Class 12th

Find the general solution of $\frac{d y}{d x} - 3 y = s i n 2 x$

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Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \frac{d y}{d x} - 3 y = s i n 2 x \\ H e r e, P = - 3 a n d Q = s i n 2 x \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d x} = e^{\int - 3 . d x} = e^{- 3 x} \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ y * I . F . = \int Q * I . F . d x + c \\ \Rightarrow y . e^{- 3 x} = \int s i n 2 x . e^{- 3 x} d x + c \\ L e t I = \int \underset{I}{s i n 2 x} . \underset{I I}{e^{- 3 x}} d x \\ \Rightarrow I = s i n 2 x . \int e^{- 3 x} d x - \int (D (s i n 2 x) . \int e^{- 3 x} d x) d x \\ \Rightarrow I = s i n 2 x . \frac{e^{- 3 x}}{- 3} - \int 2 c o s 2 x . \frac{e^{- 3 x}}{- 3} d x \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x + \frac{2}{3} \int \underset{I}{c o s 2 x} . \underset{I I}{e^{- 3 x}} d x \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x + \frac{2}{3} [c o s 2 x . \int e^{- 3 x} d x - \int (D (c o s 2 x) . \int e^{- 3 x} d x) d x] \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x + \frac{2}{3} [c o s 2 x . \frac{e^{- 3 x}}{- 3} - \int 2 s i n 2 x . \frac{e^{- 3 x}}{- 3} d x] \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x - \frac{2}{9} c o s 2 x . e^{- 3 x} - \frac{4}{9} \int s i n 2 x . e^{- 3 x} d x \\ \Rightarrow I = \frac{e^{- 3 x}}{- 3} s i n 2 x - \frac{2}{9} c o s 2 x . e^{- 3 x} - \frac{4}{9} I \\ \Rightarrow I + \frac{4}{9} I = \frac{e^{- 3 x}}{- 3} s i n 2 x - \frac{2}{9} e^{- 3 x} c o s 2 x \\ \Rightarrow \frac{13}{9} I = - \frac{1}{9} [3 e^{- 3 x} s i n 2 x + 2 e^{- 3 x} c o s 2 x] \\ \Rightarrow I = - \frac{1}{13} e^{- 3 x} [3 s i n 2 x + 2 c o s 2 x] \\ ∴ T h e e q u a t i o n b e c o m e s \\ \Rightarrow y . e^{- 3 x} = - \frac{1}{13} e^{- 3 x} [3 s i n 2 x + 2 c o s 2 x] + c \\ ∴ y = - \frac{1}{13} [3 s i n 2 x + 2 c o s 2 x] + c . e^{3 x} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s y = - [\frac{3 s i n 2 x + 2 c o s 2 x}{13}] + c . e^{3 x} \end{array}$

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New answer posted

a year ago

Solve: $\frac{d y}{d x} = c o s (x + y) + s i n (x + y)$ . [Hint: Substitute $x + y = z$ ]

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Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted

a year ago

Find the general solution of $(1 + t a n y) (d x- d y) + 2 x d y = 0$ .

0 Follower 5 Views

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n t h a t (1 + t a n y) (d x - d y) + 2 x d y = 0 \\ \Rightarrow (1 + t a n y) d x - (1 + t a n y) d y + 2 x d y = 0 \\ \Rightarrow (1 + t a n y) d x - (1 + t a n y - 2 x) d y = 0 \\ \Rightarrow (1 + t a n y) \frac{d x}{d y} = (1 + t a n y - 2 x) \Rightarrow \frac{d x}{d y} = \frac{1 + t a n y - 2 x}{1 + t a n y} \\ \Rightarrow \frac{d x}{d y} = 1 - \frac{2 x}{1 + t a n y} \Rightarrow \frac{d x}{d y} + \frac{2 x}{1 + t a n y} = 1 \\ H e r e, P = \frac{2}{1 + t a n y} a n d Q = 1 \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d y} = e^{\int \frac{2}{1 + t a n y} . d y} = e^{\int \frac{2 c o s y}{s i n y + c o s y} . d y} \\ = e^{\int \frac{s i n y + c o s y - s i n y + c o s y}{s i n y + c o s y} . d y} = e^{\int (1 + \frac{c o s y - s i n y}{s i n y + c o s y}) . d y} \\ = e^{\int 1 . d y} . e^{\int \frac{c o s y - s i n y}{s i n y + c o s y} . d y} = e^{y} . e^{l o g (s i n y + c o s y)} = e^{y} . (s i n y + c o s y) \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ x * I . F . = \int Q * I . F . d y + c \\ \Rightarrow x . e^{y} . (s i n y + c o s y) = \int 1 . e^{y} . (s i n y + c o s y) d y + c \\ \Rightarrow x . e^{y} . (s i n y + c o s y) = e^{y} . s i n y + c [? \int e^{x} [f (x) + f^{'} (x)] d x = e^{x} f (x) + c] \\ \Rightarrow x (s i n y + c o s y) = s i n y + c . e^{- y} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s x (s i n y + c o s y) = s i n y + c . e^{- y} \end{array}$

