Differential Equations

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t t h e s l o p e o f tangent t o a c u r v e a t (x, y) i s \\ \frac{d y}{d x} = \frac{y - 1}{x^{2} + x} \Rightarrow \frac{d y}{y - 1} = \frac{d x}{x^{2} + x} \\ I n t e g r a t i n g b o t h s i d e s, w e h a v e \\ \int \frac{d y}{y - 1} = \int \frac{d x}{x^{2} + x} \\ \Rightarrow \int \frac{d y}{y - 1} = \int \frac{d x}{x^{2} + x + \frac{1}{4} - \frac{1}{4}} [M a k i n g p e r f e c t s q u a r e] \\ \Rightarrow \int \frac{d y}{y - 1} = \int \frac{d x}{{(x + \frac{1}{2})}^{2} - {(\frac{1}{2})}^{2}} \\ \Rightarrow l o g | y - 1 | = \frac{1}{2 * \frac{1}{2}} l o g | \frac{x + \frac{1}{2} - \frac{1}{2}}{x + \frac{1}{2} + \frac{1}{2}} | + l o g c \\ \Rightarrow l o g | y - 1 | = l o g | \frac{x}{x + 1} | + l o g c \\ \Rightarrow l o g | y - 1 | = l o g | c (\frac{x}{x + 1}) | \\ ∴ y - 1 = \frac{c x}{x + 1} \Rightarrow (y - 1) (x + 1) = c x \\ Since, t h e l i n e i s passing t h r o u g h t h e point (1, 0), t h e n \\ (0 - 1) (1 + 1) = c (1) \Rightarrow c = 2 \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s (y - 1) (x + 1) = 2 x . \end{array}$

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New answer posted

9 months ago

Find the equation of a curve passing through $(2, 1)$ if the slope of the tangent to the curve at any point $(x, y)$ is $\frac{x^{2} + y^{2}}{2 x y} .$

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Contributor-Level 10

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

Sol:

$\begin{array}{l} G i v e n t h a t t h e s l o p e o f tangent t o a c u r v e a t (x, y) i s \\ \frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 x y} \\ I t i s a homogeneous d i f f e r e n t i a l e q u a t i o n \\ S o, p u t y = v x \Rightarrow \frac{d y}{d x} = v + x . \frac{d v}{d x} \\ \Rightarrow v + x . \frac{d v}{d x} = \frac{x^{2} + v^{2} x^{2}}{2 x . v x} \\ \Rightarrow v + x . \frac{d v}{d x} = \frac{1 + v^{2}}{2 v} \\ \Rightarrow x . \frac{d v}{d x} = \frac{1 + v^{2}}{2 v} - v \Rightarrow x . \frac{d v}{d x} = \frac{1 + v^{2} - 2 v^{2}}{2 v} \\ \Rightarrow x . \frac{d v}{d x} = \frac{1 - v^{2}}{2 v} \Rightarrow \frac{2 v}{1 - v^{2}} d v = \frac{d x}{x} \\ I n t e g r a t i n g b o t h s i d e s, w e h a v e \\ \int \frac{2 v}{1 - v^{2}} d v = \int \frac{d x}{x} \\ \Rightarrow - l o g | 1 - v^{2} | = l o g x + l o g c \\ \Rightarrow - l o g | 1 - \frac{y^{2}}{x^{2}} | = l o g x + l o g c \Rightarrow - l o g | \frac{x^{2} - y^{2}}{x^{2}} | = l o g x + l o g c \\ \Rightarrow l o g | \frac{x^{2}}{x^{2} - y^{2}} | = l o g | x c | \Rightarrow \frac{x^{2}}{x^{2} - y^{2}} = x c \\ Since, t h e c u r v e i s passing t h r o u g h t h e point (2, 1) \\ ∴ \frac{{(2)}^{2}}{{(2)}^{2} - {(1)}^{2}} = 2 c \Rightarrow \frac{4}{3} = 2 c \Rightarrow c = \frac{2}{3} \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s \frac{x^{2}}{x^{2} - y^{2}} = \frac{2}{3} x \Rightarrow 2 (x^{2} - y^{2}) = 3 x \end{array}$

