$
$\int \frac{d y}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) d x$
$\log (y - 1) = \log x - \log (x + 1) + \log c$
$\log (y - 1) = \log (\frac{c x}{x + 1})$
$y - 1 = \frac{c x}{x + 1}$
Since, curve passes through point $(1, 0)$
$\begin{array}{r} 0 - 1 = \frac{1 . c}{1 + 1} \\ c = - 2 \end{array}$
Put the above value of 'c' in equation (i) we get -
$y - 1 = \frac{- 2 x}{x + 1}$
$(y - 1) (x + 1) = - 2 x$
$(y - 1) (x + 1) + 2 x = 0$

SRM JEEE-2012

Differential Equation

87193 The solution of the equation $\frac{d^{2} y}{d^{2}} = e^{- 2 x}$ is

1

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

2

\frac{e^{- 2 x}}{4} + c x + d

3

\frac{1}{4} e^{- 2 x} + {cx}^{2} + d

4

\frac{e^{- 4 x}}{4} + c x + d

Explanation:

(B) :Given, differential equation,
$\frac{d^{2} y}{d x^{2}} = e^{- 2 x}$
Integrating both sides we get,
$\frac{dy}{dx} = \frac{e^{- 2 x}}{- 2} + C$
Again integrating both sides we get,
$y = \frac{e^{- 2 x}}{(- 2) (- 2)} + c x + d$
$y = \frac{e^{- 2 x}}{4} + c x + d$

Differential Equation

87194 The differential equation of the family of curves $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

1

2 (x^{2} - y^{2}) y^{'} = x y

2

2 (x^{2} + y^{2}) y^{'} = x y

3

(x^{2} - y^{2}) y^{'} = 2 x y

4

(x^{2} + y^{2}) y^{'} = 2 x y

Explanation:

(C): $\begin{array}{r} Given, \\ x^{2} + y^{2} - 2 a y = 0 \end{array}$
Differentiating w.r.t.x we get,
$2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0 \Rightarrow x + (y - a) \frac{d y}{d x} = 0$
$x + {y - \frac{x^{2} + y^{2}}{2 y}} \frac{dy}{dx} = 0 [∵ a = \frac{x^{2} + y^{2}}{2 y}]$
$x + {\frac{y^{2} - x^{2}}{2 y}} \frac{d y}{d x} = 0 \Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} = 2 x y$
$∵ \frac{dy}{dx} = y^{'}$
$∴ (x^{2} - y^{2}) y^{'} = 2 xy$

SRM JEEE-2015

Differential Equation

87180 The differential equation of all lines making intercept 3 on the $X$ -axis is

1

x \frac{d y}{d x} = 3 y

2

(x - 3) \frac{d y}{d x} = y

3

\frac{d y}{d x} = x - 3

4

\frac{d y}{d x} = x + 3

Explanation:

(B) : Given, differential equation of straight line. $y = m x + c$
Making intercept 3 on the $x$ axis then $y = 0$ $0 = 3 m + c$ $\begin{array}{r} ∴ c = - 3 m \\ ∴ y = mx - 3 m \end{array}$
Differentiating both side w.r.t. $x$ , we get-
$\frac{dy}{dx} = m$
Putting the value of $c$ and $m$ in equation (i), we get -
$y = m x + (- 3 m) \Rightarrow y = m (x - 3)$
$y = \frac{d y}{d x} (x - 3)$

MHT CET-2019

Differential Equation

87181 The particular solution of the differential equation $\log (\frac{d y}{d x}) = x$ , when $x = 0, y = 1$ , is

1

y = - e^{x} + 2

2

y = e^{x} + 2

3

y = e^{x}

4

y = - e^{x}

Explanation:

(C) : Given, differential equation,
$\log (\frac{d y}{d x}) = x$
Taking antilog on both sides, we get-
$\frac{d y}{d x} = e^{x} \Rightarrow d y = e^{x} d x$
Integrating on both sides, we get-
$\int d y = \int e^{x} d x \Rightarrow y = e^{x} + c$
When, $x = 0$ and $y = 1$
$1 = e^{0} + c \Rightarrow 1 = 1 + c$
$c = 0$
Putting the value of $c$ in equation (i)
Then, $y = e^{x}$

MHT CET-2019

Differential Equation

87192 The equation of the curve through the point
$(1, 0)$ and whose slope is $\frac{y - 1}{x^{2} + x}$ is

1

(y - 1) (x + 1) + 2 x = 0

2

2 x (y - 1) + x + 1 = 0

3

x (y - 1) (x + 1) + 2 = 0

4

y (x + 1) - x = 0

Explanation:

