QBManager

Differential Equation

87079 The order of the differential equation $y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x}$ is

1 1

2 3

3 2

4 4

Explanation:

(A) : Given, differential equation
$y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x} \Rightarrow y = C_{1} e^{C_{2}} \cdot e^{x} + C_{3} e^{C_{4}} \cdot e^{x}$
$y = (C_{1} e^{C_{2}} + C_{3} e^{C_{4}}) e^{x} \Rightarrow y = k e^{x}$
(Where, $k = C_{1} e^{C_{2}} + C_{3} e^{C_{4}}$ )
$\frac{dy}{dx} = {ke}^{x} \Rightarrow \frac{dy}{dx} = y \Rightarrow \frac{dy}{dx} - y = 0$
Hence, order is one.

Karnataka CET-2019]**#

Differential Equation

87080 The degree and order of the differential equation $\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$ respectively are

1 2 and 3

2 3 and 2

3 2 and 2

4 3 and 3

Explanation:

(B) : Given, differential equation -
$\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$
${(\frac{d^{2} y}{d x^{2}})}^{3} = (1 + {(\frac{d y}{d x})}^{2}) \Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{3} - {(\frac{d y}{d x})}^{2} - 1 = 0$
Hence order $= 2$
Degree $= 3$

Karnataka CET-2018

Differential Equation

87082 The order and degree of the differential
equation $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$ is

1 1,2

2 1,3

3 2,1

4 1,1

Explanation:

(A) : Given, $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$
$y \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 2 \Rightarrow x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} + 2 = 0$
Hence order $\to 1$ and degree 2

Karnataka CET-2014

Differential Equation

87106 The solution of the equation $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$ is

1

\sin^{- 1} y - \sin^{- 1} x = c

2

\sin^{- 1} y + \sin^{- 1} x = c

3

\sin^{- 1} (x y) = 2

4 None of the above

Explanation:

(A) : Given $\frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$
$\frac{dy}{\sqrt{1 - y^{2}}} = \frac{dx}{\sqrt{1 - x^{2}}}$
Integrating both sides, we get -
$\sin^{- 1} y = \sin^{- 1} x + c$
$\sin^{- 1} y - \sin^{- 1} x = c$

UPSEE-2009

Differential Equation

87135 Find the degree of the differential equation
$y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$

1 4

2 2

3 3

4 1

Explanation:

(B) : Given, differential equation,
$y_{3}^{2 / 3} + 2 + 3 y_{3} + y_{1} = 0$
Then, it is also written as-
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} + 2 + 3 (\frac{d^{2} y}{d x^{2}}) + \frac{d y}{d x} = 0$
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} = - [(\frac{d^{2} y}{d x^{2}}) \times 3 + \frac{d y}{d x} + 2]$
${(\frac{d^{3} y}{d x^{3}})}^{2} = - {[3 \times \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + 2]}^{3}$
So, the degree of differential equation is 2 and order is 3 .

AP EAMCET-21.09.2020

Differential Equation

87079 The order of the differential equation $y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x}$ is

1 1

2 3

3 2

4 4

Explanation:

(A) : Given, differential equation
$y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x} \Rightarrow y = C_{1} e^{C_{2}} \cdot e^{x} + C_{3} e^{C_{4}} \cdot e^{x}$
$y = (C_{1} e^{C_{2}} + C_{3} e^{C_{4}}) e^{x} \Rightarrow y = k e^{x}$
(Where, $k = C_{1} e^{C_{2}} + C_{3} e^{C_{4}}$ )
$\frac{dy}{dx} = {ke}^{x} \Rightarrow \frac{dy}{dx} = y \Rightarrow \frac{dy}{dx} - y = 0$
Hence, order is one.

Karnataka CET-2019]**#

Differential Equation

87080 The degree and order of the differential equation $\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$ respectively are

1 2 and 3

2 3 and 2

3 2 and 2

4 3 and 3

Explanation:

(B) : Given, differential equation -
$\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$
${(\frac{d^{2} y}{d x^{2}})}^{3} = (1 + {(\frac{d y}{d x})}^{2}) \Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{3} - {(\frac{d y}{d x})}^{2} - 1 = 0$
Hence order $= 2$
Degree $= 3$

Karnataka CET-2018

Differential Equation

87082 The order and degree of the differential
equation $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$ is

1 1,2

2 1,3

3 2,1

4 1,1

Explanation:

(A) : Given, $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$
$y \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 2 \Rightarrow x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} + 2 = 0$
Hence order $\to 1$ and degree 2

Karnataka CET-2014

Differential Equation

87106 The solution of the equation $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$ is

1

\sin^{- 1} y - \sin^{- 1} x = c

2

\sin^{- 1} y + \sin^{- 1} x = c

3

\sin^{- 1} (x y) = 2

4 None of the above

Explanation:

(A) : Given $\frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$
$\frac{dy}{\sqrt{1 - y^{2}}} = \frac{dx}{\sqrt{1 - x^{2}}}$
Integrating both sides, we get -
$\sin^{- 1} y = \sin^{- 1} x + c$
$\sin^{- 1} y - \sin^{- 1} x = c$

UPSEE-2009

Differential Equation

87135 Find the degree of the differential equation
$y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$

1 4

2 2

3 3

4 1

Explanation:

(B) : Given, differential equation,
$y_{3}^{2 / 3} + 2 + 3 y_{3} + y_{1} = 0$
Then, it is also written as-
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} + 2 + 3 (\frac{d^{2} y}{d x^{2}}) + \frac{d y}{d x} = 0$
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} = - [(\frac{d^{2} y}{d x^{2}}) \times 3 + \frac{d y}{d x} + 2]$
${(\frac{d^{3} y}{d x^{3}})}^{2} = - {[3 \times \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + 2]}^{3}$
So, the degree of differential equation is 2 and order is 3 .

