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Ncert Solutions Maths class 12th Differential Equations

State True or False for the following: (i) The integrating factor of a differential equation of the form <math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>+</mo><msub><mrow><mi>P</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>x</mi><mo>=</mo><msub><mrow><mi>Q</mi></mrow><mrow><mn>1</mn></mrow></msub><mo> </mo></mrow></math> is given by <math><mrow><msup><mrow><mi>e</mi></mrow><mrow><mo>∫</mo><msub><mrow><mi>P</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>d</mi><mi>y</mi></mrow></msup><mo>.</mo></mrow></math> (ii) solution of a differential equation of the type <math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>+</mo><msub><mrow><mi>P</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>x</mi><mo>=</mo><msub><mrow><mi>Q</mi></mrow><mrow><mn>1</mn></mrow></msub></mrow></math> is given by <math><mrow><mi>y</mi><mo>⋅</mo><mi>I</mi><mo>.</mo><mi>F</mi><mo>.</mo><mo>=</mo><mrow><mo>(</mo><mrow><mi>I</mi><mo>.</mo><mi>F</mi><mo>.</mo></mrow><mo>)</mo></mrow><msub><mrow><mi>Q</mi></mrow><mrow><mn>1</mn></mrow></msub><mi>d</mi><mi>y</mi><mo>.</mo></mrow></math> (iii) Correct substitution for solving the differential equation of the type <math><mrow><mfrac><mrow><mi>d</mi><mi>y</mi></mrow><mrow><mi>d</mi><mi>x</mi></mrow></mfrac><mo>=</mo><mi>f</mi><mrow><mo>(</mo><mrow><mi>x</mi><mo>,</mo><mi>y</mi></mrow><mo>)</mo></mrow><mo>,</mo></mrow></math> where <math><mrow><mi>f</mi><mrow><mo>(</mo><mrow><mi>x</mi><mo>,</mo><mi>y</mi></mrow><mo>)</mo></mrow></mrow></math> is a homogeneous function of zero degree, is <math><mrow><mi>y</mi><mo>=</mo><mi>v</mi><mi>x</mi></mrow></math> . (iv) Correct substitution for the solution of the differential equation of the type dx/dy = g(x, y) where g (x, y) is a homogeneous function of the degree zero is x = vy. (v) Number of arbitrary constants in the particular solution of a differential equation of order two is two. (vi) The differential equation representing the family of circles x2 + (y – a)2 = a2 will be of order two. (vii) The solution of dy/dx = (y/x)1/3 is y 2/3-x2/3 = c. (viii) Differential equation representing the family of curves  y ex (A cos x + B sin x is d2 y/dx2 – 2dy/dx + 2y = 0 (ix) The solution of the differential equation dy/dx = x + 2y/x is x + y = kx2.  (x) Solution of xdy/dx = y +x tan y/x is sin (y/x) =cx         (xi) The differential equation of all non- horizontal lines in a plane is d2 y/dx2=0

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Ncert Solutions Maths class 12th Differential Equations

Fill in the blanks for the following (I to xi)

(i) The degree of the differential equation $\frac{d^{2} y}{d x^{2}} + e^{\frac{d y}{d x^{}}} + 0$ is _____.

(ii) The degree of the differential equation is _.

(iii) The number of arbitrary constants in the general solution of a differential equation of order three is _.

(iv) $\frac{d y}{d x} + \frac{y}{x l o g x} = \frac{1}{x}$ is an equation is of the type _____.

(v) The general solution of a differential equation of the type $\frac{d y}{d x} + P_{1} x = Q_{1}$ is given by _____.

(vi) The solution of $\frac{x d y}{d x} + 2 y = x^{2}$ is _.

(vii) The solution of $(1 + x^{2}) \frac{d y}{d x} + 2 x y - 4 x^{2} = 0$ is _.

(viii) The solution of $y d x + (x + x y) d y = 0$ is _.

(ix) The general solution of $\frac{d y}{d x} + y = s i n x$ is _.

(x) The solution of $c o t y d x = x d y$ is _.

(xi) The integrating factor of $\frac{d y}{d x} + y = \frac{1 + y}{x} = 0$ is _____.

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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 Fill in the Blanks Type Questions as classified in NCERT Exemplar

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

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Ncert Solutions Maths class 12th Differential Equations

Find the differential equation of the system of concentric circles with center $(1, 2)$ .

