QBManager

Differential Equation

87203 Solution of differential equation
$x d y - y d x = 0$ represents

1 a rectangular hyperbola

2 parabola whose vertex is at origin

3 straight line passing through origin

4 a circle whose centre is origin

Explanation:

(C) :We have differential equation-
$x d y - y d x = 0$
$x d y = y d x$
$\frac{dy}{y} = \frac{dx}{x}$
Integrating both side, we get-
$\int \frac{d y}{y} = \int \frac{d x}{x}$
$\log y = \log x + \log c$
$y = cx$
Which is a straight line passing through the origin.

Karnataka CET-2021

Differential Equation

87204 The solution of $\frac{d y}{d x} - 1 = e^{x - y}$ is

1

e^{x + y} + x = c

2

e^{- (x + y)} + x = c

3

e^{- (x + y)} = x + c

4

e^{- x + y} = x + c

Explanation:

(D) : Given that the differential equation -
$\frac{d y}{d x} - 1 = e^{x - y} \Rightarrow \frac{d y}{d x} - 1 = \frac{e^{x}}{e^{y}}$
$e^{y} d y - e^{y} d x = e^{x} d x$
$e^{y - x} (dy - dx) = dx$
Integrating both sides, $\int e^{y - x} (d y - d x) = \int d x + c$
Put $y - x = t$
$d (y - x) = dt$
$∴ \int e^{t} dt = x + c$
$e^{y - x} = x + c$

COMEDK-2011

Differential Equation

87239 The solution of the differential equation
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$ is

1

\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C

2

\log (\frac{y}{x}) = \frac{1}{x} + \frac{1}{y} + C

3

\log (x y) = \frac{1}{x} + \frac{1}{y} + C

4

\log (xy) + \frac{1}{x} + \frac{1}{y} = C

Explanation:

(A) : Given,
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$
$x^{2} (1 - y) d y + (y^{2} + x y^{2}) d x = 0$
Separating the variable, we get-
$\frac{(1 - y) d y}{y^{2}} + \frac{(1 + x) d x}{x^{2}} = 0$
$(\frac{1}{y^{2}} - \frac{1}{y}) d y + (\frac{1}{x^{2}} + \frac{1}{x}) d x = 0$
Integrating both side, we get-
$- \frac{1}{y} - \log y - \frac{1}{x} + \log x = \log C$
$\log x - \log y = \frac{1}{x} + \frac{1}{y} + C$
$\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C$

JCECE-2014

Differential Equation

87240 Find the differential equation of curves $y = A e^{x} + {B e}^{- x}$ for different values of $A$ and $B$

1

\frac{d^{2} y}{d x^{2}} - 2 y = 0

2

\frac{d^{2} y}{d x^{2}} = y

3

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

4

\frac{d^{2} y}{d x^{2}} + y = 0

Explanation:

(B) : Given,
$y = A e^{x} + B e^{- x}$
Differentiating w.r.t $x$ , we get-
$\frac{d y}{d x} = A e^{x} - B e^{- x}$
Again differentiating w.r.t $x$ , we get-
$\frac{d^{2} y}{d x^{2}} = A e^{x} - (-) B (e^{- x}) \Rightarrow \frac{d^{2} y}{d x^{2}} = A e^{x} + B e^{- x}$
$\frac{d^{2} y}{d x^{2}} = y$

JCECE-2009

Differential Equation

87203 Solution of differential equation
$x d y - y d x = 0$ represents

1 a rectangular hyperbola

2 parabola whose vertex is at origin

3 straight line passing through origin

4 a circle whose centre is origin

Explanation:

(C) :We have differential equation-
$x d y - y d x = 0$
$x d y = y d x$
$\frac{dy}{y} = \frac{dx}{x}$
Integrating both side, we get-
$\int \frac{d y}{y} = \int \frac{d x}{x}$
$\log y = \log x + \log c$
$y = cx$
Which is a straight line passing through the origin.

