Ncert Solutions Maths class 12th

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The general solution of the differential equation $(e^{x} + 1) y d y = (y + 1) e^{x} d x$ is:

(A) $y + 1 = k (e^{x} + 1)$

(B) $y + 1 = e^{x} + 1 + k$

(C) $y = l o g {k (y + 1) (e^{x} + 1)}$

(D) y $= (e^{x} + 1 / y + 1) +k$

10 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ (e^{x} + 1) y d y = (y + 1) e^{x} d x \\ \Rightarrow \frac{y}{y + 1} d y = \frac{e^{x}}{e^{x} + 1} d x \\ I n t e g r a t i n g b o t h s i d e s, w e h a v e \\ \Rightarrow \int \frac{y}{y + 1} d y = \int \frac{e^{x}}{e^{x} + 1} d x \\ \Rightarrow \int \frac{y + 1 - 1}{y + 1} d y = \int \frac{e^{x}}{e^{x} + 1} d x \\ \Rightarrow \int 1 . d y + \int \frac{1}{y + 1} d y = \int \frac{e^{x}}{e^{x} + 1} d x \\ \Rightarrow y - l o g | y + 1 | = l o g | e^{x} + 1 | + l o g k \\ \Rightarrow y = l o g | y + 1 | + l o g | e^{x} + 1 | + l o g k \\ \Rightarrow y = l o g | k (y + 1) (e^{x} + 1) | \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The solution of the differential equation $s i n x \frac{d y}{d x} + y = x$ is:

(A) $x (y + cosx) = sinx + c$

(B) $x (y - cosx) = sinx + c$

(C) $x ycos x = sinx + c$

(D) $x (y + cosx) = cosx + c$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \frac{d y}{d x} + \frac{y}{x} = s i n x \\ Since i t i s l i n e a r d i f f e r e n t i a l e q u a t i o n w h e r e P = \frac{1}{x} a n d Q = s i n x \\ ∴ I . F . = e^{\int P . d x} = e^{\int \frac{1}{x} . d x} = e^{l o g x} = x \\ S o, t h e s o l u t i o n i s \\ y \times I . F . = \int Q \times I . F . d x + c \\ \Rightarrow y \times x = \int s i n x . x d x + c \Rightarrow y \times x = \int x s i n x . d x + c \\ \Rightarrow y x = x . \int s i n x d x - \int (D (x) \int s i n x d x) d x + c \\ \Rightarrow y x = x (- c o s x) - \int - c o s x d x + c \\ \Rightarrow y x = - x c o s x + \int c o s x d x + c \\ \Rightarrow y x = - x c o s x + s i n x + c \\ \Rightarrow y x + x c o s x = s i n x + c \\ \Rightarrow x (y + c o s x) = s i n x + c \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The general solution of $\frac{d y}{d x} + y t a n x = s e c x$ is:

(A) $y secx =tan x + c$

(B) $y tanx = secx + c$

(C) tan $x = y tanx + c$

(D) $x sec x =tan y + c$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ \frac{d y}{d x} + y t a n x = s e c x \\ Since i t i s l i n e a r d i f f e r e n t i a l e q u a t i o n w h e r e P = t a n x a n d Q = s e c x \\ ∴ I . F . = e^{\int P . d x} = e^{\int t a n x . d x} = e^{l o g s e c x} = s e c x \\ S o, t h e s o l u t i o n i s \\ y \times I . F . = \int Q \times I . F . d x + c \\ \Rightarrow y \times s e c x = \int s e c x . s e c x d x + c \\ \Rightarrow y \times s e c x = \int s e c^{2} x . d x + c \\ \Rightarrow y s e c x = t a n x + c \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

Which of the following is the general solution of $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 0$ ?

(A) $y = (A x + B) e^{x}$

(B) $y = (A x + B) e^{- x}$

(C) $y = A e^{x} + B e^{- x}$

(D) $y = A cosx + B sinx$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 0 \\ Since t h e a b o v e e q u a t i o n i s o f o r d e r a n d f i r s t \\ ∴ D^{2} y - 2 D y + y = 0, w h e r e D = \frac{d}{d x} \\ \Rightarrow (D^{2} - 2 D + 1) = 0 \\ ∴ a u x i l i a r y e q u a t i o n i s m^{2} - 2 m + 1 = 0 \\ \Rightarrow {(m - 1)}^{2} = 0 \Rightarrow m = 1, 1 \\ I f t h e r o o t s o f A u x i l i a r y e q u a t i o n a r e r e a l a n d e q u a l s a y (m) t h e n, \\ C F = (c_{1} x + c_{2}) . e^{m x} \\ ∴ C F = (A x + B) . e^{x} \\ S o, y = (A x + B) . e^{x} \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The differential equation of the family of curves $y^{2} = 4 a (x + a)$ is:

(A) $y^{2} 4 \frac{d y}{d x} = (x + \frac{d y}{d x})$

(B) $2 y \frac{d y}{d x} = 4 a$

(C) $y \frac{d^{2} y}{d x^{2}} + {\frac{d y}{d x}}^{2} = 0$

(D) $2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} - y$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n o f f a m i l y o f c u r v e s i s \\ y^{2} = 4 a (x + a) \\ \Rightarrow y^{2} = 4 a x + 4 a^{2} \dots (i) \\ D i f f e r e n t i a t i n g b o t h s i d e s, w . r . t . x, w e g e t \\ 2 y . \frac{d y}{d x} = 4 a \\ \Rightarrow y . \frac{d y}{d x} = 2 a \Rightarrow \frac{y}{2} . \frac{d y}{d x} = a \\ N o w, p u t t i n g t h e v a l u e o f a i n e q n . (i) w e g e t \\ y^{2} = 4 x (\frac{y}{2} . \frac{d y}{d x}) + 4 {(\frac{y}{2} . \frac{d y}{d x})}^{2} \\ \Rightarrow y^{2} = 2 x y \frac{d y}{d x} + y^{2} {(\frac{d y}{d x})}^{2} \\ \Rightarrow y = 2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} \\ \Rightarrow 2 x . \frac{d y}{d x} + y . {(\frac{d y}{d x})}^{2} - y = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The order and degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{} + \frac{d^{2} y}{d x^{2}} = 0$ are:

