Differential Equations

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Find the general solution of $\frac{d y}{d x} + a y = e^{m x}$ .

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solve the differential equation $\frac{d y}{d x} + 2 x y = y$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solve the differential equation $(x^{2} - 1) \frac{d y}{d x} + 2 x y = \frac{1}{x^{2} - 1}$ .

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Given that $\frac{d y}{d x} = e^{- 2 y}$ and $y = 0$ when $x = 5$ . Find the value of $x$ when $y = 3$ .

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s \frac{d y}{d x} = e^{- 2 y} \\ \Rightarrow \frac{d y}{e^{- 2 y}} = d x \Rightarrow e^{2 y} . d y = d x \\ I n t e g r a t i n g b o t h s i d e s, w e g e t \\ \int e^{2 y} . d y = \int d x \Rightarrow \frac{1}{2} e^{2 y} = x + c \\ P u t y = 0 a n d x = 5 \\ \Rightarrow \frac{1}{2} e^{2 y} = 5 + c \Rightarrow c = \frac{1}{2} - 5 = - \frac{9}{2} \\ ∴ T h e e q u a t i o n b e c o m e s \frac{1}{2} e^{2 y} = x - \frac{9}{2} \\ N o w p u t t i n g y = 3, w e g e t \\ \frac{1}{2} e^{6} = x - \frac{9}{2} \Rightarrow x = \frac{1}{2} e^{6} + \frac{9}{2} \\ H e n c e, t h e r e q u i r e d v a l u e o f x = \frac{1}{2} (e^{6} + 9) . \end{array}$

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New question posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

<p><strong>Given that <math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><msup><mrow><mi>e</mi></mrow><mrow><mo>−</mo><mn>2</mn><mi>y</mi></mrow></msup></mrow></math> and <math><mrow><mi>y</mi><mo>=</mo><mn>0</mn></mrow></math> when <math><mrow><mi>x</mi><mo>=</mo><mn>5</mn></mrow></math> . Find the value of <math><mrow><mi>x</mi></mrow></math> when <math><mrow><mi>y</mi><mo>=</mo><mn>3</mn></mrow></math> </strong></p>

5 Views|Posted a year ago

Asked by Alok Kumar Singh

New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Find the differential equation of all non-vertical lines in a plane.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Find the solution of $\frac{d y}{d x} = 2^{y - x}$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solve: $x \frac{d y}{d x} = y (l o g y - l o g x + 1) .$

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Find the equation of a curve passing through the point $(1, 1)$ . If the tangent drawn at any point $P (x, y)$ on the curve meets the coordinate axes at $A$ and $B$ such that $P$ is the midpoint of $A B$ .

5 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Find the equation of a curve passing through the origin if the slope of the tangent to the curve at any point $(x, y)$ is equal to the square of the difference of the abscissa and ordinate of the point.

6 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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