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:

$\begin{array}{l} F a m i l y o f c o n c e n t r i c c i r c l e s w i t h c e n t r e (1, 2) a n d r a d i u s' r' i s \\ {(x - 1)}^{2} + {(y - 2)}^{2} = r^{2} \\ D i f f e r e n t i a t i n g b o t h s i d e s w . r . t . x, w e g e t \\ 2 (x - 1) + 2 (y - 2) \frac{d y}{d x} = 0 \\ \Rightarrow (x - 1) + (y - 2) \frac{d y}{d x} = 0 \\ W h i c h i s t h e r e q u i r e d e q u a t i o n . \end{array}$

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Ncert Solutions Maths class 12th Differential Equations

Solve the differential equation $(1 + y^{2}) t a n^{- 1} x d x + 2 y (1 + x^{2}) d y = 0 .$

13 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:

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

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Ncert Solutions Maths class 12th Differential Equations

Form the differential equation by eliminating $A$ and $B$ in $A x^{2} + B y^{2} = 1$ .

8 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:

$\begin{array}{l} G i v e n e q u a t i o n i s A x^{2} + B y^{2} = 1 \\ D i f f e r e n t i a t i n g w . r . t . x, w e g e t \\ \Rightarrow 2 A . x + 2 B y \frac{d y}{d x} = 0 \\ \Rightarrow A . x + B y \frac{d y}{d x} = 0 \\ \Rightarrow B y . \frac{d y}{d x} = - A x \Rightarrow \frac{y}{x} . \frac{d y}{d x} = - \frac{A}{B} \\ D i f f e r e n t i a t i n g b o t h s i d e s a g a i n w . r . t . x, w e h a v e \\ \frac{y}{x} . \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} (\frac{x . \frac{d y}{d x} - y . 1}{x^{2}}) = 0 \\ \Rightarrow \frac{y x^{2}}{x} . \frac{d^{2} y}{d x^{2}} + x . {(\frac{d y}{d x})}^{2} - y . \frac{d y}{d x} = 0 \\ \Rightarrow x y . \frac{d^{2} y}{d x^{2}} + x . {(\frac{d y}{d x})}^{2} - y . \frac{d y}{d x} = 0 \\ \Rightarrow x y . y^{''} + x . {(y^{'})}^{2} - y . y^{'} = 0 \\ H e n c e, t h e r e q u i r e d e q u a t i o n i s x y . y^{''} + x . {(y^{'})}^{2} - y . y^{'} = 0 \end{array}$

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Ncert Solutions Maths class 12th Differential Equations

Solve the differential equation $d y = c o s x (2 - y c o s e c x),$ given that $y = 2$ when $x = \frac{π}{2}$ .

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:

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

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Ncert Solutions Maths class 12th Differential Equations

Solve: $2 (y + 3) - x y \frac{d y}{d x} = 0,$ given that $y (1) = - 2$ .

9 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:

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

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Ncert Solutions Maths class 12th Differential Equations

Solve: $(x + y) (d x - d y) = d x + d y$ .Hint: Substitute $x + y = z$ after separating $d x$ and $d y$ .

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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:

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

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Ncert Solutions Maths class 12th Differential Equations

Find the general solution of $y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$ .

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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:

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

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Ncert Solutions Maths class 12th Differential Equations

Find the general solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0 .$

6 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:

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

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Ncert Solutions Maths class 12th

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Find the differential equation of the system of concentric circles with center $(1, 2)$ .

Solve the differential equation $(1 + y^{2}) t a n^{- 1} x d x + 2 y (1 + x^{2}) d y = 0 .$

Form the differential equation by eliminating $A$ and $B$ in $A x^{2} + B y^{2} = 1$ .

Solve the differential equation $d y = c o s x (2 - y c o s e c x),$ given that $y = 2$ when $x = \frac{π}{2}$ .

Solve: $2 (y + 3) - x y \frac{d y}{d x} = 0,$ given that $y (1) = - 2$ .

Solve: $(x + y) (d x - d y) = d x + d y$ .Hint: Substitute $x + y = z$ after separating $d x$ and $d y$ .

Find the general solution of $y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$ .

Find the general solution of the differential equation $(1 + y^{2}) + (x - e^{t a n^{- 1} y}) \frac{d y}{d x} = 0 .$