Karnataka CET-2021

Differential Equation

87204 The solution of $\frac{d y}{d x} - 1 = e^{x - y}$ is

1

e^{x + y} + x = c

2

e^{- (x + y)} + x = c

3

e^{- (x + y)} = x + c

4

e^{- x + y} = x + c

Explanation:

(D) : Given that the differential equation -
$\frac{d y}{d x} - 1 = e^{x - y} \Rightarrow \frac{d y}{d x} - 1 = \frac{e^{x}}{e^{y}}$
$e^{y} d y - e^{y} d x = e^{x} d x$
$e^{y - x} (dy - dx) = dx$
Integrating both sides, $\int e^{y - x} (d y - d x) = \int d x + c$
Put $y - x = t$
$d (y - x) = dt$
$∴ \int e^{t} dt = x + c$
$e^{y - x} = x + c$

COMEDK-2011

Differential Equation

87239 The solution of the differential equation
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$ is

1

\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C

2

\log (\frac{y}{x}) = \frac{1}{x} + \frac{1}{y} + C

3

\log (x y) = \frac{1}{x} + \frac{1}{y} + C

4

\log (xy) + \frac{1}{x} + \frac{1}{y} = C

Explanation:

(A) : Given,
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$
$x^{2} (1 - y) d y + (y^{2} + x y^{2}) d x = 0$
Separating the variable, we get-
$\frac{(1 - y) d y}{y^{2}} + \frac{(1 + x) d x}{x^{2}} = 0$
$(\frac{1}{y^{2}} - \frac{1}{y}) d y + (\frac{1}{x^{2}} + \frac{1}{x}) d x = 0$
Integrating both side, we get-
$- \frac{1}{y} - \log y - \frac{1}{x} + \log x = \log C$
$\log x - \log y = \frac{1}{x} + \frac{1}{y} + C$
$\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C$

JCECE-2014

Differential Equation

87240 Find the differential equation of curves $y = A e^{x} + {B e}^{- x}$ for different values of $A$ and $B$

1

\frac{d^{2} y}{d x^{2}} - 2 y = 0

2

\frac{d^{2} y}{d x^{2}} = y

3

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

4

\frac{d^{2} y}{d x^{2}} + y = 0

Explanation:

(B) : Given,
$y = A e^{x} + B e^{- x}$
Differentiating w.r.t $x$ , we get-
$\frac{d y}{d x} = A e^{x} - B e^{- x}$
Again differentiating w.r.t $x$ , we get-
$\frac{d^{2} y}{d x^{2}} = A e^{x} - (-) B (e^{- x}) \Rightarrow \frac{d^{2} y}{d x^{2}} = A e^{x} + B e^{- x}$
$\frac{d^{2} y}{d x^{2}} = y$

JCECE-2009

Differential Equation

87203 Solution of differential equation
$x d y - y d x = 0$ represents

1 a rectangular hyperbola

2 parabola whose vertex is at origin

3 straight line passing through origin

4 a circle whose centre is origin

Explanation:

(C) :We have differential equation-
$x d y - y d x = 0$
$x d y = y d x$
$\frac{dy}{y} = \frac{dx}{x}$
Integrating both side, we get-
$\int \frac{d y}{y} = \int \frac{d x}{x}$
$\log y = \log x + \log c$
$y = cx$
Which is a straight line passing through the origin.

Karnataka CET-2021

Differential Equation

87204 The solution of $\frac{d y}{d x} - 1 = e^{x - y}$ is

1

e^{x + y} + x = c

2

e^{- (x + y)} + x = c

3

e^{- (x + y)} = x + c

4

e^{- x + y} = x + c

Explanation:

(D) : Given that the differential equation -
$\frac{d y}{d x} - 1 = e^{x - y} \Rightarrow \frac{d y}{d x} - 1 = \frac{e^{x}}{e^{y}}$
$e^{y} d y - e^{y} d x = e^{x} d x$
$e^{y - x} (dy - dx) = dx$
Integrating both sides, $\int e^{y - x} (d y - d x) = \int d x + c$
Put $y - x = t$
$d (y - x) = dt$
$∴ \int e^{t} dt = x + c$
$e^{y - x} = x + c$

COMEDK-2011

Differential Equation

87239 The solution of the differential equation
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$ is

1

\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C

2

\log (\frac{y}{x}) = \frac{1}{x} + \frac{1}{y} + C

3

\log (x y) = \frac{1}{x} + \frac{1}{y} + C

4

\log (xy) + \frac{1}{x} + \frac{1}{y} = C

Explanation:

(A) : Given,
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$
$x^{2} (1 - y) d y + (y^{2} + x y^{2}) d x = 0$
Separating the variable, we get-
$\frac{(1 - y) d y}{y^{2}} + \frac{(1 + x) d x}{x^{2}} = 0$
$(\frac{1}{y^{2}} - \frac{1}{y}) d y + (\frac{1}{x^{2}} + \frac{1}{x}) d x = 0$
Integrating both side, we get-
$- \frac{1}{y} - \log y - \frac{1}{x} + \log x = \log C$
$\log x - \log y = \frac{1}{x} + \frac{1}{y} + C$
$\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C$

