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Class 12th Differential Equations

The general solution of $\frac{d y}{d x} = 2 x e^{x - y}$ is:

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

(B) $e^{- y} + e^{x^{2}} = c$

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

(D) $e^{x^{2 + y}} = c$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The family $y = A x + A^{3}$ of curves corresponds to a differential equation of order:

(A) 3

(B) 2

(C) 1

(D) Not defined

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The differential equation of the family of curves $x^{2} + y^{2} - 2 a y = 0,$ where a is an arbitrary constant, is:

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

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

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

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The solution of $\frac{d y}{d x} + y = e^{x}$ is:

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

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

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

(D) $x = \frac{e^{y}}{y} + \frac{k}{y}$

0 Follower 16 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The solution of the differential equation $c o s x s i n y d x + s i n x c o s y d y = 0$ is:

(A) $\frac{Sinx}{Siny} = c$

(B) $sinxsin y = c$

(C) $sinx + siny = c$

(D) $cosx cos y = c$

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

$y = A e^{m x} + B e^{- m x}$ satisfies which of the following differential equations?

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

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

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

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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The integrating factor of the differential equation $\frac{d y}{d x} + y = \frac{1 + y}{x}$ is:

(A) $x / e^{x^{x}}$

(B) $e^{x} / x$

(C) $x e^{x}$

(D) $e^{x}$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

a year ago

Class 12th Differential Equations

The solution of the differential equation $\frac{d y}{d x} + \frac{1 + y^{2}}{1 + x^{2}}$ is:

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

(B) $y - x = k (1 + x y)$

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

(D) $tan(x y) = k$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The integrating factor of the differential equation $\frac{d y}{d x} + y t a n x - s e c x = 0$ is:

(A) Cos $x$

(B) $Secx$

(C) $e^{cosx}$

(D) $e^{secx}$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

a year ago

Class 12th Differential Equations

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

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

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

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

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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