Class 12th

98. Form the differential equation representing the family of curves ${(x - a)}^{2} + 2 y^{2} = a^{2}$ where a an arbitrary constant.

0 Follower 5 Views

Contributor-Level 10

Equation of the given family of curves is ${(x - a)}^{2} + 2 y^{2} = a^{2}$

$\begin{array}{l} {(x - a)}^{2} + 2 y^{2} = a^{2} \\ \Rightarrow x^{2} + a^{2} - 2 a x + 2 y^{2} = a^{2} \\ \Rightarrow 2 y^{2} = 2 a x - x^{2} . . . . . . . . . . (1) \end{array}$

Differentiating with respect to x, we get:

$\begin{array}{l} 2 y \frac{d y}{d x} = \frac{2 a - 2 x}{2} \\ \Rightarrow \frac{d y}{d x} = \frac{a - x}{2 y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 a - 2 x^{2}}{4 x y} . . . . . . . . . . (2) \end{array}$

From equation (*1), we get:

$2 a x = 2 y^{2} + x^{2}$

On substituting this value in equation (3), we get:

$\begin{array}{l} \frac{d y}{d x} = \frac{2 y^{2} + x^{2} - 2 x^{2}}{4 x y} \\ \Rightarrow \frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y} \end{array}$

Hence, the differential equation of the family of curves is given as $\frac{d y}{d x} = \frac{2 y^{2} - x^{2}}{4 x y}$

0 0

New answer posted

11 months ago

19. Which of the following is true about the value of refractive index of quartz glass?

(i) Same in all directions

(ii) Different in different directions

(iii) Cannot be measured

(iv) Always zero

0 Follower 2 Views

Contributor-Level 10

19. Option (i) Same in all directions is correct since quartz glass is an amorphous solid showing isotropic properties and hence exhibits the same values of refractive index when measured along different directions.

0 0

New answer posted

11 months ago

97. For each of the exercises given below verify that the given function (implicit or explicit) is a solution of the corresponding differential equation:

$\begin{array}{l} (i) y a e^{2} + b e^{- x} + x^{2} : x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 = 0 \\ (i i) y = e^{x} (a c o s x + b s i n x) : \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + 2 y = 0 \\ (i i i) y = x s i n 3 x : \frac{d^{2} y}{d x^{2}} + 9 y - 6 c o s 3 x = 0 \\ (i v) x^{2} = 2 y^{2} l o g y : (x^{2} + y^{2}) \frac{d y}{d x} - x y = 0 \end{array}$

0 Follower 15 Views

Contributor-Level 10

(i) $y a e^{2} + b e^{- x} + x^{2}$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a \frac{d}{d x} (e^{x}) + b \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (x^{2}) \\ \Rightarrow \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\frac{d^{2} y}{d x^{2}} = a e^{x} - b e^{- x} + 2 x$

Now, on substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the differential equation, we get:

L.H.S

$\begin{array}{l} x \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} - x y + x^{2} - 2 \\ = x (a e^{x} - b e^{- x} + 2) + 2 (a e^{x} - b e^{- x} + 2 x) - x (a e^{x} + b e^{- x} + x^{2}) + x^{2} - 2 \\ = (x a e^{x} - b x e^{- x} + 2 x) + (2 a e^{x} - 2 b e^{- x} + 4 x) - (a x e^{x} + b x e^{- x} + x^{3}) + x^{2} - 2 \\ = 2 a e^{x} - 2 b e^{- x} + x^{2} + 6 x - 2 \\ \neq 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation. (ii).

(ii) $y = e^{x} (a c o s x + b s i n x) = a e^{x} c o s x + b e^{x} s i n x$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a . \frac{d}{d x} (e^{x} c o s x) + b . \frac{d}{d x} (e^{x} s i n x) \\ \Rightarrow \frac{d y}{d x} = a (e^{x} c o s x - e^{x} s i n x) + b . (e^{x} s i n x + e^{x} c o s x) \\ \Rightarrow \frac{d y}{d x} = (a + b) e^{x} c o s x + (b - a) e^{x} s i n x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = (a + b) . \frac{d}{d x} (e^{x} c o s x) (b - a) \frac{d}{d x} (e^{x} s i n x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = (a + b) . [e^{x} c o s x - e^{x} s i n x] + (b - a) [e^{x} s i n x + e^{x} c o s x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [(a + b) (c o s x - s i n x) + (b - a) (s i n x + c o s x)] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = e^{x} [a c o s x - a s i n x + b c o s x - b s i n x + b s i n x + b c o s x - a s i n x - a c o s x] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = [2 e^{x} (b c o s x - a s i n x)] \end{array}$

