Differential Equations

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New answer posted•a year ago

Differential Equations Class 12th

Let y = y(x) be a solution curve of the differential equation $(y + 1) t a n^{2} x d x + t a n x d y + y d x = 0,$ $x \in (0, \frac{π}{2}) .$   If $\underset{x \to 0^{+}}{l i m} x y (x) = 1,$  then the value of y  $(\frac{π}{4}) i s :$

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

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

IF = tan x

Solution : y tan x = x – tan x + C

$\underset{x \to 0^{+}}{l i m} x y = 1 \Rightarrow C = 1$

For $x = \frac{π}{4}, y = \frac{π}{4}$

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New answer posted•a year ago

Differential Equations Class 12th

If y = y(x) is an implicit function of x such that log_e (x + y) = 4xy, then $\frac{d^{2} y}{d x^{2}} a t x = 0$  is equal to

7 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

(x + y) = 4xy

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

$\frac{d^{2} y}{d x^{2}} = \frac{(1 - 4 x^{2} - 4 x y) (4 y + 4 x y' + 8 y y') - (4 x y + 4 y^{2} - 1) (- 8 x - 4 y - 4 x y')}{{(1 - 4 x^{2} - 4 x y)}^{2}}$

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New answer posted•a year ago

Differential Equations Class 12th

If the solution curve y = y(x) of the differential equation y²dx + (x² – xy + y²)dy = 0, which passes through the point (1, 1) and intersects the line y = $\sqrt[]{3} x$  at the point $(α, \sqrt[]{3} α),$  then value of $l o g_{e} (\sqrt[]{3} α)$  is equal to:

2 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$y^{2} d x + (x^{2} - x y + y^{2}) d y = 0$

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

Put x = vy $\Rightarrow v + y \frac{d v}{d y} + v^{2} - v + 1 = 0$

$\Rightarrow \frac{y d v}{d y} + v^{2} + 1 = 0$

$\Rightarrow \frac{π}{6} + l n | y | = \frac{π}{4}$

$l n | y | = \frac{π}{12}$

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New answer posted•a year ago

Differential Equations Class 12th

Let y = y(x) be the solution curve of the differential equation $\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{(x + 3)}{x + 1}, x > - 1,$ which passes through the point (0, 1). Then y(1) is equal to

3 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$\frac{d y}{d x} + (\frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6}) y = \frac{x + 3}{x + 1}, x > - 1$

IF = $e^{\int p d x} = \frac{{(x + 1)}^{2} (x + 2)}{x + 3}$

$\int P d x = \int \frac{2 x^{2} + 11 x + 13}{x^{3} + 6 x^{2} + 11 x + 6} d n = \int (\frac{2}{x + 1} + \frac{1}{x + 2} - \frac{1}{x + 3}) d x$

$= l n ({(x + 1)}^{2} \cdot (x + 2) / (x + 3))$

$\frac{2 x^{2} + 11 x + 13}{(x + 1) (x + 2) (x + 3)} = \frac{A}{x + 1} + \frac{B}{x + 2} + \frac{C}{x + 3}$

$2 x^{2} + 11 x + 13 = A (x + 2) (x + 3) + B (x + 1) (x + 3) + C (x + 1) (x + 2)$

x = -1

⇒ 4 = 2A ⇒ A = 2

x = -2

⇒ -1 = -B Þ B = 1

x₂ – 3 ⇒ -2 = 2c

c = -1

$y \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} = \int \frac{x + 3}{x + 1} \cdot \frac{{(x + 1)}^{2} (x + 2)}{x + 3} d x$

= $\int (x + 1) (x + 2) d x$

= $\frac{x^{3}}{3} + \frac{3 x^{2}}{2} + 2 x + c$

$(0, 1) \Rightarrow 1 \cdot \frac{2}{3} = c$

x = 1 $y \cdot (3) = \frac{1}{3} + \frac{3}{2} + 2 + \frac{2}{3} = \frac{3}{2} + 3 = \frac{9}{2}$

y = $\frac{3}{2}$

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New answer posted•a year ago

