Let y - yx be the solution of the differential equation x-2e y-1-x-2 -y-1 dx-x-2dy- y1-1- If the domain of y - yx is an open interval - - - then - - - is equal to

Let y + 1 = Y and x + 2 = X Such that dy = dY and dx = dX Hence, we have: ( Xe^Y/X+Y )dX=XdY ⇒XdY YdXX^2=e^Y/XXdX ⇒ e^ Y/Xd ( YX )=dXX ⇒ e^ Y/X=ln |X|+c ⇒ e^ ...

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Byju's Answer

Standard XII

Mathematics

Linear Differential Equations of First Order

Let y = yx be...

Question

Let $y = y (x)$ be the solution of the differential equation $((x + 2) e^{(\frac{y + 1}{x + 2})} + (y + 1)) d x = (x + 2) d y, y (1) = 1.$ If the domain of $y = y (x)$ is an open interval $(α, β),$ then $| α + β |$ is equal to

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Solution

Let $y + 1 = Y$ and $x + 2 = X$
Such that $d y = d Y$ and $d x = d X$
Hence, we have: $(X e^{\frac{Y}{X}} + Y) d X = X d Y$
$\Rightarrow \frac{X d Y - Y d X}{X^{2}} = \frac{e^{\frac{Y}{X}}}{X} d X$
$\Rightarrow e^{- \frac{Y}{X}} d (\frac{Y}{X}) = \frac{d X}{X}$
$\Rightarrow - e^{- \frac{Y}{X}} = ln | X | + c$
$\Rightarrow - e^{- (\frac{y + 1}{x + 2})} = ln | x + 2 | + c$
$∵ (1, 1)$ satisfy this equation
So, $c = - e^{- \frac{2}{3}} - ln 3$
Now $y = - 1 - (x + 2) ln (ln (∣ ∣ ∣ \frac{3}{x + 2} ∣ ∣ ∣) + e^{- \frac{2}{3}})$

Domain:
$ln ∣ ∣ ∣ \frac{3}{x + 2} ∣ ∣ ∣ > - e^{- \frac{2}{3}}$
$\Rightarrow \frac{3}{| x + 2 |} > e^{- e^{- \frac{2}{3}}}$
$\Rightarrow | x + 2 | < 3 e^{e^{- \frac{2}{3}}}$
$\Rightarrow - 3 e^{e^{- \frac{2}{3}}} - 2 < x < 3 e^{e^{- \frac{2}{3}}} - 2$
So, $α + β = - 4$
$\Rightarrow | α + β | = 4$

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Similar questions

Q. Let

y = y (x)

be a solution of the differential equation,

\sqrt{1 - x^{2}} \frac{d y}{d x} + \sqrt{1 - y^{2}} = 0, | x | < 1

.
If

y (\frac{1}{2}) = \frac{\sqrt{3}}{2}

, then

y (\frac{- 1}{\sqrt{2}})

is equal to:

Q. If

y = y (x)

is the solution curve of the differential equation

x^{2} d y + (y - \frac{1}{x}) d x = 0; x > 0,

and

y (1) = 1,

then

y (\frac{1}{2})

is equal to:

Q. Let the curve

y = y (x)

be the solution of the differential equation.

\frac{d y}{d x} = 2 (x + 1) .

. If the numerical value of area bounded by the curve

y = y (x)

and

x -

axis is

\frac{4 \sqrt{8}}{3}

, then the value of

y (1)

is equal to

Q. Assertion :Let a solution

y = y (x)

of the differential equation

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

satisfy

y (2) = 2 / \sqrt{3}

y (x) = sec ({sec}^{- 1} x - \frac{π}{6})

Reason:

y (x)

is given by

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

Q.
Let a solution

y = y (x)

of the differential equation

x \sqrt{x^{2} - 1} d y - y \sqrt{y^{2} - 1} d x = 0

satisfy

y (2) = \frac{2}{\sqrt{3}}

STATEMENT-1:

y (x) = sec (s e c^{- 1} x - \frac{π}{6})

STATEMENT-2 :

y (x)

is given by

\frac{1}{y} = \frac{2 \sqrt{3}}{x} - \sqrt{1 - \frac{1}{x^{2}}}

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Let y=y(x) be the solution of the differential equation ((x+2)e(y+1x+2)+(y+1))dx=(x+2)dy, y(1)=1. If the domain of y=y(x) is an open interval (α,β), then |α+β| is equal to

Let $y = y (x)$ be the solution of the differential equation $((x + 2) e^{(\frac{y + 1}{x + 2})} + (y + 1)) d x = (x + 2) d y, y (1) = 1.$ If the domain of $y = y (x)$ is an open interval $(α, β),$ then $| α + β |$ is equal to