(A) : Given,
slope $= (\frac{d y}{d x}) = \frac{y - 1}{x^{2} + x} \Rightarrow \frac{d y}{y - 1} = \frac{d x}{x^{2} + x}$
Integrating on both sides we get,
$\int \frac{d y}{y - 1} = \int \frac{d x}{x^{2} + x} \Rightarrow \int \frac{d y}{y - 1} = \int \frac{d x}{x (x + 1)}$
$\int \frac{d y}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) d x$
$\log (y - 1) = \log x - \log (x + 1) + \log c$
$\log (y - 1) = \log (\frac{c x}{x + 1})$
$y - 1 = \frac{c x}{x + 1}$
Since, curve passes through point $(1, 0)$
$\begin{array}{r} 0 - 1 = \frac{1 . c}{1 + 1} \\ c = - 2 \end{array}$
Put the above value of 'c' in equation (i) we get -
$y - 1 = \frac{- 2 x}{x + 1}$
$(y - 1) (x + 1) = - 2 x$
$(y - 1) (x + 1) + 2 x = 0$

SRM JEEE-2012

Differential Equation

87193 The solution of the equation $\frac{d^{2} y}{d^{2}} = e^{- 2 x}$ is

1

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

2

\frac{e^{- 2 x}}{4} + c x + d

3

\frac{1}{4} e^{- 2 x} + {cx}^{2} + d

4

\frac{e^{- 4 x}}{4} + c x + d

Explanation:

(B) :Given, differential equation,
$\frac{d^{2} y}{d x^{2}} = e^{- 2 x}$
Integrating both sides we get,
$\frac{dy}{dx} = \frac{e^{- 2 x}}{- 2} + C$
Again integrating both sides we get,
$y = \frac{e^{- 2 x}}{(- 2) (- 2)} + c x + d$
$y = \frac{e^{- 2 x}}{4} + c x + d$

Differential Equation

87194 The differential equation of the family of curves $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

1

2 (x^{2} - y^{2}) y^{'} = x y

2

2 (x^{2} + y^{2}) y^{'} = x y

3

(x^{2} - y^{2}) y^{'} = 2 x y

4

(x^{2} + y^{2}) y^{'} = 2 x y

Explanation:

(C): $\begin{array}{r} Given, \\ x^{2} + y^{2} - 2 a y = 0 \end{array}$
Differentiating w.r.t.x we get,
$2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0 \Rightarrow x + (y - a) \frac{d y}{d x} = 0$
$x + {y - \frac{x^{2} + y^{2}}{2 y}} \frac{dy}{dx} = 0 [∵ a = \frac{x^{2} + y^{2}}{2 y}]$
$x + {\frac{y^{2} - x^{2}}{2 y}} \frac{d y}{d x} = 0 \Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} = 2 x y$
$∵ \frac{dy}{dx} = y^{'}$
$∴ (x^{2} - y^{2}) y^{'} = 2 xy$

SRM JEEE-2015

Differential Equation

87180 The differential equation of all lines making intercept 3 on the $X$ -axis is

1

x \frac{d y}{d x} = 3 y

2

(x - 3) \frac{d y}{d x} = y

3

\frac{d y}{d x} = x - 3

4

\frac{d y}{d x} = x + 3

Explanation:

(B) : Given, differential equation of straight line. $y = m x + c$
Making intercept 3 on the $x$ axis then $y = 0$ $0 = 3 m + c$ $\begin{array}{r} ∴ c = - 3 m \\ ∴ y = mx - 3 m \end{array}$
Differentiating both side w.r.t. $x$ , we get-
$\frac{dy}{dx} = m$
Putting the value of $c$ and $m$ in equation (i), we get -
$y = m x + (- 3 m) \Rightarrow y = m (x - 3)$
$y = \frac{d y}{d x} (x - 3)$

MHT CET-2019

Differential Equation

87181 The particular solution of the differential equation $\log (\frac{d y}{d x}) = x$ , when $x = 0, y = 1$ , is

1

y = - e^{x} + 2

2

y = e^{x} + 2

3

y = e^{x}

4

y = - e^{x}

Explanation:

