Differential Equations

Get insights from 323 questions on Differential Equations, answered by students, alumni, and experts. You may also ask and answer any question you like about Differential Equations

323
Questions
0
Discussions
0
Active Users
1
Followers

Have a question about Differential Equations? Ask our experts

Ask Your Question

Keep it short & simple. Type complete word. Avoid abusive language.

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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:

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

Upvote

Differential Equations

Questions

Discussions

Active Users

Followers