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2 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Suppose y = y(x) be the solution curve to the differential equation $\frac{d y}{d x} - y = 2 - e^{- x}$ such that $\underset{x \to \infty}{l i m} y (x)$ is finite. If a and b are respectively the x and y intercepts of the tangent to the curve at x = 0, then the value of a 4b is equal to…………

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alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} - y = 2 - e^{- x}$

$I . F . = e^{- \int d x} = e^{- x}$

$∴$ soln

$y e^{- x} = \int (2 e^{- x} - e^{- 2 x}) d x$

$y = - 2 + \frac{e^{- x}}{2} + C e^{x}$

as for x $\infty$ y finite c = 0

$∴ y = \frac{e^{- x}}{2} - 2$

$\begin{array}{l} \Rightarrow x + 2 y = - 3 \\ \Rightarrow a = - 3 b = \frac{- 3}{2} \end{array}$

$∴ a = 4 b = - 3 + 6 = 3$

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2 months ago

Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter One

$I f \frac{d y}{d x}$ + 2y tan x = sin x, 0 < x < $\frac{π}{2}$ and y $(\frac{π}{3})$ = 0, then the maximum value of y(x) is:

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Vishal Baghel

Contributor-Level 10

$\frac{d y}{d x} + 2 y t a n x = s i n x, ∴ I . F . e^{\int 2 t a n x d x} = s e c^{2} x$

= cos x – 2 cos² x= $- 2 {(c o s x - \frac{1}{4})}^{2} + \frac{1}{8} ∴ y_{m a x} = \frac{1}{8}$

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Differential Equations Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Le the solution curve y = f(x) of the differential equation $\frac{d x}{d y} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt[]{1} - x^{2}}, x \in (- 1, 1)$ pass through the origin. Then $\int_{- \frac{\sqrt[]{3}}{2}}^{\frac{\sqrt[]{3}}{2}} f (x) d x$ is equal to

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alok kumar singh

Contributor-Level 10

$\frac{d y}{d x} + \frac{x y}{x^{2} - 1} = \frac{x^{4} + 2 x}{\sqrt[]{1 - x^{2}}},$ $I . F . e^{\int \frac{x d x}{x^{2} - 1}} = \sqrt[]{| x^{2} - 1 |} = \sqrt[]{1 - x^{2}}$ $(? x \in (- 1, 1))$

Solution of differential equation is $y \sqrt[]{1 - x^{2}} = \int (x^{4} + 2 x) d x = \frac{x^{5}}{5} + x^{2} + c$

Curve is passing through origin, c = 0 $y = \frac{x^{5} + 5 x^{2}}{5 \sqrt[]{1 - x^{2}}}$

$∴ \int_{- \frac{\sqrt[]{3}}{2}}^{\frac{\sqrt[]{3}}{2}} \frac{x^{5} + 5 x^{2}}{5 \sqrt[]{1 - x^{2}}} d x = \frac{π}{3} - \frac{\sqrt[]{3}}{4}$

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2 months 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:

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Vishal Baghel

Contributor-Level 10

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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2 months 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:

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Vishal Baghel

Contributor-Level 10

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

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

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

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

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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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……….

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Payal Gupta

Contributor-Level 10

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New answer posted

2 months 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

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Payal Gupta

Contributor-Level 10

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

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

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

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2 months 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

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Payal Gupta

Contributor-Level 10

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

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

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

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

2 months 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:

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Vishal Baghel

Contributor-Level 10

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

2 months 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.

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Payal Gupta

Contributor-Level 10

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

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

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

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

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

$∴ k = \frac{1}{320}$

$∴ k^{- 1} = 320$

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