AP EAMCET-21.09.2020

Differential Equation

87079 The order of the differential equation $y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x}$ is

1 1

2 3

3 2

4 4

Explanation:

(A) : Given, differential equation
$y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x} \Rightarrow y = C_{1} e^{C_{2}} \cdot e^{x} + C_{3} e^{C_{4}} \cdot e^{x}$
$y = (C_{1} e^{C_{2}} + C_{3} e^{C_{4}}) e^{x} \Rightarrow y = k e^{x}$
(Where, $k = C_{1} e^{C_{2}} + C_{3} e^{C_{4}}$ )
$\frac{dy}{dx} = {ke}^{x} \Rightarrow \frac{dy}{dx} = y \Rightarrow \frac{dy}{dx} - y = 0$
Hence, order is one.

Karnataka CET-2019]**#

Differential Equation

87080 The degree and order of the differential equation $\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$ respectively are

1 2 and 3

2 3 and 2

3 2 and 2

4 3 and 3

Explanation:

(B) : Given, differential equation -
$\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$
${(\frac{d^{2} y}{d x^{2}})}^{3} = (1 + {(\frac{d y}{d x})}^{2}) \Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{3} - {(\frac{d y}{d x})}^{2} - 1 = 0$
Hence order $= 2$
Degree $= 3$

Karnataka CET-2018

Differential Equation

87082 The order and degree of the differential
equation $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$ is

1 1,2

2 1,3

3 2,1

4 1,1

Explanation:

(A) : Given, $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$
$y \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 2 \Rightarrow x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} + 2 = 0$
Hence order $\to 1$ and degree 2

Karnataka CET-2014

Differential Equation

87106 The solution of the equation $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$ is

1

\sin^{- 1} y - \sin^{- 1} x = c

2

\sin^{- 1} y + \sin^{- 1} x = c

3

\sin^{- 1} (x y) = 2

4 None of the above

Explanation:

(A) : Given $\frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$
$\frac{dy}{\sqrt{1 - y^{2}}} = \frac{dx}{\sqrt{1 - x^{2}}}$
Integrating both sides, we get -
$\sin^{- 1} y = \sin^{- 1} x + c$
$\sin^{- 1} y - \sin^{- 1} x = c$

UPSEE-2009

Differential Equation

87135 Find the degree of the differential equation
$y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$

1 4

2 2

3 3

4 1

Explanation:

(B) : Given, differential equation,
$y_{3}^{2 / 3} + 2 + 3 y_{3} + y_{1} = 0$
Then, it is also written as-
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} + 2 + 3 (\frac{d^{2} y}{d x^{2}}) + \frac{d y}{d x} = 0$
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} = - [(\frac{d^{2} y}{d x^{2}}) \times 3 + \frac{d y}{d x} + 2]$
${(\frac{d^{3} y}{d x^{3}})}^{2} = - {[3 \times \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + 2]}^{3}$
So, the degree of differential equation is 2 and order is 3 .

AP EAMCET-21.09.2020

Differential Equation

87079 The order of the differential equation $y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x}$ is

1 1

2 3

3 2

4 4

Explanation:

(A) : Given, differential equation
$y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x} \Rightarrow y = C_{1} e^{C_{2}} \cdot e^{x} + C_{3} e^{C_{4}} \cdot e^{x}$
$y = (C_{1} e^{C_{2}} + C_{3} e^{C_{4}}) e^{x} \Rightarrow y = k e^{x}$
(Where, $k = C_{1} e^{C_{2}} + C_{3} e^{C_{4}}$ )
$\frac{dy}{dx} = {ke}^{x} \Rightarrow \frac{dy}{dx} = y \Rightarrow \frac{dy}{dx} - y = 0$
Hence, order is one.

