Differential Equations

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

2 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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

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

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

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

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

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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

2 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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

16 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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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) $cosx cos y = c$

9 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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

$y = A e^{m x} + B e^{- m x}$ 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) $\frac{d^{2} y}{d x^{2}} + m^{2} y = 0$

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:

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

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New answer posted•a year ago

Differential Equations 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) $e^{x}$

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:

Upvote

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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) $tan(x y) = k$

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:

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

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