Differential Equations

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New answer posted•a year 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) $- x$

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:

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

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New answer posted•a year 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

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:

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

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New answer posted•a year 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) $tanx \cdot tany = k$

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:

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

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New answer posted•a year 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) $sinx$

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:

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

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New answer posted•a year 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

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:

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

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New answer posted•a year 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) $\frac{d^{2} y}{d x^{2}} - α y = 0$

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:

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

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New answer posted•a year 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) $\frac{d^{2} y}{d x^{2}} + 2 y = 0$

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:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The order and degree of the differential equation $\frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{\frac{1}{4}} + x^{\frac{1}{5}} = 0$ respectively, are

(A) 2 and not defined

(B) 2 and 2

(C) 2 and 3

(D) 3 and 3

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:

$\begin{array}{l} G i v e n d i f f e r e n t i a l e q u a t i o n i s \\ \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{\frac{1}{4}} + x^{\frac{1}{5}} = 0 \\ \Rightarrow \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{\frac{1}{4}} = - x^{\frac{1}{5}} \\ Since, t h e o f \frac{d y}{d x} i s i n f r a c t i o n . \\ S o, t h e degree o f t h e g i v e n d i f f e r e n t i a l e q u a t i o n i s n o t d e f i n e d a s t h e o r d e r i s 2 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{\frac{3}{2}} = \frac{d^{2} y}{d x^{2}}$ is

(A) 4

(B) $\frac{3}{2}$

(C) Not defined

(D) 2

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:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ {[1 + {(\frac{d y}{d x})}^{2}]}^{3 / 2} = (\frac{d^{2} y}{d x^{2}}) \\ S q u a r i n g b o t h s i d e s, w e h a v e \\ {[1 + {(\frac{d y}{d x})}^{2}]}^{3} = {(\frac{d^{2} y}{d x^{2}})}^{2} \\ S o, t h e degree o f t h e g i v e n d i f f e r e n t i a l e q u a t i o n i s 2 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Choose the correct answer from the given four options in each of the Exercises from 1 to 42 (M.C.Q)

Q1.The degree of the differential equation ${(\frac{d^{2} y}{d x^{2}})}^{2} + {(\frac{d y}{d x})}^{2} = x s i n x (\frac{d y}{d x})$ is:

(A) 1

(B) 2

(C) 3

(D) Not defined

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e degree o f g i v e n d i f f e r e n t i a l e q u a t i o n i s n o t d e f i n e d b e c a u s e t h e v a l u e o f \\ s i n (\frac{d y}{d x}) o n expansionwillbeincreasing p o w e r o f (\frac{d y}{d x}) . \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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