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Differential Equation

87156 The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0$ are
respectively

1 2,4

2 2,3

3 2,2

4 3,4

5 4,3

Explanation:

(C) : Given that,
$\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0 \Rightarrow \sqrt{x^{2} + {(\frac{d y}{d x})}^{\frac{3}{2}}} = - \frac{d^{2} y}{d x^{2}}$
On squaring on both side-
$x^{2} + {[\frac{d y}{d x}]}^{3 / 2} = {[- \frac{d^{2} y}{d x^{2}}]}^{2} \Rightarrow x^{2} + {[\frac{d y}{d x}]}^{\frac{3}{2}} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
$∴$ Order $= 2$ and degree $= 2$

Kerala CEE-2020

Differential Equation

87157 The degree of the differential equation
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}} is$

1 1

2 2

3 3

4 4

5 5

Explanation:

(B) : Given that,
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}}$
On squaring both sides, we get-
${[1 + {(\frac{d y}{d x})}^{2}]}^{3} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
So, the degree of differential equation is 2 .

Kerala CEE-2019

Differential Equation

87160 The differential equation representing the family of curves $y^{2} = a (a x + b)$ , where $a$ and $b$ are arbitrary constants, is of

1 order 1, degree 1

2 order 1, degree 3

3 order 2, degree 3

4 order 1, degree 4

5 order 2, degree 1

Explanation:

(E) : Given that,
$y^{2} = a (a x + b)$
$y^{2} = a^{2} x + a b$
On differentiating with respect to $x$ , we get-
$2 y \frac{d y}{d x} = a^{2} \Rightarrow y \frac{d y}{d x} = \frac{a^{2}}{2}$
$y \cdot \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} \cdot \frac{d y}{d x} = 0$
Hence, order and degree of the given equation is 2 and 1 respectively.

Kerala CEE-2016

Differential Equation

87161 The order and degree of the differential equation
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = (\frac{d^{3} y}{d x^{3}}) are respectively.$

1 2 and 2

2 2 and 1

3 3 and 2

4 3 and 3

5 2 and 4

Explanation:

(C) : Given that the differential equation-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = \frac{d^{3} y}{d x^{3}}$
On squaring both sides, we get-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3} = {(\frac{d^{3} y}{d x^{3}})}^{2}$
Hence, order is 3 and degree is 2.

Kerala CEE-2015

Differential Equation

87083 If ' $m$ ' and ' $n$ ' are the order and degree of the differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x, then$

1

m = 3, n = 5

2

m = 3, n = 1

3

m = 3, n = 3

4

m = 3, n = 2

Explanation:

(D) : Given differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x$
$\Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = \sin x$
${(\frac{d^{3} y}{d x^{3}})}^{2} + 4 {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d^{2} y}{d x^{2}})}^{5} (\frac{d^{3} y}{d x^{3}}) = \sin x (\frac{d^{3} y}{d x^{3}})$
Hence, order $(m) = 3$
Degree $(n) = 2$

Karnataka CET-2013

Differential Equation

87156 The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0$ are
respectively

1 2,4

2 2,3

3 2,2

4 3,4

5 4,3

Explanation:

(C) : Given that,
$\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0 \Rightarrow \sqrt{x^{2} + {(\frac{d y}{d x})}^{\frac{3}{2}}} = - \frac{d^{2} y}{d x^{2}}$
On squaring on both side-
$x^{2} + {[\frac{d y}{d x}]}^{3 / 2} = {[- \frac{d^{2} y}{d x^{2}}]}^{2} \Rightarrow x^{2} + {[\frac{d y}{d x}]}^{\frac{3}{2}} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
$∴$ Order $= 2$ and degree $= 2$

Kerala CEE-2020

Differential Equation

87157 The degree of the differential equation
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}} is$

1 1

2 2

3 3

4 4

5 5

Explanation:

(B) : Given that,
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}}$
On squaring both sides, we get-
${[1 + {(\frac{d y}{d x})}^{2}]}^{3} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
So, the degree of differential equation is 2 .