Differential Equations Class 12th

If the solution curve of the differential equation $\frac{d y}{d x} = \frac{x + y - 2}{x - y}$ passes through the points (2, 1) and (k + 1, 2), k > 0, then

2 Views|Posted a year ago

Asked by Payal Gupta

Answered by

Payal Gupta | Contributor-Level 10

a year ago

$\frac{d y}{d x} = \frac{x + y - 2}{x - y}$

Let x – 1 = X, y – 1 = Y

then DE: $\frac{d Y}{d X} = \frac{X + Y}{X - Y} = \frac{1 + \frac{Y}{X}}{1 - \frac{Y}{X}}$

Put y = vx

then $\frac{d Y}{d X} = V + X \frac{d V}{d X}$

V + $X \frac{d V}{d X} = \frac{1 + V}{1 - V}$

$X \frac{d V}{d X} = \frac{1 + V}{1 - V} - V$

$= \frac{1 + V^{2}}{1 - V}$

$\frac{1 - V}{1 + V^{2}} d V = \frac{d X}{X}$

$\frac{V - 1}{V^{2} + 1} d V + \frac{d X}{X} = 0$

$\frac{1}{2} l n | V^{2} + 1 | - t a n^{- 1} V + l n | X | = c$

$l n \sqrt[]{V^{2} + 1} \cdot X - t a n^{- 1} V = c$

$l n (\sqrt[]{1 + {(\frac{Y - 1}{X -})}^{2}} \cdot | X - 1 |) t a n^{- 1} \frac{Y - 1}{X - 1} = c$

$(2, 1) \Rightarrow l n (\sqrt[]{1 + 0} \cdot 1) - 0 = c$

c = 0

$l n \sqrt[]{{(X - 1)}^{2} + {(Y - 1)}^{2}} = t a n^{- 1} \frac{Y - 1}{X - 1}$

point (k + 1, 2) $l n \sqrt[]{k^{2} + 1} = t a n^{- 1} \frac{1}{k}$

$\Rightarrow \frac{1}{2} l n (k^{2} + 1) = t a n^{- 1} \frac{1}{k}$

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New answer posted•a year ago

Differential Equations Class 12th

If y = y(x) is the solution of the differential $2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$ such that y(e) = $\frac{e}{3},$ then y(1) equal to

7 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$2 x^{2} \frac{d y}{d x} - 2 x y + 3 y^{2} = 0$

$\frac{d y}{d x} = \frac{y}{x} - \frac{3}{2} {(\frac{y}{x})}^{2}$

Put y = vx

$\frac{d y}{d x} = v + x \frac{d v}{d x}$

$P o i n t (e, \frac{e}{3})$

$\Rightarrow \frac{- e}{(\frac{e}{3})} + \frac{3}{2} l n e = C$

$C = \frac{3}{2} - 3 = \frac{- 3}{2}$

$\frac{3}{2} l n | x | \frac{- x}{y} + \frac{3}{2} = 0$

$x = 1 \Rightarrow y = \frac{2}{3}$

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New answer posted•a year ago

Differential Equations Class 12th

If $b_{n} = \int_{0}^{\frac{π}{2}} \frac{c o s^{2} n x}{s i n x}$ dx, $n \in N,$  then

2 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$b_{n} = \int_{0}^{π / 2} \frac{c o s^{2} n x}{s i n x} d x$

$= \int_{0}^{π / 2} \frac{1 + c o s 2 n x}{2 s i n x} d x$

$b_{n} - b_{n - 1} = \int_{0}^{π / 2} \frac{c o s (2 n x) - c o s (2 (n - 1) x)}{2 s i n x} d x$

$= {\frac{c o s (2 n - 1) x}{(2 n - 1)}]}_{0}^{π / 2} = \frac{1}{2 n - 1} [0 - 1] = \frac{- 1}{2 n - 1}$

$b_{n - 1} - b_{n} = \frac{1}{2 n - 1}$

$b_{1} - b_{2} = \frac{1}{3} (A) - \frac{1}{5}, . . . . . .$

$b_{2} - b_{3} = \frac{1}{3} (B) - 5, - 7, - 9, . . . . .$

$b_{3} - b_{4} = \frac{1}{7} (C)$

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New answer posted•a year ago

