Let y - yx be a solution curve of the differential equation y-1tan 2x dx-tan x dy-ydx-0- x - 0-2 - If lim x- 0-xyx-1- then the value of y -4 is

(y+1)tan ^2xdx+tan xdy+ydx=0 ⇒(y+1)( ^2x 1)dx+tan xd(y+1)+y dx=0 ⇒(y+1) ^2xdx+tan xd(y+1) (y+1)dx+ydx=0 ⇒(y+1)d(tan x)+tan x d(y+1)=dx ⇒∫ d((y+1)tan x)=∫ dx ⇒(y ...

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Byju's Answer

Standard XII

Mathematics

Definite Integral as Limit of Sum

Let y = yx be...

Question

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) {tan}^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2}) .$ If $lim x \to 0^{+} x y (x) = 1,$ then the value of $y (\frac{π}{4}) is$

\frac{π}{4}

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\frac{π}{4} + 1

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\frac{π}{4} - 1

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- \frac{π}{4}

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Solution

The correct option is A $\frac{π}{4}$
$(y + 1) {tan}^{2} x d x + tan x d y + y d x = 0$
$\Rightarrow (y + 1) ({sec}^{2} x - 1) d x + tan x d (y + 1) + y d x = 0$
$\Rightarrow (y + 1) {sec}^{2} x d x + tan x d (y + 1) - (y + 1) d x + y d x = 0$
$\Rightarrow (y + 1) d (tan x) + tan x d (y + 1) = d x$
$\Rightarrow \int d ((y + 1) tan x) = \int d x$
$\Rightarrow (y + 1) tan x = x + C$
$\Rightarrow y = \frac{x + C}{tan x} - 1$
$lim x \to 0^{+} x y (x) = 1$
$\Rightarrow lim x \to 0 \frac{x}{tan x} (C + x) - x = 1$
$\Rightarrow C = 1$
$y = \frac{x + 1}{tan x} - 1$
$\Rightarrow y (\frac{π}{4}) = \frac{π}{4}$

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Let y=y(x) be a solution curve of the differential equation (y+1)tan2x dx+tanx dy+ydx=0, x ∈(0,π2). If lim x→0+xy(x)=1, then the value of y(π4) is

Let $y = y (x)$ be a solution curve of the differential equation $(y + 1) {tan}^{2} x d x + tan x d y + y d x = 0, x \in (0, \frac{π}{2}) .$ If $lim x \to 0^{+} x y (x) = 1,$ then the value of $y (\frac{π}{4}) is$