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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solve: $x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2} .$

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Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Find the equation of a curve passing through the origin and satisfying the differential equation $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2} .$

8 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s (1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2} \\ \Rightarrow \frac{d y}{d x} + \frac{2 x}{1 + x^{2}} . y = \frac{4 x^{2}}{1 + x^{2}} \\ ∴ P = \frac{2 x}{1 + x^{2}} a n d Q = \frac{4 x^{2}}{1 + x^{2}} \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d x} = e^{\int \frac{2 x}{1 + x^{2}} . d x} = e^{l o g (1 + x^{2})} = 1 + x^{2} \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ y \times I . F . = \int Q \times I . F . d x + c \\ \Rightarrow y . (1 + x^{2}) = \int \frac{4 x^{2}}{1 + x^{2}} \times (1 + x^{2}) d x + c \\ \Rightarrow y . (1 + x^{2}) = \int 4 x^{2} d x + c \\ \Rightarrow y . (1 + x^{2}) = \frac{4}{3} x^{3} + c \dots (i) \\ Since t h e c u r v e i s passing t h r o u g h i . e ., (0, 0) \\ P u t y = 0 a n d x = 0 i n e q n . (i) \\ \Rightarrow 0 (1 + 0) = \frac{4}{3} {(0)}^{3} + c \Rightarrow c = 0 \\ S o t h e e q u a t i o n b e c o m e s \\ ∴ y (1 + x^{2}) = \frac{4}{3} x^{3} \Rightarrow y = \frac{4 x^{3}}{3 (1 + x^{2})} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s y = \frac{4 x^{3}}{3 (1 + x^{2})} . \end{array}$

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Form the differential equation of all circles which pass through the origin and whose centers lie on the y-axis.

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Form the differential equation having $y = {(s i n^{- 1} x)}^{2} + A c o s^{- 1} x + B$ , where $A$ and $B$ are arbitrary constants, as its general solution.

12 Views|Posted a year ago

Asked by Alok Kumar Singh

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Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

If $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then show that $y (1) = - \frac{1}{2}$ .

11 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s (1 + t) \frac{d y}{d t} - t y = 1 \\ \Rightarrow \frac{d y}{d t} - (\frac{t}{1 + t}) y = \frac{1}{1 + t} \\ ∴ P = - \frac{t}{1 + t} a n d Q = \frac{1}{1 + t} \\ I n t e g r a t i n g f a c t o r I . F = e^{\int P d t} = e^{\int - \frac{t}{1 + t} . d t} = e^{- \int \frac{1 + t - 1}{1 + t} . d t} \\ = e^{- \int (1 - \frac{1}{1 + t}) . d t} = e^{- [t - l o g (1 + t)]} \\ = e^{- t + l o g (1 + t)} = e^{- t} . e^{l o g (1 + t)} = e^{- t} (1 + t) \\ ∴ S o l u t i o n o f t h e e q u a t i o n i s \\ y \times I . F . = \int Q . I . F . d t + c \\ \Rightarrow y . e^{- t} (1 + t) = \int \frac{1}{(1 + t)} . e^{- t} (1 + t) d t + c \\ \Rightarrow y . e^{- t} (1 + t) = \int e^{- t} d t + c \\ \Rightarrow y . e^{- t} (1 + t) = - e^{- t} + c \\ P u t t = 0 a n d y = - 1 [? y (0) = - 1] \\ \Rightarrow - 1 . e^{0} . 1 = - e^{0} + c \\ \Rightarrow - 1 = - 1 + c \Rightarrow c = 0 \\ S o t h e e q u a t i o n b e c o m e s \\ ∴ y . e^{- t} (1 + 1) = - e^{- t} \\ \Rightarrow 2 y = - 1 \Rightarrow y = - \frac{1}{2} \\ H e n c e, y (1) = - \frac{1}{2} i s v e r i f i e d . \end{array}$

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

If $y (x)$ is a solution of $(\frac{2 + s i n x}{1 + y}) \frac{d y}{d x} + y = - c o s x$ and $y (0) = 1$ , then find the value of $y (\frac{π}{2})$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

$\begin{array}{l} G i v e n e q u a t i o n i s (\frac{2 + s i n x}{1 + y}) \frac{d y}{d x} = - c o s x \\ \Rightarrow (\frac{2 + s i n x}{c o s x}) \frac{d y}{d x} = - (1 + y) \Rightarrow \frac{d y}{(1 + y)} = - (\frac{c o s x}{2 + s i n x}) d x \\ I n t e g r a t i n g b o t h s i d e s, w e g e t \\ \int \frac{d y}{(1 + y)} = - \int (\frac{c o s x}{2 + s i n x}) d x \\ \Rightarrow l o g | 1 + y | = - l o g | 2 + s i n x | + l o g c \\ \Rightarrow l o g | 1 + y | + l o g | 2 + s i n x | = l o g c \\ \Rightarrow l o g (1 + y) (2 + s i n x) = l o g c \\ \Rightarrow (1 + y) (2 + s i n x) = c \\ P u t x = 0 a n d y = 1, w e g e t \\ \Rightarrow (1 + 1) (2 + s i n 0) = c \Rightarrow c = 4 \\ ∴ e q u a t i o n i s (1 + y) (2 + s i n x) = 4 \\ N o w p u t x = \frac{π}{2} \\ ∴ (1 + y) (2 + s i n \frac{π}{2}) = 4 \Rightarrow (1 + y) (2 + 1) = 4 \\ \Rightarrow 1 + y = \frac{4}{3} \Rightarrow y = \frac{4}{3} - 1 = \frac{1}{3} \\ S o, y (\frac{π}{2}) = \frac{1}{3} \\ H e n c e, t h e r e q u i r e d s o l u t i o n i s y (\frac{π}{2}) = \frac{1}{3} . \end{array}$

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Find the general solution of $(x + 2 y^{3}) \frac{d y}{d x} = y$ .

5 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solve the differential equation $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}$ , when $y = 0$ , $x = 0$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solve: $y d x - x d y = x^{2} y d x$ .

6 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Solve the differential equation $\frac{d y}{d x} + 1 + y = e^{x + y}$ .

4 Views|Posted a year ago

Asked by Alok Kumar Singh

S

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

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

Sol:

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

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Solve: $x^{2} \frac{d y}{d x} = x^{2} + x y + y^{2} .$

Find the equation of a curve passing through the origin and satisfying the differential equation $(1 + x^{2}) \frac{d y}{d x} + 2 x y = 4 x^{2} .$

Form the differential equation of all circles which pass through the origin and whose centers lie on the y-axis.

Form the differential equation having $y = {(s i n^{- 1} x)}^{2} + A c o s^{- 1} x + B$ , where $A$ and $B$ are arbitrary constants, as its general solution.

If $y (t)$ is a solution of $(1 + t) \frac{d y}{d t} - t y = 1$ and $y (0) = - 1$ , then show that $y (1) = - \frac{1}{2}$ .

If $y (x)$ is a solution of $(\frac{2 + s i n x}{1 + y}) \frac{d y}{d x} + y = - c o s x$ and $y (0) = 1$ , then find the value of $y (\frac{π}{2})$ .

Find the general solution of $(x + 2 y^{3}) \frac{d y}{d x} = y$ .

Solve the differential equation $\frac{d y}{d x} = 1 + x + y^{2} + x y^{2}$ , when $y = 0$ , $x = 0$ .

Solve: $y d x - x d y = x^{2} y d x$ .

Solve the differential equation $\frac{d y}{d x} + 1 + y = e^{x + y}$ .