JCECE-2014

Differential Equation

87240 Find the differential equation of curves $y = A e^{x} + {B e}^{- x}$ for different values of $A$ and $B$

1

\frac{d^{2} y}{d x^{2}} - 2 y = 0

2

\frac{d^{2} y}{d x^{2}} = y

3

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

4

\frac{d^{2} y}{d x^{2}} + y = 0

Explanation:

(B) : Given,
$y = A e^{x} + B e^{- x}$
Differentiating w.r.t $x$ , we get-
$\frac{d y}{d x} = A e^{x} - B e^{- x}$
Again differentiating w.r.t $x$ , we get-
$\frac{d^{2} y}{d x^{2}} = A e^{x} - (-) B (e^{- x}) \Rightarrow \frac{d^{2} y}{d x^{2}} = A e^{x} + B e^{- x}$
$\frac{d^{2} y}{d x^{2}} = y$

JCECE-2009

Differential Equation

87203 Solution of differential equation
$x d y - y d x = 0$ represents

1 a rectangular hyperbola

2 parabola whose vertex is at origin

3 straight line passing through origin

4 a circle whose centre is origin

Explanation:

(C) :We have differential equation-
$x d y - y d x = 0$
$x d y = y d x$
$\frac{dy}{y} = \frac{dx}{x}$
Integrating both side, we get-
$\int \frac{d y}{y} = \int \frac{d x}{x}$
$\log y = \log x + \log c$
$y = cx$
Which is a straight line passing through the origin.

Karnataka CET-2021

Differential Equation

87204 The solution of $\frac{d y}{d x} - 1 = e^{x - y}$ is

1

e^{x + y} + x = c

2

e^{- (x + y)} + x = c

3

e^{- (x + y)} = x + c

4

e^{- x + y} = x + c

Explanation:

(D) : Given that the differential equation -
$\frac{d y}{d x} - 1 = e^{x - y} \Rightarrow \frac{d y}{d x} - 1 = \frac{e^{x}}{e^{y}}$
$e^{y} d y - e^{y} d x = e^{x} d x$
$e^{y - x} (dy - dx) = dx$
Integrating both sides, $\int e^{y - x} (d y - d x) = \int d x + c$
Put $y - x = t$
$d (y - x) = dt$
$∴ \int e^{t} dt = x + c$
$e^{y - x} = x + c$

COMEDK-2011

Differential Equation

87239 The solution of the differential equation
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$ is

1

\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C

2

\log (\frac{y}{x}) = \frac{1}{x} + \frac{1}{y} + C

3

\log (x y) = \frac{1}{x} + \frac{1}{y} + C

4

\log (xy) + \frac{1}{x} + \frac{1}{y} = C

Explanation:

(A) : Given,
$(x^{2} - y x^{2}) \frac{d y}{d x} + y^{2} + x y^{2} = 0$
$x^{2} (1 - y) d y + (y^{2} + x y^{2}) d x = 0$
Separating the variable, we get-
$\frac{(1 - y) d y}{y^{2}} + \frac{(1 + x) d x}{x^{2}} = 0$
$(\frac{1}{y^{2}} - \frac{1}{y}) d y + (\frac{1}{x^{2}} + \frac{1}{x}) d x = 0$
Integrating both side, we get-
$- \frac{1}{y} - \log y - \frac{1}{x} + \log x = \log C$
$\log x - \log y = \frac{1}{x} + \frac{1}{y} + C$
$\log (\frac{x}{y}) = \frac{1}{x} + \frac{1}{y} + C$

JCECE-2014

Differential Equation

87240 Find the differential equation of curves $y = A e^{x} + {B e}^{- x}$ for different values of $A$ and $B$

1

\frac{d^{2} y}{d x^{2}} - 2 y = 0

2

\frac{d^{2} y}{d x^{2}} = y

3

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

4

\frac{d^{2} y}{d x^{2}} + y = 0

Explanation:

(B) : Given,
$y = A e^{x} + B e^{- x}$
Differentiating w.r.t $x$ , we get-
$\frac{d y}{d x} = A e^{x} - B e^{- x}$
Again differentiating w.r.t $x$ , we get-
$\frac{d^{2} y}{d x^{2}} = A e^{x} - (-) B (e^{- x}) \Rightarrow \frac{d^{2} y}{d x^{2}} = A e^{x} + B e^{- x}$
$\frac{d^{2} y}{d x^{2}} = y$

JCECE-2009

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