Kerala CEE-2019

Differential Equation

87160 The differential equation representing the family of curves $y^{2} = a (a x + b)$ , where $a$ and $b$ are arbitrary constants, is of

1 order 1, degree 1

2 order 1, degree 3

3 order 2, degree 3

4 order 1, degree 4

5 order 2, degree 1

Explanation:

(E) : Given that,
$y^{2} = a (a x + b)$
$y^{2} = a^{2} x + a b$
On differentiating with respect to $x$ , we get-
$2 y \frac{d y}{d x} = a^{2} \Rightarrow y \frac{d y}{d x} = \frac{a^{2}}{2}$
$y \cdot \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} \cdot \frac{d y}{d x} = 0$
Hence, order and degree of the given equation is 2 and 1 respectively.

Kerala CEE-2016

Differential Equation

87161 The order and degree of the differential equation
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = (\frac{d^{3} y}{d x^{3}}) are respectively.$

1 2 and 2

2 2 and 1

3 3 and 2

4 3 and 3

5 2 and 4

Explanation:

(C) : Given that the differential equation-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = \frac{d^{3} y}{d x^{3}}$
On squaring both sides, we get-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3} = {(\frac{d^{3} y}{d x^{3}})}^{2}$
Hence, order is 3 and degree is 2.

Kerala CEE-2015

Differential Equation

87083 If ' $m$ ' and ' $n$ ' are the order and degree of the differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x, then$

1

m = 3, n = 5

2

m = 3, n = 1

3

m = 3, n = 3

4

m = 3, n = 2

Explanation:

(D) : Given differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x$
$\Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = \sin x$
${(\frac{d^{3} y}{d x^{3}})}^{2} + 4 {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d^{2} y}{d x^{2}})}^{5} (\frac{d^{3} y}{d x^{3}}) = \sin x (\frac{d^{3} y}{d x^{3}})$
Hence, order $(m) = 3$
Degree $(n) = 2$

Karnataka CET-2013

Differential Equation

87156 The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0$ are
respectively

1 2,4

2 2,3

3 2,2

4 3,4

5 4,3

Explanation:

(C) : Given that,
$\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0 \Rightarrow \sqrt{x^{2} + {(\frac{d y}{d x})}^{\frac{3}{2}}} = - \frac{d^{2} y}{d x^{2}}$
On squaring on both side-
$x^{2} + {[\frac{d y}{d x}]}^{3 / 2} = {[- \frac{d^{2} y}{d x^{2}}]}^{2} \Rightarrow x^{2} + {[\frac{d y}{d x}]}^{\frac{3}{2}} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
$∴$ Order $= 2$ and degree $= 2$

Kerala CEE-2020

Differential Equation

87157 The degree of the differential equation
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}} is$

1 1

2 2

3 3

4 4

5 5

Explanation:

(B) : Given that,
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}}$
On squaring both sides, we get-
${[1 + {(\frac{d y}{d x})}^{2}]}^{3} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
So, the degree of differential equation is 2 .

Kerala CEE-2019

Differential Equation

87160 The differential equation representing the family of curves $y^{2} = a (a x + b)$ , where $a$ and $b$ are arbitrary constants, is of

1 order 1, degree 1

2 order 1, degree 3

3 order 2, degree 3

4 order 1, degree 4

5 order 2, degree 1

Explanation:

(E) : Given that,
$y^{2} = a (a x + b)$
$y^{2} = a^{2} x + a b$
On differentiating with respect to $x$ , we get-
$2 y \frac{d y}{d x} = a^{2} \Rightarrow y \frac{d y}{d x} = \frac{a^{2}}{2}$
$y \cdot \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} \cdot \frac{d y}{d x} = 0$
Hence, order and degree of the given equation is 2 and 1 respectively.