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New answer posted

a year ago

Solve: $(\frac{d y}{d x}) + x y = x (s i n x + l o g x)$ .

0 Follower 2 Views

Contributor-Level 10

This is a Long Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} T h e g i v e n d i f f e r e n t i a l e q u a t i o n i s y + \frac{d}{d x} (x y) = x (s i n x + l o g x) \\ \Rightarrow y + x . \frac{d y}{d x} + y = x (s i n x + l o g x) \\ \Rightarrow x \frac{d y}{d x} = x (s i n x + l o g x) - 2 y \\ \Rightarrow \frac{d y}{d x} = (s i n x + l o g x) - \frac{2 y}{x} \\ \Rightarrow \frac{d y}{d x} + \frac{2}{x} y = (s i n x + l o g x) \\ H e r e, P = \frac{2}{x} a n d Q = (s i n x + l o g x) \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d x} = e^{\int \frac{2}{x} . d x} = e^{2 l o g x} = e^{l o g x^{2}} = x^{2} \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ y * I . F . = \int Q * I . F . d x + c \\ \Rightarrow y . x^{2} = \int (s i n x + l o g x) x^{2} d x + c \dots (1) \\ L e t I = \int (s i n x + l o g x) x^{2} d x \\ = \int \underset{I}{x^{2}} \underset{I I}{s i n x} d x + \int \underset{I I}{x^{2}} \underset{I}{l o g x} d x \\ = [x^{2} . \int s i n x d x - \int (D (x^{2}) . \int s i n x d x) d x] + [l o g x . \int x^{2} d x - \int (D (l o g x) . \int x^{2} d x) d x] \\ = [x^{2} (- c o s x) - 2 \int - x c o s x d x] + [l o g x . \frac{x^{3}}{3} - \int \frac{1}{x} . \frac{x^{3}}{3} d x] \\ = [- x^{2} c o s x + 2 (x s i n x - \int 1 . s i n x d x)] + [\frac{x^{3}}{3} l o g x - \frac{1}{3} \int x^{2} d x] \\ = - x^{2} c o s x + 2 x s i n x + 2 c o s x + \frac{x^{3}}{3} l o g x - \frac{1}{9} x^{3} \\ N o w f r o m e q (1) w e g e t, \\ y . x^{2} = - x^{2} c o s x + 2 x s i n x + 2 c o s x + \frac{x^{3}}{3} l o g x - \frac{1}{9} x^{3} + c \\ ∴ y = - c o s x + \frac{2 s i n x}{x} + \frac{2 c o s x}{x^{2}} + \frac{x l o g x}{3} - \frac{1}{9} x + c . x^{- 2} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s \\ y = - c o s x + \frac{2 s i n x}{x} + \frac{2 c o s x}{x^{2}} + \frac{x l o g x}{3} - \frac{1}{9} x + c . x^{- 2} \end{array}$

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New answer posted

a year ago

The least value of the function f(x) = ax + ( $\frac{b}{x}$ ), where a > 0, b > 0, x > 0, is ___.

0 Follower 6 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

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New answer posted

a year ago

The function f(x) = $\frac{2 x^{2} + 1}{x^{4}}$ , x > 0 decreases in the interval ___.

