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New answer posted

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Class 12th Differential Equations

The general solution of $\frac{d y}{d x} + y t a n x = s e c x$ is:

(A) $y secx =tan x + c$

(B) $y tanx = secx + c$

(C) tan $x = y tanx + c$

(D) $x sec x =tan y + c$

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

Which of the following is the general solution of $\frac{d^{2} y}{d x^{2}} - 2 \frac{d y}{d x} + y = 0$ ?

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

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

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

(D) $y = A cosx + B sinx$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The differential equation of the family of curves $y^{2} = 4 a (x + a)$ is:

(A) $y^{2} 4 \frac{d y}{d x} = (x + \frac{d y}{d x})$

(B) $2 y \frac{d y}{d x} = 4 a$

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

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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n e q u a t i o n o f f a m i l y o f c u r v e s i s \\ y^{2} = 4 a (x + a) \\ \Rightarrow y^{2} = 4 a x + 4 a^{2} \dots (i) \\ D i f f e r e n t i a t i n g b o t h s i d e s, w . r . t . x, w e g e t \\ 2 y . \frac{d y}{d x} = 4 a \\ \Rightarrow y . \frac{d y}{d x} = 2 a \Rightarrow \frac{y}{2} . \frac{d y}{d x} = a \\ N o w, p u t t i n g t h e v a l u e o f a i n e q n . (i) w e g e t \\ y^{2} = 4 x (\frac{y}{2} . \frac{d y}{d x}) + 4 {(\frac{y}{2} . \frac{d y}{d x})}^{2} \\ \Rightarrow y^{2} = 2 x y \frac{d y}{d x} + y^{2} {(\frac{d y}{d x})}^{2} \\ \Rightarrow y = 2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} \\ \Rightarrow 2 x . \frac{d y}{d x} + y . {(\frac{d y}{d x})}^{2} - y = 0 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 12th Differential Equations

The order and degree of the differential equation ${[1 + {(\frac{d y}{d x})}^{2}]}^{} + \frac{d^{2} y}{d x^{2}} = 0$ are:

(A) 2, 3/2

(B) 2, 3

(C) 2, 1

(D) 3, 4

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The order and degree of the differential equation ${(\frac{d^{3} y}{d x^{3}})}^{2} - 3 {(\frac{d^{2} y}{d x^{2}})}^{3} + {(\frac{d y}{d x})}^{4} = y^{4}$ are:

(A) 1, 4

(B) 3, 4

(C) 2, 4

(D) 3, 2

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ {[\frac{d^{3} y}{d x^{3}}]}^{2} - 3 \frac{d^{2} y}{d x^{2}} + 2 {(\frac{d y}{d x})}^{4} = y^{4} \\ H e r e, t h e h i g h e s t d e r i v a t i v e i s \frac{d^{3} y}{d x^{3}} . \\ ∴ o r d e r o f t h e d i f f e r e n t i a l e q u a t i o n i s 3 \\ a n d t h e p o w e r o f t h e h i g h e s t d e r i v a t i v e i s 2 \\ ∴ i t s degreeis 2 \\ H e n c e, t h e c o r r e c t o p t i o n i s (d) . \end{array}$

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New answer posted

a year ago

Class 12th Differential Equations

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

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

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

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

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

0 Follower 3 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The differential equation for which $y = a c o s x + b s i n x$ is a solution, is:

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

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

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

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

0 Follower 4 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

The solution of the equation $(2 y - 1) d x - (2 x + 3) d y = 0$ is:

(A) $\frac{2 x - 1}{2 y + 3} = k$

(B) $\frac{2 y + 1}{2 x - 3} = k$

(C) $\frac{2 x + 3}{2 y - 1} = k$

(D) $\frac{2 x - 1}{2 y - 1} = k$

0 Follower 6 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s \\ (2 y - 1) d x - (2 x + 3) d y = 0 \\ \Rightarrow (2 x + 3) d y = (2 y - 1) d x \\ \Rightarrow \frac{d y}{2 y - 1} = \frac{d x}{2 x + 3} \\ I n t e g r a t i n g b o t h s i d e s, w e g e t \\ \int \frac{d y}{2 y - 1} = \int \frac{d x}{2 x + 3} \\ \Rightarrow \frac{1}{2} l o g | 2 y - 1 | = \frac{1}{2} l o g | 2 x + 3 | + l o g c \\ \Rightarrow l o g | 2 y - 1 | = l o g | 2 x + 3 | + 2 l o g c \\ \Rightarrow l o g | 2 y - 1 | - l o g | 2 x + 3 | = l o g c^{2} \\ \Rightarrow l o g | \frac{2 y - 1}{2 x + 3} | = l o g c^{2} \\ \Rightarrow \frac{2 y - 1}{2 x + 3} = c^{2} \Rightarrow \frac{2 x + 3}{2 y - 1} = \frac{1}{c^{2}} \\ \Rightarrow \frac{2 x + 3}{2 y - 1} = k, w h e r e k = \frac{1}{c^{2}} \\ H e n c e, t h e c o r r e c t o p t i o n i s (c) . \end{array}$

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New answer posted

a year ago

Class 12th Differential Equations

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

0 Follower 5 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Class 12th Differential Equations

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

0 Follower 2 Views

alok kumar singh

Contributor-Level 10

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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