Let y - y x be a curve satisfying the differential equation y d 2 y - dx 2-2 dy - dx 2- If the curve passes through 2-2 and 8- 1-2- then the value of y 1-9 is

Given, nbsp;y(d^2ydx^2) = 2(dydx)^2 ⇒y”y'=2y'y Integrating both sides, ln y'= 2ln y + ln a ⇒y'y^2=a nbsp; ⇒∫1y^2 dy = nbsp;∫ a dx nbsp; ⇒ 1y = ax+b Sinc ...

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Byju's Answer

Standard XII

Mathematics

Variable Separable Method

Let y = y x b...

Question

Let $y = y (x)$ be a curve satisfying the differential equation $y (\frac{d^{2} y}{d x^{2}}) = 2 {(\frac{d y}{d x})}^{2} .$ If the curve passes through $(2, 2)$ and $(8, \frac{1}{2}),$ then the value of $y (\frac{1}{9})$ is

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Solution

Given, $y (\frac{d^{2} y}{d x^{2}}) = 2 {(\frac{d y}{d x})}^{2}$
$\Rightarrow \frac{y^{''}}{y^{'}} = \frac{2 y^{'}}{y}$
Integrating both sides,
$ln y^{'} = 2 ln y + ln a$
$\Rightarrow \frac{y^{'}}{y^{2}} = a$
$\Rightarrow \int \frac{1}{y^{2}} d y = \int a d x$
$\Rightarrow \frac{- 1}{y} = a x + b$
Since the curve passes through $(2, 2)$ and $(8, \frac{1}{2})$ , so
$2 a + b = \frac{- 1}{2}$
$8 a + b = - 2$
Solving the above equations, we get $a = \frac{- 1}{4}, b = 0$
Hence, $y = \frac{4}{x}$
$y (\frac{1}{9}) = 36$

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Let y=y(x) be a curve satisfying the differential equation y(d2ydx2)=2(dydx)2. If the curve passes through (2,2) and (8,12), then the value of y(19) is

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