0 Follower 4 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e f (x) = \frac{2 x^{2} - 1}{x^{4}} \\ \Rightarrow f^{'} (x) = \frac{x^{4} (4 x) - (2 x^{2} - 1) . 4 x^{3}}{x^{8}} \\ \Rightarrow f^{'} (x) = \frac{4 x^{5} - (2 x^{2} - 1) . 4 x^{3}}{x^{8}} = \frac{4 x^{3} [x^{2} - 2 x^{2} + 1]}{x^{8}} = \frac{4 [- x^{2} + 1]}{x^{5}} \\ F o r d e creasing t h e f u n c t i o n f^{'} (x) < 0 \\ ∴ \frac{4 (- x^{2} + 1)}{x^{5}} < 0 \Rightarrow - x^{2} + 1 < 0 \Rightarrow x^{2} > 1 \\ ∴ x > \pm 1 \Rightarrow x \in (1, \infty) \\ H e n c e, t h e r e q u i r e d i n t e r v a l i s (1, \infty) . \end{array}$

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New answer posted

a year ago

The values of a for which the function f(x) = sinx – ax + b increases on R are __.

0 Follower 105 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

$\begin{array}{l} W e h a v e f (x) = s i n x - a x + b \Rightarrow f^{'} (x) = c o s x - a \\ F o r increasing t h e f u n c t i o n f^{'} (x) > 0 \\ ∴ c o s x - a > 0 \\ Since c o s x \in [- 1, 1] \\ ∴ a < - 1 \Rightarrow a \in (- \infty, - 1) \\ H e n c e, t h e v a l u e o f a \in (- \infty, - 1) . \end{array}$

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New answer posted

a year ago

The equation of the normal to the curve y = tanx at (0, 0) is ____.

0 Follower 5 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol.

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

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New answer posted

a year ago

The curves y = 4x² + 2x – 8 and y = x³ – x + 13 touch each other at the point ___.

0 Follower 4 Views

Contributor-Level 10

This is a Objective Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} W e h a v e y = 4 x^{2} + 2 x - 8 \dots (i) \\ a n d y = x^{3} - x + 13 \dots (i i) \\ D i f f e r e n t i a t i n g e q . (i) w . r . t . x, w e h a v e \\ \frac{d y}{d x} = 8 x + 2 \Rightarrow m_{1} = 8 x + 2 [m_{1} i s s l o p e o f c u r v e (i)] \\ D i f f e r e n t i a t i n g e q . (i i) w . r . t . x, w e g e t \\ \frac{d y}{d x} = 3 x^{2} - 1 \Rightarrow m_{2} = 3 x^{2} - 1 [m_{2} i s s l o p e o f c u r v e (i i)] \\ I f t h e t w o c u r v e s t o u c h e a c h o t h e r, t h e n m_{1} = m_{2} \\ ∴ 8 x + 2 = 3 x^{2} - 1 \\ \Rightarrow 3 x^{2} - 8 x - 3 = 0 \Rightarrow 3 x^{2} - 9 x + x - 3 = 0 \\ \Rightarrow 3 x (x - 3) + 1 (x - 3) = 0 \Rightarrow (x - 3) (3 x + 1) = 0 \\ ∴ x = 3, \frac{- 1}{3} \\ P u t t i n g x = 3 i n e q (i), w e g e t \\ y = 4 {(3)}^{2} + 2 (3) - 8 = 36 + 6 - 8 = 34 \\ S o, t h e r e q u i r e d point i s (3, 34) \\ N o w f o r x = \frac{- 1}{3} \\ y = 4 {(\frac{- 1}{3})}^{2} + 2 (\frac{- 1}{3}) - 8 = 4 * \frac{1}{9} - \frac{2}{3} - 8 \\ = \frac{4}{9} - \frac{2}{3} - 8 = \frac{4 - 6 - 72}{9} = \frac{- 74}{9} \\ ∴ O t h e r r e q u i r e d point i s (\frac{- 1}{3}, \frac{- 74}{9}) . \\ H e n c e, t h e r e \end{array}$

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New answer posted

a year ago

The maximum value of (1/x)^x is:

(A) e

(B) $e^{e}$

(C) $\frac{1}{e^{e}}$

(D) $e^{\frac{1}{e}}$

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