Now, on substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ in the L.H.S of the given differential equation, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} + 2 \frac{d y}{d x} + 2 y \\ = 2 e^{x} (b c o s x - a s i n x) - 2 e^{x} [(a + b) c o s x + (b - a) s i n x] + 2 e^{x} (a c o s x + b s i n x) \\ = e^{x} [(2 b c o s x - 2 a s i n x) - (2 a c o s x + 2 b c o s x) - (2 b s i n x - 2 a s i n x) + (2 a c o s x + 2 b s i n x)] \\ = e^{x} [(2 b - 2 a - 2 b + 2 a) c o s x] + e^{x} [(- 2 a - 2 b + 2 a + 2 b) s i n x] \\ = 0 \end{array}$

Therefore, Function given by equation (i) is solution of differential equation (ii)

(iii)&nb

...more

(i) $y a e^{2} + b e^{- x} + x^{2}$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = a \frac{d}{d x} (e^{x}) + b \frac{d}{d x} (e^{- x}) + \frac{d}{d x} (x^{2}) \\ \Rightarrow \frac{d y}{d x} = a e^{x} - b e^{- x} + 2 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\frac{d^{2} y}{d x^{2}} = a e^{x} - b e^{- x} + 2 x$

Now, on substituting the values of $\frac{d y}{d x}$ and $\frac{d^{2} y}{d x^{2}}$ in the differential equation, we get:

L.H.S

Therefore, Function given by equation (i) is a solution of differential equation. (ii).

(ii) $y = e^{x} (a c o s x + b s i n x) = a e^{x} c o s x + b e^{x} s i n x$

Differentiating both sides with respect to x, we get:

Again, differentiating both sides with respect to x, we get:

Now, on substituting the values of $\frac{d^{2} y}{d x^{2}}$ and $\frac{d y}{d x}$ in the L.H.S of the given differential equation, we get:

Therefore, Function given by equation (i) is solution of differential equation (ii)

(iii) $y = x s i n 3 x$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d y}{d x} = \frac{d}{d x} (x s i n 3 x) = s i n 3 x + x . c o s 3 x . 3 \\ \Rightarrow \frac{d y}{d x} = s i n 3 x + 3 x c o s 3 x \end{array}$

Again, differentiating both sides with respect to x, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} = \frac{d}{d x} (s i n 3 x) + 3 \frac{d}{d x} (x c o s 3 x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = 3 c o s 3 x + 3 [c o s 3 x + x (- s i n 3 x) . 3] \\ \Rightarrow \frac{d^{2} y}{d x^{2}} 6 c o s 3 x - 9 x s i n 3 x \end{array}$

Substituting the value of $\frac{d^{2} y}{d x^{2}}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l} \frac{d^{2} y}{d x^{2}} + 9 y - 6 c o s 3 x \\ = (6 . c o s 3 x - 9 x s i n 3 x) + 9 x s i n 3 x - 6 c o s 3 x \\ = 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

(iv) $x^{2} = 2 y^{2} l o g y$

Differentiating both sides with respect to x, we get:

$\begin{array}{l} 2 x = 2 . \frac{d}{d x} = [y^{2} l o g y] \\ \Rightarrow x = [2 y . l o g y . \frac{d y}{d x} + y^{2} . \frac{1}{y} . \frac{d y}{d x}] \\ \Rightarrow x = \frac{d y}{d x} (2 y l o g y + y) \\ \Rightarrow \frac{d y}{d x} = \frac{x}{y (1 + 2 l o g y)} \end{array}$

Substituting the value of $\frac{d y}{d x}$ in the L.H.S. of the given differential equation, we get:

$\begin{array}{l} (x^{2} + y^{2}) \frac{d y}{d x} - x y \\ = (2 y^{2} l o g y + y^{2}) . \frac{x}{y (1 + 2 l o g y)} - x y \\ = y^{2} (1 + 2 l o g y) . \frac{x}{y (1 + 2 l o g y)} - x y \\ = x y - x y \\ = 0 \end{array}$

Therefore, Function given by equation (i) is a solution of differential equation (ii).

less

0 0

New answer posted

11 months ago

18. Which of the following arrangements shows schematic alignment of magnetic moments of antiferromagnetic substances?

0 Follower 6 Views

Contributor-Level 10

18. Option (iv) is correct since in antiferromagnetic substances the domains are oppositely oriented and hence they cancel out each other's magnetic moments.

0 0

New answer posted

11 months ago

Class 12th Solid State

17. Which of the following is an amorphous solid?

(i) Graphite (C)

(ii) Quartz glass (SiO₂)

(iii) Chrome alum

(iv) Silicon carbide (SiC)

0 Follower 8 Views

Contributor-Level 10

17. Option (ii) Quartz glass (SiO₂) is correct since quartz glass (SiO₂)is amorphous in nature as there is no long range ordered arrangement of the constituent particles being present in it and hence it is an amorphous solid.