Karnataka CET-2019]**#

Differential Equation

87080 The degree and order of the differential equation $\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$ respectively are

1 2 and 3

2 3 and 2

3 2 and 2

4 3 and 3

Explanation:

(B) : Given, differential equation -
$\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$
${(\frac{d^{2} y}{d x^{2}})}^{3} = (1 + {(\frac{d y}{d x})}^{2}) \Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{3} - {(\frac{d y}{d x})}^{2} - 1 = 0$
Hence order $= 2$
Degree $= 3$

Karnataka CET-2018

Differential Equation

87082 The order and degree of the differential
equation $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$ is

1 1,2

2 1,3

3 2,1

4 1,1

Explanation:

(A) : Given, $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$
$y \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 2 \Rightarrow x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} + 2 = 0$
Hence order $\to 1$ and degree 2

Karnataka CET-2014

Differential Equation

87106 The solution of the equation $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$ is

1

\sin^{- 1} y - \sin^{- 1} x = c

2

\sin^{- 1} y + \sin^{- 1} x = c

3

\sin^{- 1} (x y) = 2

4 None of the above

Explanation:

(A) : Given $\frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$
$\frac{dy}{\sqrt{1 - y^{2}}} = \frac{dx}{\sqrt{1 - x^{2}}}$
Integrating both sides, we get -
$\sin^{- 1} y = \sin^{- 1} x + c$
$\sin^{- 1} y - \sin^{- 1} x = c$

UPSEE-2009

Differential Equation

87135 Find the degree of the differential equation
$y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$

1 4

2 2

3 3

4 1

Explanation:

(B) : Given, differential equation,
$y_{3}^{2 / 3} + 2 + 3 y_{3} + y_{1} = 0$
Then, it is also written as-
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} + 2 + 3 (\frac{d^{2} y}{d x^{2}}) + \frac{d y}{d x} = 0$
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} = - [(\frac{d^{2} y}{d x^{2}}) \times 3 + \frac{d y}{d x} + 2]$
${(\frac{d^{3} y}{d x^{3}})}^{2} = - {[3 \times \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + 2]}^{3}$
So, the degree of differential equation is 2 and order is 3 .

AP EAMCET-21.09.2020

Differential Equation

87079 The order of the differential equation $y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x}$ is

1 1

2 3

3 2

4 4

Explanation:

(A) : Given, differential equation
$y = C_{1} e^{C_{2} + x} + C_{3} e^{C_{4} + x} \Rightarrow y = C_{1} e^{C_{2}} \cdot e^{x} + C_{3} e^{C_{4}} \cdot e^{x}$
$y = (C_{1} e^{C_{2}} + C_{3} e^{C_{4}}) e^{x} \Rightarrow y = k e^{x}$
(Where, $k = C_{1} e^{C_{2}} + C_{3} e^{C_{4}}$ )
$\frac{dy}{dx} = {ke}^{x} \Rightarrow \frac{dy}{dx} = y \Rightarrow \frac{dy}{dx} - y = 0$
Hence, order is one.

Karnataka CET-2019]**#

Differential Equation

87080 The degree and order of the differential equation $\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$ respectively are

1 2 and 3

2 3 and 2

3 2 and 2

4 3 and 3

Explanation:

(B) : Given, differential equation -
$\frac{d^{2} y}{d x^{2}} = \sqrt[3]{1 + {(\frac{d y}{d x})}^{2}}$
${(\frac{d^{2} y}{d x^{2}})}^{3} = (1 + {(\frac{d y}{d x})}^{2}) \Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{3} - {(\frac{d y}{d x})}^{2} - 1 = 0$
Hence order $= 2$
Degree $= 3$

Karnataka CET-2018

Differential Equation

87082 The order and degree of the differential
equation $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$ is

1 1,2

2 1,3

3 2,1

4 1,1

Explanation:

(A) : Given, $y = x \frac{d y}{d x} + \frac{2}{\frac{d y}{d x}}$
$y \frac{d y}{d x} = x {(\frac{d y}{d x})}^{2} + 2 \Rightarrow x {(\frac{d y}{d x})}^{2} - y \frac{d y}{d x} + 2 = 0$
Hence order $\to 1$ and degree 2

Karnataka CET-2014

Differential Equation

87106 The solution of the equation $\frac{d y}{d x} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$ is

1

\sin^{- 1} y - \sin^{- 1} x = c

2

\sin^{- 1} y + \sin^{- 1} x = c

3

\sin^{- 1} (x y) = 2

4 None of the above

Explanation:

(A) : Given $\frac{dy}{dx} = \sqrt{\frac{1 - y^{2}}{1 - x^{2}}}$
$\frac{dy}{\sqrt{1 - y^{2}}} = \frac{dx}{\sqrt{1 - x^{2}}}$
Integrating both sides, we get -
$\sin^{- 1} y = \sin^{- 1} x + c$
$\sin^{- 1} y - \sin^{- 1} x = c$

UPSEE-2009

Differential Equation

87135 Find the degree of the differential equation
$y_{3}^{2 / 3} + 2 + 3 y_{2} + y_{1} = 0$

1 4

2 2

3 3

4 1

Explanation:

(B) : Given, differential equation,
$y_{3}^{2 / 3} + 2 + 3 y_{3} + y_{1} = 0$
Then, it is also written as-
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} + 2 + 3 (\frac{d^{2} y}{d x^{2}}) + \frac{d y}{d x} = 0$
${(\frac{d^{3} y}{d x^{3}})}^{2 / 3} = - [(\frac{d^{2} y}{d x^{2}}) \times 3 + \frac{d y}{d x} + 2]$
${(\frac{d^{3} y}{d x^{3}})}^{2} = - {[3 \times \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + 2]}^{3}$
So, the degree of differential equation is 2 and order is 3 .

AP EAMCET-21.09.2020