(A) 2, 3/2

(B) 2, 3

(C) 2, 1

(D) 3, 4

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ [1 + {(\frac{d y}{d x})}^{2}] = \frac{d^{2} y}{d x^{2}} \\ H e r e, t h e h i g h e s t d e r i v a t i v e i s 2, \\ ∴ o r d e r = 2 \\ a n d t h e p o w e r o f t h e h i g h e s t d e r i v a t i v e i s 1 \\ ∴ degree = 1 \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The order and degree of the differential equation ${(\frac{d^{3} y}{d x^{3}})}^{2} - 3 {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{4} = y^{4}$ are:

(A) 1, 4

(B) 3, 4

(C) 2, 4

(D) 3, 2

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ {[\frac{d^{3} y}{d x^{3}}]}^{2} - 3 \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{4} = y^{4} \\ H e r e, t h e h i g h e s t d e r i v a t i v e i s \frac{d^{3} y}{d x^{3}} . \\ ∴ o r d e r o f t h e d i f f e r e n t i a l e q u a t i o n i s 3 \\ a n d t h e p o w e r o f t h e h i g h e s t d e r i v a t i v e i s 2 \\ ∴ i t s degreeis 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The solution of $\frac{d y}{d x} + y = e^{- x}, y (0) = 0$ is:

(A) $y = e^{- x} (x - 1)$

(B) $y = x e^{x}$

(C) $y = x e^{- x} + 1$

(D) $y = x e^{- x}$

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \frac{d y}{d x} + y = e^{- x} Since$
$\begin{array}{l} it i s l i n e a r d i f f e r e n t i a l e q u a t i o n w h e r e P = 1 a n d Q = e^{- x} \\ ∴ I . F . = e^{\int P . d x} = e^{\int 1 . d x} = e^{x} \\ S o, t h e s o l u t i o n i s \\ y \times I . F . = \int Q \times I . F . d x + c \Rightarrow y \times e^{x} = \int e^{- x} . e^{x} d x + c \\ \Rightarrow y \times e^{x} = \int e^{0} . d x + c \Rightarrow y . e^{x} = \int 1 . d x + c \\ \Rightarrow y . e^{x} = x + c \\ P u t y = 0 a n d x = 0 \\ ∴ 0 = (0 + c) ∴ c = 0 \\ ∴ e q u a t i o n i s y . e^{x} = x \\ S o, y = x . e^{- x} \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The differential equation for which $y = a c o s x + b s i n x$ is a solution, is:

(A) $\frac{d^{2} y}{d x^{2}} + y = 0$

(B) $\frac{d^{2} y}{d x^{2}} - y = 0$

(C) $\frac{d^{2} y}{d x^{2}} + (a + b) y = 0$

(D) $\frac{d^{2} y}{d x^{2}} + (a - b) y = 0$

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n i s y = a c o s x + b s i n x \\ ∴ \frac{d y}{d x} = - a s i n x + b c o s x \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = - a c o s x - b s i n x \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = - (a c o s x + b s i n x) \\ \Rightarrow \frac{d^{2} y}{d x^{2}} = - y \Rightarrow \frac{d^{2} y}{d x^{2}} + y = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Ncert Solutions Maths class 12th Differential Equations

The solution of the equation $(2 y - 1) d x - (2 x + 3) d y = 0$ is:

(A) $\frac{2 x - 1}{2 y + 3} = k$

(B) $\frac{2 y + 1}{2 x - 3} = k$

(C) $\frac{2 x + 3}{2 y - 1} = k$

(D) $\frac{2 x - 1}{2 y - 1} = k$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ (2 y - 1) d x - (2 x + 3) d y = 0 \\ \Rightarrow (2 x + 3) d y = (2 y - 1) d x \\ \Rightarrow \frac{d y}{2 y - 1} = \frac{d x}{2 x + 3} \\ I n t e g r a t i n g b o t h s i d e s, w e g e t \\ \int \frac{d y}{2 y - 1} = \int \frac{d x}{2 x + 3} \\ \Rightarrow \frac{1}{2} l o g | 2 y - 1 | = \frac{1}{2} l o g | 2 x + 3 | + l o g c \\ \Rightarrow l o g | 2 y - 1 | = l o g | 2 x + 3 | + 2 l o g c \\ \Rightarrow l o g | 2 y - 1 | - l o g | 2 x + 3 | = l o g c^{2} \\ \Rightarrow l o g | \frac{2 y - 1}{2 x + 3} | = l o g c^{2} \\ \Rightarrow \frac{2 y - 1}{2 x + 3} = c^{2} \Rightarrow \frac{2 x + 3}{2 y - 1} = \frac{1}{c^{2}} \\ \Rightarrow \frac{2 x + 3}{2 y - 1} = k, w h e r e k = \frac{1}{c^{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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Ncert Solutions Maths class 12th

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