0 0

New answer posted

11 months ago

96. For each of the differential equations given below, indicate its order and degree (if defined):

$\begin{array}{l} (i) \frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x \\ (i i) {(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x \\ (i i i) \frac{d^{4} y}{d x^{4}} - s i n \frac{d^{3} y}{d x^{3}} = 0 \end{array}$

0 Follower 2 Views

Contributor-Level 10

(i) Given: Differential equation $\frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x$

The highest order derivative present in this differential equation is $\frac{d^{2} y}{d x^{2}}$ and hence order of this differential equation if 2.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d^{2} y}{d x^{2}}$ is 1.

Therefore, Order = 2, Degree = 1

(ii) Given: Differential equation ${(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x$

The highest order derivative present in this differential equation is $\frac{d y}{d x}$ and hence order of this differential equation if 1.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivativ

...more

(i) Given: Differential equation $\frac{d^{2} y}{d x^{2}} + 5 x {(\frac{d y}{d x})}^{2} - 6 y = l o g x$

The highest order derivative present in this differential equation is $\frac{d^{2} y}{d x^{2}}$ and hence order of this differential equation if 2.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d^{2} y}{d x^{2}}$ is 1.

Therefore, Order = 2, Degree = 1

(ii) Given: Differential equation ${(\frac{d y}{d x})}^{3} - 4 {(\frac{d y}{d x})}^{2} + 7 y = s i n x$

The highest order derivative present in this differential equation is $\frac{d y}{d x}$ and hence order of this differential equation if 1.

The given differential equation is a polynomial equation in derivatives and highest power of the highest order derivative $\frac{d y}{d x}$ is 3.

Therefore, Order = 1, Degree = 3

(iii) Given: Differential equation $\frac{d^{4} y}{d x^{4}} - s i n \frac{d^{3} y}{d x^{3}} = 0$

The highest order derivative present in this differential equation is $\frac{d^{4} y}{d x^{4}}$ and hence order of this differential equation if 4.

The given differential equation is not a polynomial equation in derivatives therefore, degree of this differential equation is not defined.

Therefore, Order = 4, Degree not defined

less

0 0

New answer posted

11 months ago

16. Which of the following is not a characteristic of a crystalline solid?

(i) Definite and characteristic heat of fusion.

(ii) Isotropic nature.

(iii) A regular periodically repeated pattern of arrangement of constituent particles in the entire crystal.

(iv) A true solid

0 Follower 2 Views

Contributor-Level 10

16. Option (ii) Isotropic nature is correct since crystalline solids exhibit anisotropic properties like refractive index, electrical resistance etc. Since these are found to have different values when measured along different directions in the same crystal and hence they are not isotropic in nature.

0 0

New answer posted

11 months ago

95. Choose the correct answer:

The integrating factor of the differential equation $(1 - y^{2}) \frac{d x}{d y} + y x = a y (- 1 < < 1)$ .

0 Follower 4 Views

Contributor-Level 10

The given D.E. is

$(1 - y^{2}) \frac{d x}{d y} + y x = \propto y (- 1 < y < 1)$

$\frac{d x}{d y} + \frac{y}{1 - y^{2}} * x = \frac{\propto y}{1 - y^{2}}$ which is of form

$\begin{array}{l} \frac{d x}{d y} + P x = Q & P = \frac{y}{1 - y^{2}} & Q = \frac{\propto y}{1 - y^{2}} \\ \int p d x = \int \frac{y}{1 - y^{2}} d x = - \frac{1}{2} \int \frac{- 2 y}{1 - y^{2} d x} \end{array}$

$\begin{array}{l} = - \frac{1}{2} l o g | 1 - y^{2} | \\ = l o g {[1 - y^{2}]}^{- \frac{1}{2}} \end{array}$

$∴ I . F = e^{\int P d x} = e^{l o g {[1 - y^{2}]}^{- \frac{1}{2}}} = {[1 - y^{2}]}^{- \frac{1}{2}}$

$∴$ option (D ) is correct.

0 0

New answer posted

11 months ago

15. Which of the following conditions favours the existence of a substance in the solid state?

(i) High temperature

(ii) Low temperature

(iii) High thermal energy

(iv) Weak cohesive forces

0 Follower 1 View

Contributor-Level 10

15. Option (ii) Low temperature is correct since at sufficiently low temperature, the thermal energy is low, so the intermolecular forces bring the molecules of a substance closer so that they cling to one another and occupy fixed positions. They keep on vibrating about their fixed positions. Such conditions favours the existence of the substance in solid state.

0 0

New answer posted

11 months ago

94. Choose the correct answer:

The integrating factor of the differential equation $\frac{d y}{d x} - y = 2 x^{2}$ is:

(A) e^–x

(B) e^–y

(C) $\frac{1}{x}$

(D) x

0 Follower 2 Views