Kerala CEE-2016

Differential Equation

87161 The order and degree of the differential equation
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = (\frac{d^{3} y}{d x^{3}}) are respectively.$

1 2 and 2

2 2 and 1

3 3 and 2

4 3 and 3

5 2 and 4

Explanation:

(C) : Given that the differential equation-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = \frac{d^{3} y}{d x^{3}}$
On squaring both sides, we get-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3} = {(\frac{d^{3} y}{d x^{3}})}^{2}$
Hence, order is 3 and degree is 2.

Kerala CEE-2015

Differential Equation

87083 If ' $m$ ' and ' $n$ ' are the order and degree of the differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x, then$

1

m = 3, n = 5

2

m = 3, n = 1

3

m = 3, n = 3

4

m = 3, n = 2

Explanation:

(D) : Given differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x$
$\Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = \sin x$
${(\frac{d^{3} y}{d x^{3}})}^{2} + 4 {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d^{2} y}{d x^{2}})}^{5} (\frac{d^{3} y}{d x^{3}}) = \sin x (\frac{d^{3} y}{d x^{3}})$
Hence, order $(m) = 3$
Degree $(n) = 2$

Karnataka CET-2013

Differential Equation

87156 The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0$ are
respectively

1 2,4

2 2,3

3 2,2

4 3,4

5 4,3

Explanation:

(C) : Given that,
$\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0 \Rightarrow \sqrt{x^{2} + {(\frac{d y}{d x})}^{\frac{3}{2}}} = - \frac{d^{2} y}{d x^{2}}$
On squaring on both side-
$x^{2} + {[\frac{d y}{d x}]}^{3 / 2} = {[- \frac{d^{2} y}{d x^{2}}]}^{2} \Rightarrow x^{2} + {[\frac{d y}{d x}]}^{\frac{3}{2}} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
$∴$ Order $= 2$ and degree $= 2$

Kerala CEE-2020

Differential Equation

87157 The degree of the differential equation
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}} is$

1 1

2 2

3 3

4 4

5 5

Explanation:

(B) : Given that,
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}}$
On squaring both sides, we get-
${[1 + {(\frac{d y}{d x})}^{2}]}^{3} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
So, the degree of differential equation is 2 .

Kerala CEE-2019

Differential Equation

87160 The differential equation representing the family of curves $y^{2} = a (a x + b)$ , where $a$ and $b$ are arbitrary constants, is of

1 order 1, degree 1

2 order 1, degree 3

3 order 2, degree 3

4 order 1, degree 4

5 order 2, degree 1

Explanation:

(E) : Given that,
$y^{2} = a (a x + b)$
$y^{2} = a^{2} x + a b$
On differentiating with respect to $x$ , we get-
$2 y \frac{d y}{d x} = a^{2} \Rightarrow y \frac{d y}{d x} = \frac{a^{2}}{2}$
$y \cdot \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} \cdot \frac{d y}{d x} = 0$
Hence, order and degree of the given equation is 2 and 1 respectively.

Kerala CEE-2016

Differential Equation

87161 The order and degree of the differential equation
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = (\frac{d^{3} y}{d x^{3}}) are respectively.$

1 2 and 2

2 2 and 1

3 3 and 2

4 3 and 3

5 2 and 4

Explanation:

(C) : Given that the differential equation-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = \frac{d^{3} y}{d x^{3}}$
On squaring both sides, we get-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3} = {(\frac{d^{3} y}{d x^{3}})}^{2}$
Hence, order is 3 and degree is 2.