(C) : Given, differential equation,
$\log (\frac{d y}{d x}) = x$
Taking antilog on both sides, we get-
$\frac{d y}{d x} = e^{x} \Rightarrow d y = e^{x} d x$
Integrating on both sides, we get-
$\int d y = \int e^{x} d x \Rightarrow y = e^{x} + c$
When, $x = 0$ and $y = 1$
$1 = e^{0} + c \Rightarrow 1 = 1 + c$
$c = 0$
Putting the value of $c$ in equation (i)
Then, $y = e^{x}$

MHT CET-2019

Differential Equation

87192 The equation of the curve through the point
$(1, 0)$ and whose slope is $\frac{y - 1}{x^{2} + x}$ is

1

(y - 1) (x + 1) + 2 x = 0

2

2 x (y - 1) + x + 1 = 0

3

x (y - 1) (x + 1) + 2 = 0

4

y (x + 1) - x = 0

Explanation:

(A) : Given,
slope $= (\frac{d y}{d x}) = \frac{y - 1}{x^{2} + x} \Rightarrow \frac{d y}{y - 1} = \frac{d x}{x^{2} + x}$
Integrating on both sides we get,
$\int \frac{d y}{y - 1} = \int \frac{d x}{x^{2} + x} \Rightarrow \int \frac{d y}{y - 1} = \int \frac{d x}{x (x + 1)}$
$\int \frac{d y}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) d x$
$\log (y - 1) = \log x - \log (x + 1) + \log c$
$\log (y - 1) = \log (\frac{c x}{x + 1})$
$y - 1 = \frac{c x}{x + 1}$
Since, curve passes through point $(1, 0)$
$\begin{array}{r} 0 - 1 = \frac{1 . c}{1 + 1} \\ c = - 2 \end{array}$
Put the above value of 'c' in equation (i) we get -
$y - 1 = \frac{- 2 x}{x + 1}$
$(y - 1) (x + 1) = - 2 x$
$(y - 1) (x + 1) + 2 x = 0$

SRM JEEE-2012

Differential Equation

87193 The solution of the equation $\frac{d^{2} y}{d^{2}} = e^{- 2 x}$ is

1

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

2

\frac{e^{- 2 x}}{4} + c x + d

3

\frac{1}{4} e^{- 2 x} + {cx}^{2} + d

4

\frac{e^{- 4 x}}{4} + c x + d

Explanation:

(B) :Given, differential equation,
$\frac{d^{2} y}{d x^{2}} = e^{- 2 x}$
Integrating both sides we get,
$\frac{dy}{dx} = \frac{e^{- 2 x}}{- 2} + C$
Again integrating both sides we get,
$y = \frac{e^{- 2 x}}{(- 2) (- 2)} + c x + d$
$y = \frac{e^{- 2 x}}{4} + c x + d$

Differential Equation

87194 The differential equation of the family of curves $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

1

2 (x^{2} - y^{2}) y^{'} = x y

2

2 (x^{2} + y^{2}) y^{'} = x y

3

(x^{2} - y^{2}) y^{'} = 2 x y

4

(x^{2} + y^{2}) y^{'} = 2 x y

Explanation:

(C): $\begin{array}{r} Given, \\ x^{2} + y^{2} - 2 a y = 0 \end{array}$
Differentiating w.r.t.x we get,
$2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0 \Rightarrow x + (y - a) \frac{d y}{d x} = 0$
$x + {y - \frac{x^{2} + y^{2}}{2 y}} \frac{dy}{dx} = 0 [∵ a = \frac{x^{2} + y^{2}}{2 y}]$
$x + {\frac{y^{2} - x^{2}}{2 y}} \frac{d y}{d x} = 0 \Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} = 2 x y$
$∵ \frac{dy}{dx} = y^{'}$
$∴ (x^{2} - y^{2}) y^{'} = 2 xy$

SRM JEEE-2015

Differential Equation

87180 The differential equation of all lines making intercept 3 on the $X$ -axis is

1

x \frac{d y}{d x} = 3 y

2

(x - 3) \frac{d y}{d x} = y

3

\frac{d y}{d x} = x - 3

4

\frac{d y}{d x} = x + 3

Explanation:

(B) : Given, differential equation of straight line. $y = m x + c$
Making intercept 3 on the $x$ axis then $y = 0$ $0 = 3 m + c$ $\begin{array}{r} ∴ c = - 3 m \\ ∴ y = mx - 3 m \end{array}$
Differentiating both side w.r.t. $x$ , we get-
$\frac{dy}{dx} = m$
Putting the value of $c$ and $m$ in equation (i), we get -
$y = m x + (- 3 m) \Rightarrow y = m (x - 3)$
$y = \frac{d y}{d x} (x - 3)$

