Differential Equations

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

a year ago

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

Sol:

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

Upvote

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 $\frac{x^{2} + y^{2}}{2 x y} .$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

a year ago

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

Sol:

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

Upvote

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

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

a year ago

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

Sol:

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

Upvote

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

2 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

a year ago

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

Sol:

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

Upvote

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

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

a year ago

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

Sol:

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

Upvote

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

3 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

a year ago

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

Sol:

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

Upvote

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

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

5 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

a year ago

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,

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

Therefore, option (c) is correct.

Upvote

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

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

7 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

a year ago

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,

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

Hence, the correct answer is C.

Upvote

Choose the correct answer:

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

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

2 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by

a year ago

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.

Upvote

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?

13 Views|Posted a year ago

Asked by Vishal Baghel

S

Answered by