Differential Equations Class 12th

The general solution of the differential equation (x – y²)dx + y(5x + y²)dy = 0 is:

4 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$(x - y^{2}) d x + y (5 x + y^{2}) d y = 0$

$y \frac{d y}{d x} = \frac{y^{2} - x}{5 x + y^{2}}$

Let y² = t

$\frac{1}{2} . \frac{d t}{d x} = \frac{t - x}{5 x = t}$

Now substitute, t = yx

$\frac{d t}{d x} = v + x \frac{d v}{d x}$

$| \frac{{(v + 1)}^{4}}{{(v + 2)}^{3}} | = \frac{C}{x}$

$| {(y^{2} + x)}^{4} | = C | {(y^{2} + 2 x)}^{3} |$

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New answer posted•a year ago

Differential Equations Class 12th

The slope of the tangent to a curve C : y = y(x) at any point (x, y) on it is $\frac{2 e^{2 x} - 6 e^{- x} + 9}{2 + 9 e^{- 2 x}} .$  If C passes through the points $(0, \frac{1}{2} + \frac{π}{2 \sqrt[]{2}}) a n d (α, \frac{1}{2} e^{2 a}),$  then e^a is equal to:

6 Views|Posted a year ago

Asked by Vishal Baghel

Answered by

Vishal Baghel | Contributor-Level 10

a year ago

$\frac{d y}{d x} = \frac{2 e^{2 x} - 6 e^{- x} + 9}{2 + 9 e^{- 2 x}} = e^{2 x} - \frac{6 e^{- x}}{2 + 9 e^{- 2 x}}$

$\int d y = \int e^{2 x} d x - 3 \int \frac{e^{- z}}{\underset{p u t e^{- x} = t}{\underset{?}{1 + (\frac{3 e^{- x}}{\sqrt[]{2}})}}} d x$

= $y = \frac{e^{2 x}}{2} = \sqrt[]{2} t a n^{- 1} (\frac{3 e^{- x}}{\sqrt[]{2}}) + C$

It is given that the curve passes through

$(0, \frac{1}{2} + \frac{π}{2 \sqrt[]{2}})$

$\frac{3}{\sqrt[]{2}} e^{α} - \frac{3}{\sqrt[]{2}} = e^{α} + \frac{9}{2}$

$e^{α} = \frac{\frac{9}{2} + \frac{3}{\sqrt[]{2}}}{\frac{3}{\sqrt[]{2}} - 1} = \frac{3}{\sqrt[]{2}} (\frac{3 + \sqrt[]{2}}{3 - \sqrt[]{2}})$

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New answer posted•a year ago

Differential Equations Class 12th

Let y = y(x) be the solution curve of the differential equation $s i n (2 x^{2}) l o g_{e} (t a n x^{2}) d y + (4 x y - 4 \sqrt[]{2} x s i n (x^{2} - \frac{π}{4})) d x = 0,$ $0 < x < \sqrt[]{\frac{π}{2},}$ which passes through the point $(\sqrt[]{\frac{π}{6}, 1})$ . Then $| y (\sqrt[]{\frac{π}{3}}) |$  is equal to………….

3 Views|Posted a year ago

Asked by Alok Kumar Singh

Answered by

Alok Kumar Singh | Contributor-Level 10

a year ago

$s i n (2 x^{2}) . 10 9_{e} (t a n x^{2}) d y + 4 x y d x = 4 \sqrt[]{2} . x . (s i n x^{2} c o s \frac{π}{4} - c o s x^{2} s i n \frac{π}{4}) d x$

$\Rightarrow l n (t a n x^{2}) d y + \frac{4 x}{s i n (2 x^{2})} y d x = \frac{4 x (s i n x^{2} - c o s x^{2})}{s i n (2 x^{2})} d x$

Integrate

$\Rightarrow y . l n (t a n x^{2}) = 2 . l n (\frac{s i n x^{2} + c o s x^{2} - 1}{s i n x^{2} + c o s x^{2} + 1}) + C$

$x = \sqrt[]{\frac{π}{6}}, y = 1$ calculate C.

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