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Class 12th Differential Equations

The degree of the differential equation $\frac{d^{2} y}{d x^{2}} {(\frac{d y}{d x})}^{3} + 6 y^{5} = 0$ is:

(A) 1

(B) 2

(C) 3

(D) 5

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e degree o f t h e g i v e n D i f f e r e n t i a l e q u a t i o n i s 1 a s t h e p o w e r o f t h e h i g h e s t o r d e r i s 1 . \\ H e n c e, t h e c o r r e c t o p t i o n i s (a) . \end{array}$

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New answer posted

a year ago

Class 12th Differential Equations

The general solution of $e^{x} c o s y d x - e^{x} s i n y d y = 0$ is:

(A) $e^{x} cosy = k$

(B) $e^{x} siny = k$

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

(D) $e^{x} = ksin y$

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The differential equation $y \frac{d y}{d x} + x = c$ represents:

(A) Family of hyperbolas

(B) Family of parabolas

(C) Family of ellipses

(D) Family of circles

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

$t a n^{- 1} x + t a n^{- 1} y = c$ is the general solution of the differential equation:

(A) $\frac{d y}{d x} + \frac{1 + y^{2}}{1 + x^{2}}$

(B) $\frac{d y}{d x} + \frac{1 + x^{2}}{1 + y^{2}}$

(C) $(1 + x^{2}) d y + (1 + y^{2}) d x = 0$
(D) $(1 + x^{2}) d x + (1 + y^{2}) d y = 0$

0 Follower 9 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

Integrating factor of the differential equation $(1 - x^{2}) \frac{d y}{d x} - x y = 1$

(A) $- x$

(B) $x + 1$

(C) $1 - x^{2}$

(D) $log(1 - x^{2})$

0 Follower 14 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol: $\begin{array}{l} \end{array}$

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New answer posted

a year ago

Class 12th Differential Equations

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

0 Follower 8 Views

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

a year ago

Class 12th Differential Equations

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 6 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

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 18 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

a year ago

Class 12th Differential Equations

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 5 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

a year ago

Class 12th Differential Equations

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 4 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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