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3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Which of the following is a second-order differential equation?

(A) ${(y^{'})}^{2} + x = y^{2}$

(B) $y^{'} y^{'} + y = sinx$

(C) $y^{'} + {(y^{'})}^{2} + y = 0$

(D) $y^{'} = y^{2}$

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alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} second o r d e r d i f f e r e n t i a l e q u a t i o n i s y^{'} y^{''} + y = s i n x \\ H e n c e, t h e c o r r e c t o p t i o n i s (b) . \end{array}$

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New answer posted

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The number of solutions of $\frac{d y}{d x} + \frac{y + 1}{x - 1} = 1$ when $y (1) = 2$ is:

(A) None

(B) One

(C) Two

(D) Infinite

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solution of $\frac{d y}{d x} - y = 1, y (0) = 1$ is given by

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

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

(C) $x y = - 1$

(D) $y = 2 e^{x} - 1$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Integrating factor of $x \frac{d y}{d x} - y = x^{4} - 3 x$ is:

(A) $x$

(B) $logx$

(C) $\frac{1}{x}$

(D) $- x$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ x \frac{d y}{d x} - y = x^{4} - 3 x \\ \frac{d y}{d x} - \frac{y}{x} = x^{3} - 3 \\ H e r e, P = - \frac{1}{x} a n d Q = x^{3} - 3 \\ S o, integrating f a c t o r = e^{\int P d x} = e^{\int - \frac{1}{x} d x} = e^{- l o g x} = e^{l o g \frac{1}{x}} = \frac{1}{x} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Family $y = A x + A^{3}$ of curves is represented by the differential equation of degree:

(A) 1

(B) 2

(C) 3

(D) 4

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s y = A x + A^{3} \\ D i f f e r e n t i a t i n g b o t h s i d e s, w e g e t \\ \frac{d y}{d x} = A w h i c h h a s degree 1 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

$S o l u t i o n o f t h e d i f f e r e n t i a l e q u a t i o n t a n y s e c^{2} x d x + t a n x s e c^{2} y d y = 0$ is:

(A) $tanx + tany = k$

(B) $tanx -tan y = k$

(C) $\frac{tanx}{tany} = k$

(D) $tanx \cdot tany = k$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Integrating factor of the differential equation $c o s x \frac{d y}{d x} + y s i n x = 1$ is:

(A) $cosx$

(B) $tanx$

(C) $secx$

(D) $sinx$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solution of the differential equation $x d y - y d x = 0$ represents:

(A) A rectangular hyperbola

(B) A parabola whose vertex is at the origin

(C) A straight line passing through the origin

(D) A circle whose center is at the origin

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The differential equation for $y = A c o s α x + B s i n α x$ , where A and B are arbitrary constants, is

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

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

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

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

3 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

If $y = e^{- x} (A c o s x + B s i n x)$ , then $y$ is a solution of

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

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

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

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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