Differential Equations

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The differential equation of the family circles passing through the points (0, 2) and (0, 2) is:

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

Equation of circle passing through (0, 2) and (0, 2) is

$x^{2} + (y^{2} - 4) + λ x = 0, (λ \in R)$

Divided by x we get

$\frac{x^{2} + (y^{2} - 4)}{x} + λ = 0$

Differentiating w.r.t. x

$\frac{x [2 x + 2 y . \frac{d y}{d x}] - [x^{2} + y^{2} - 4] . 1}{x^{2}} = 0$

$\Rightarrow 2 x y . \frac{d y}{d x} + (x^{2} - y^{2} + 4) = 0$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}, x > 1$ passing through the point $(2, \sqrt[]{\frac{1}{3}}) .$ Then $\sqrt[]{7} y (8)$ is equal to:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{d y}{d x} + \frac{1}{x^{2} - 1} y = {(\frac{x - 1}{x + 1})}^{1 / 2}$

$\frac{d y}{d x} + p y = Q$

$I . F = e^{\int P d x} = {(\frac{x - 1}{x +})}^{\frac{1}{2}}$

$x - 2 l o g_{e} | x + 1 | + C$

Curves passes through $(2, \frac{1}{\sqrt[]{3}})$

$\Rightarrow C = 2 l o g_{e} 3 - \frac{5}{3}$

$a t x = 8, \sqrt[]{7} y (8) = 19 - 6 l o g_{e} 3$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

A ray of light passing through the point P(2, 3) reflects on the x-axis at point A and the reflected ray passes through the point Q(5, 4). Let R be the point that divides the line segment AQ internally into the ration 2 : 1. Let the co-ordinates of the foot of the perpendicular M from R on the bisector of the angle PAQ be (a, b). Then the value of 7a + 3b is equal to……….

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let y = y(x) be the solution of the differential equation $x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0, x > 1, w i t h y (2) = - 2 .$ Then y(3) is equal to

10 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} y - y - 4 x^{3}) = 0$

$\Rightarrow x (1 - x^{2}) \frac{d y}{d x} + (3 x^{2} - 1) y = 4 x^{3}$

$\Rightarrow y = 2 x - 1 (x^{3} - x) ∴ y (3) = - 18$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let the solution curve y = y(x) of the differential equation $[\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{\frac{y}{x}}] x \frac{d y}{d x} = x + [\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{\frac{y}{x}}] y$ pass through the points (1, 0) and (2α, α), α > 0. Then a is equal to

5 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$[\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{y / x}] x \frac{d y}{d x} = x + [\frac{x}{\sqrt[]{x^{2} - y^{2}}} + e^{y / x}] y$

$\Rightarrow e^{y / x} [x d y - y d x] = x d x + \frac{x}{\sqrt[]{x^{2} - y^{2}}} (y d x - x d y)$

$\Rightarrow e^{y / x} d (y / x) = \frac{d x}{x} - \frac{d (y / x)}{\sqrt[]{1 - {(y / x)}^{2}}}$

Integrating

$e^{y / x} = l n x - s i n^{- 1} (\frac{y}{x}) + c$

Passes (1, 0)

1 = c

$\Rightarrow α = \frac{1}{2} e x p (\sqrt[]{e} - 1 + \frac{π}{6})$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let the solution curve y = y(x) of the differential equation $(1 + e^{2 x}) (\frac{d y}{d x} + y) = 1$ pass through the point $(0, \frac{π}{2}) .$ Then, $\underset{x \to \infty}{l i m} e^{x} y (x)$ is equal to:

5 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

IF =e^x

$y e^{x} = \int \frac{e^{x}}{1 + e^{2 x}} d x$

$\int \frac{d t}{1 + t^{2}}$

= tan^-1 (t) + c

$\underset{x \to \infty}{l i m} e^{x} y = \underset{x \to \infty}{l i m} t a n^{- 1} e^{x} + \frac{π}{4} = \frac{3 π}{4}$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let y = y(x) be the solution of the differential equation $(1 - x^{2}) d y = (x y + (x^{3} + 2) \sqrt[]{1 - x^{2}}) d x, - 1 < x < 1, a n d y (0) = 0 .$ If

$\int_{\frac{- 1}{2}}^{\frac{1}{2}} \sqrt[]{1 - x^{2}} y (x) d x = k,$ then k^-1is equal to.

8 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$(1 - x^{2}) d y = (x y + (x^{3} + 2) \sqrt[]{1 - x^{2}}) d x$

$∴ \frac{d y}{d x} - \frac{x}{1 - x^{2}} y = \frac{x^{3} + 3}{\sqrt[]{1 - x^{2}}}$

$∴ l . F . = e^{\int \frac{X}{1 - x^{2}} d x} = \sqrt[]{1 - x^{2}}$

$∴ y (x) = \frac{x^{4} + 12 x}{4 \sqrt[]{1 - x^{2}}}$

$∴ \int_{\frac{- 1}{2}}^{\frac{1}{2}} \sqrt[]{1 - x^{2}} y (x) d x = \int_{\frac{- 1}{2}}^{\frac{1}{2}} (\frac{x^{4} + 12 x}{4}) d x$

$∴ k = \frac{1}{320}$

$∴ k^{- 1} = 320$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

If the solution curve of the differential equation $((t a n^{- 1} y) - x) d y = (1 + y^{2}) d x$ passes through the point (1, 0), then the abscissa of the point on the curve whose ordinate is tan(1), is

16 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$(t a n^{- 1} y) - x) d y = (1 + y^{2}) d x$

$\frac{d x}{d y} + \frac{x}{1 + y^{2}} = \frac{t a n^{- 1} y}{1 + y^{2}}$

$l . F = e^{\int \frac{1}{1 + y^{2}} d y} = e^{t a n^{- 1} y}$

$x . e^{t a n^{- 1} y} \int \frac{e^{t a n^{- 1} y} t a n^{- 1} y}{1 + y^{2}} d y$

Let

$e^{t a n^{- 1} y} = t$

$= x e^{t a n^{- 1} y} = e^{t a n^{- 1} y} y - e^{t a n^{- 1} y} + c . . . (i)$

$?$ It passes through (1, 0) = c = 2

Now put y = tan 1, then

ex = e – e + 2

$\Rightarrow x = \frac{2}{6}$

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The solution of $\frac{d y}{d x} + \frac{2 x y}{1 + x^{2}} = \frac{1}{1 + x^{2}}$ is:

(A) $y (1 + x^{2}) = c +tan^{- 1} x$

(B) $\frac{y}{1 + x^{2}} = c +tan^{- 1} x$

(C) $ylog (1 + x^{2}) = c +tan^{- 1} x$

(D) $y (1 + x^{2}) = c +sin^{- 1} x$

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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New answer posted•a year ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

The solution of the differential equation $\frac{d y}{d x} = e^{x - y} + x^{2} e^{- y}$ is:

(A) $y = e^{y - y} - x^{2} e^{- y} + c$

(B) $e^{y} - e^{x} = x^{3} / 3 + c$

(C) $e^{x} + e^{y} = x^{3} / 3 + c$

(D) $e^{x} - e^{y} = x^{3} / 3 + c$

8 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

This is an Objective Type Questions as classified in NCERT Exemplar

Sol:

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

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