Tags Class 12th

Class 12th

Get insights from 12k questions on Class 12th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 12th

Follow Ask Question

12k

Questions

Discussions

Active Users

Followers

New answer posted

7 months ago

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) $x sec x =tan y + c$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

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) $y = A cosx + B sinx$

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

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

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

Class 12th Differential Equations

The general solution of the differential equation $\frac{d y}{d x} = e^{x^{2} / 2} + x y$ is:

(A) $y = c e^{- \frac{x^{2}}{2}}$

(B) $y = c e^{\frac{x^{2}}{2}}$

(C) $y = (x + c) e^{\frac{x^{2}}{2}}$

(D) $y = c - x e^{\frac{x^{2}}{2}}$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

New answer posted

7 months ago

Class 12th Differential Equations

The curve for which the slope of the tangent at any point is equal to the ratio of the abscissa to the ordinate of the point is:

(A) An ellipse

(B) A parabola

(C) A circle

(D) A rectangular hyperbola

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

0 0

Get authentic answers from experts, students and alumni that you won't find anywhere else

On Shiksha, get access to

66k Colleges
1.2k Exams
684k Reviews
1800k Answers

Community GuidelinesUser Points System

QuestionsDiscussionsTags

Need guidance on career and education? Ask our experts

Your Question

Class 12th

Questions

Discussions

Active Users

Followers

All

Discussions

Unanswered Questions

Share Your College Life Experience