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Variable Separable Method

Let y = f x b...

Question

Let $y = f (x)$ be a non-negative function defined on $[0, 2]$ satisfying the differential equation $y^{3} y^{''} + 1 = 0$ . If $f^{'} (1) = 0$ and $f (1) = 1$ , then

the maximum value of

f (x)

1.

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the minimum value of

f (x)

0.

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the solution

y = f (x)

of the given differential equation represents a semi-circle with centre

(1, 0) .

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the area bounded by the curve

y = f (x)

and

x - axis

\frac{π}{4} sq. units .

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Solution

The correct options are
A the maximum value of $f (x)$ is $1.$
B the minimum value of $f (x)$ is $0.$
C the solution $y = f (x)$ of the given differential equation represents a semi-circle with centre $(1, 0) .$
$y^{3} y^{''} = - 1$
$\Rightarrow y^{'} y^{''} = - \frac{y^{'}}{y^{3}}$
Integrating both the sides,
$\frac{(y^{'})^{2}}{2} = \frac{1}{2 y^{2}} + C$
$f^{'} (1) = 0, f (1) = 1 \Rightarrow C = - \frac{1}{2}$
$\Rightarrow \frac{(y^{'})^{2}}{2} = \frac{1}{2 y^{2}} - \frac{1}{2}$
$\Rightarrow y^{'} = \pm \frac{\sqrt{1 - y^{2}}}{y}$
$\Rightarrow \frac{y}{\sqrt{1 - y^{2}}} y^{'} = \pm 1$
Integrating both the sides,
$\int \frac{y}{\sqrt{1 - y^{2}}} y^{'} d x = \pm \int d x$

Put $1 - y^{2} = z^{2} \Rightarrow - 2 y y^{'} d x = 2 z d z$
$\Rightarrow - \int \frac{z d z}{z} = \pm x + c$
$\Rightarrow - z = \pm x + c$
$\Rightarrow - \sqrt{1 - y^{2}} = \pm x + c$

$y (1) = 1 \Rightarrow c = \mp 1$
$\Rightarrow - \sqrt{1 - y^{2}} = \pm x \mp 1$
$\Rightarrow - \sqrt{1 - y^{2}} = x - 1 or - x + 1$
Squaring both the sides, we get
$1 - y^{2} = x^{2} - 2 x + 1$
$\Rightarrow y^{2} = 2 x - x^{2} . . . (1)$
$\Rightarrow y = \sqrt{2 x - x^{2}} = f (x)$

From eqn $(1)$ ,
$x^{2} + y^{2} - 2 x = 0$
$\Rightarrow (x - 1)^{2} + y^{2} = 1, x \in [0, 2]$

Clearly, $x = 1$ is the point of maxima.
Maximum value of $f (x) = \sqrt{2 - 1} = 1$

Minimum value of $f (x) = 0$ at $x = 0, 2$

Area bounded by the curve $y = f (x)$ and $x - axis$ is :
$\frac{π \times 1^{2}}{2} = \frac{π}{2} sq. units .$

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Let y=f(x) be a non-negative function defined on [0,2] satisfying the differential equation y3y′′+1=0. If f′(1)=0 and f(1)=1, then

Let $y = f (x)$ be a non-negative function defined on $[0, 2]$ satisfying the differential equation $y^{3} y^{''} + 1 = 0$ . If $f^{'} (1) = 0$ and $f (1) = 1$ , then