Kerala CEE-2015

Differential Equation

87083 If ' $m$ ' and ' $n$ ' are the order and degree of the differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x, then$

1

m = 3, n = 5

2

m = 3, n = 1

3

m = 3, n = 3

4

m = 3, n = 2

Explanation:

(D) : Given differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x$
$\Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = \sin x$
${(\frac{d^{3} y}{d x^{3}})}^{2} + 4 {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d^{2} y}{d x^{2}})}^{5} (\frac{d^{3} y}{d x^{3}}) = \sin x (\frac{d^{3} y}{d x^{3}})$
Hence, order $(m) = 3$
Degree $(n) = 2$

Karnataka CET-2013

Differential Equation

87156 The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0$ are
respectively

1 2,4

2 2,3

3 2,2

4 3,4

5 4,3

Explanation:

(C) : Given that,
$\frac{d^{2} y}{d x^{2}} + \sqrt{x^{2} + {(\frac{d y}{d x})}^{3 / 2}} = 0 \Rightarrow \sqrt{x^{2} + {(\frac{d y}{d x})}^{\frac{3}{2}}} = - \frac{d^{2} y}{d x^{2}}$
On squaring on both side-
$x^{2} + {[\frac{d y}{d x}]}^{3 / 2} = {[- \frac{d^{2} y}{d x^{2}}]}^{2} \Rightarrow x^{2} + {[\frac{d y}{d x}]}^{\frac{3}{2}} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
$∴$ Order $= 2$ and degree $= 2$

Kerala CEE-2020

Differential Equation

87157 The degree of the differential equation
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}} is$

1 1

2 2

3 3

4 4

5 5

Explanation:

(B) : Given that,
${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = 1 \frac{d^{2} y}{d x^{2}}$
On squaring both sides, we get-
${[1 + {(\frac{d y}{d x})}^{2}]}^{3} = {[\frac{d^{2} y}{d x^{2}}]}^{2}$
So, the degree of differential equation is 2 .

Kerala CEE-2019

Differential Equation

87160 The differential equation representing the family of curves $y^{2} = a (a x + b)$ , where $a$ and $b$ are arbitrary constants, is of

1 order 1, degree 1

2 order 1, degree 3

3 order 2, degree 3

4 order 1, degree 4

5 order 2, degree 1

Explanation:

(E) : Given that,
$y^{2} = a (a x + b)$
$y^{2} = a^{2} x + a b$
On differentiating with respect to $x$ , we get-
$2 y \frac{d y}{d x} = a^{2} \Rightarrow y \frac{d y}{d x} = \frac{a^{2}}{2}$
$y \cdot \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} \cdot \frac{d y}{d x} = 0$
Hence, order and degree of the given equation is 2 and 1 respectively.

Kerala CEE-2016

Differential Equation

87161 The order and degree of the differential equation
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = (\frac{d^{3} y}{d x^{3}}) are respectively.$

1 2 and 2

2 2 and 1

3 3 and 2

4 3 and 3

5 2 and 4

Explanation:

(C) : Given that the differential equation-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = \frac{d^{3} y}{d x^{3}}$
On squaring both sides, we get-
${[2 \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2}]}^{3} = {(\frac{d^{3} y}{d x^{3}})}^{2}$
Hence, order is 3 and degree is 2.

Kerala CEE-2015

Differential Equation

87083 If ' $m$ ' and ' $n$ ' are the order and degree of the differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x, then$

1

m = 3, n = 5

2

m = 3, n = 1

3

m = 3, n = 3

4

m = 3, n = 2

Explanation:

(D) : Given differential equation
${(y^{''})}^{5} + 4 \frac{{(y^{''})}^{3}}{y^{'''}} + y^{'''} = \sin x$
$\Rightarrow {(\frac{d^{2} y}{d x^{2}})}^{5} + 4 \frac{{(\frac{d^{2} y}{d x^{2}})}^{3}}{(\frac{d^{3} y}{d x^{3}})} + \frac{d^{3} y}{d x^{3}} = \sin x$
${(\frac{d^{3} y}{d x^{3}})}^{2} + 4 {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d^{2} y}{d x^{2}})}^{5} (\frac{d^{3} y}{d x^{3}}) = \sin x (\frac{d^{3} y}{d x^{3}})$
Hence, order $(m) = 3$
Degree $(n) = 2$

Karnataka CET-2013

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