MHT CET-2019

Differential Equation

87181 The particular solution of the differential equation $\log (\frac{d y}{d x}) = x$ , when $x = 0, y = 1$ , is

1

y = - e^{x} + 2

2

y = e^{x} + 2

3

y = e^{x}

4

y = - e^{x}

Explanation:

(C) : Given, differential equation,
$\log (\frac{d y}{d x}) = x$
Taking antilog on both sides, we get-
$\frac{d y}{d x} = e^{x} \Rightarrow d y = e^{x} d x$
Integrating on both sides, we get-
$\int d y = \int e^{x} d x \Rightarrow y = e^{x} + c$
When, $x = 0$ and $y = 1$
$1 = e^{0} + c \Rightarrow 1 = 1 + c$
$c = 0$
Putting the value of $c$ in equation (i)
Then, $y = e^{x}$

MHT CET-2019

Differential Equation

87192 The equation of the curve through the point
$(1, 0)$ and whose slope is $\frac{y - 1}{x^{2} + x}$ is

1

(y - 1) (x + 1) + 2 x = 0

2

2 x (y - 1) + x + 1 = 0

3

x (y - 1) (x + 1) + 2 = 0

4

y (x + 1) - x = 0

Explanation:

(A) : Given,
slope $= (\frac{d y}{d x}) = \frac{y - 1}{x^{2} + x} \Rightarrow \frac{d y}{y - 1} = \frac{d x}{x^{2} + x}$
Integrating on both sides we get,
$\int \frac{d y}{y - 1} = \int \frac{d x}{x^{2} + x} \Rightarrow \int \frac{d y}{y - 1} = \int \frac{d x}{x (x + 1)}$
$\int \frac{d y}{y - 1} = \int (\frac{1}{x} - \frac{1}{x + 1}) d x$
$\log (y - 1) = \log x - \log (x + 1) + \log c$
$\log (y - 1) = \log (\frac{c x}{x + 1})$
$y - 1 = \frac{c x}{x + 1}$
Since, curve passes through point $(1, 0)$
$\begin{array}{r} 0 - 1 = \frac{1 . c}{1 + 1} \\ c = - 2 \end{array}$
Put the above value of 'c' in equation (i) we get -
$y - 1 = \frac{- 2 x}{x + 1}$
$(y - 1) (x + 1) = - 2 x$
$(y - 1) (x + 1) + 2 x = 0$

SRM JEEE-2012

Differential Equation

87193 The solution of the equation $\frac{d^{2} y}{d^{2}} = e^{- 2 x}$ is

1

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

2

\frac{e^{- 2 x}}{4} + c x + d

3

\frac{1}{4} e^{- 2 x} + {cx}^{2} + d

4

\frac{e^{- 4 x}}{4} + c x + d

Explanation:

(B) :Given, differential equation,
$\frac{d^{2} y}{d x^{2}} = e^{- 2 x}$
Integrating both sides we get,
$\frac{dy}{dx} = \frac{e^{- 2 x}}{- 2} + C$
Again integrating both sides we get,
$y = \frac{e^{- 2 x}}{(- 2) (- 2)} + c x + d$
$y = \frac{e^{- 2 x}}{4} + c x + d$

Differential Equation

87194 The differential equation of the family of curves $x^{2} + y^{2} - 2 a y = 0$ , where $a$ is an arbitrary constant is

1

2 (x^{2} - y^{2}) y^{'} = x y

2

2 (x^{2} + y^{2}) y^{'} = x y

3

(x^{2} - y^{2}) y^{'} = 2 x y

4

(x^{2} + y^{2}) y^{'} = 2 x y

Explanation:

(C): $\begin{array}{r} Given, \\ x^{2} + y^{2} - 2 a y = 0 \end{array}$
Differentiating w.r.t.x we get,
$2 x + 2 y \frac{d y}{d x} - 2 a \frac{d y}{d x} = 0 \Rightarrow x + (y - a) \frac{d y}{d x} = 0$
$x + {y - \frac{x^{2} + y^{2}}{2 y}} \frac{dy}{dx} = 0 [∵ a = \frac{x^{2} + y^{2}}{2 y}]$
$x + {\frac{y^{2} - x^{2}}{2 y}} \frac{d y}{d x} = 0 \Rightarrow (x^{2} - y^{2}) \frac{d y}{d x} = 2 x y$
$∵ \frac{dy}{dx} = y^{'}$
$∴ (x^{2} - y^{2}) y^{'} = 2 xy$

SRM JEEE-2015