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New answer posted

9 months ago

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

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

9 months 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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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

9 months ago

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

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

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

9 months ago

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

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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

10 months ago

113. The general solution of the differential equation $e^{x} d y + (y e^{x} + 2 x) d x = 0$ is:

$\begin{array}{l} (A) x e^{y} + x^{2} = C \\ (B) x e^{y} + y^{2} = C \\ (C) y e^{x} + x^{2} = C \\ (D) y e^{y} + x^{2} = C \end{array}$

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Contributor-Level 10

The given differential equation is:

$\begin{array}{l} e^{x} d y + (y e^{x} + 2 x) d x = 0 \\ \Rightarrow e^{x} \frac{d y}{d x} + y e^{x} + 2 x = 0 \\ \Rightarrow \frac{d y}{d x} + y = - 2 x e^{- x} \end{array}$

This is a linear differential equation of the form

$\begin{array}{l} \frac{d y}{d x} + P y = Q, w h e r e P = 1 & Q = - 2 x e^{- x} \\ N o w, I . F . = e^{\int P d x} = e^{\int d x} = e^{x} \end{array}$

The general solution of the given differential equation is given by,

$\begin{array}{l} y (I . F .) = \int (Q * I . F .) d x + C \\ \Rightarrow y e^{x} = \int (- 2 x e^{- x} . e^{x}) d x + C \\ \Rightarrow y e^{x} = - \int 2 x d x + C \\ \Rightarrow y e^{x} = - x^{2} + C \\ \Rightarrow y e^{x} + x^{2} = C \end{array}$

Therefore, option (c) is correct.

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New answer posted

10 months ago

112. The general equation of a differential equation of the type $\frac{d x}{d y} + P_{1} x = Q, i s :$

$\begin{array}{l} (A) y e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \\ (B) y . e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C \\ (C) x e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \\ (D) x . e^{\int P_{1} d x} = \int (Q_{1} e^{\int P_{1} d x}) d x + C \end{array}$

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Contributor-Level 10

The integrating factor of the given differential equation $\frac{d x}{d y} + P_{1} x = Q_{1}$

The general solution of the differential equation is given by,

$\begin{array}{l} x (I . F .) = \int (Q * I . F .) d y + C \\ \Rightarrow x . e^{\int P_{1} d y} = \int (Q_{1} e^{\int P_{1} d y}) d y + C \end{array}$

Hence, the correct answer is C.

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New answer posted

10 months ago

Choose the correct answer:

111. The general solution of the differential equation $\frac{y d x - x d y}{y} = 0$ is:

$\begin{array}{l} (A) x y = c \\ (B) x = c y^{2} \\ (C) y = c x \\ (D) y = c x^{2} \end{array}$

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Contributor-Level 10

The given differential equation is:

$\begin{array}{l} \frac{y d x - x d y}{y} = 0 \\ \Rightarrow \frac{y d x - x d y}{x y} = 0 \\ \Rightarrow \frac{1}{x} d x - \frac{1}{y} d y = 0 \end{array}$

Integration both sides, we get:

$\begin{array}{l} l o g | x | - l o g | y | = l o g k \\ \Rightarrow l o g | \frac{x}{y} | = l o g k \\ \Rightarrow \frac{x}{y} = k \\ \Rightarrow y = \frac{1}{k} x \\ \Rightarrow y = C x, w h e r e, C = \frac{1}{k} \end{array}$

Therefore, option (C) is correct.

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New answer posted

10 months ago

110. The population of a village increases continuously at the rate proportional to the number of its inhabitants present at any time. If the population of the village was 20,000 in 1999 and 25,000 in the year 2004, what will be the population of the village in 2009?

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