First slide

Applications of differential equations

Question

If the slope of the tangent to the curve is maximum at x = 1 and curve has minim value 1 at x = 0, then the curve which also satisfies the equation $\frac{d^{3} y}{d x^{3}} = 4 x - 3$ is

Difficult

A

$y + 2 x + \frac{d y}{d x} = 0$
B

$y = 1 + \frac{x^{2}}{2} - \frac{x^{3}}{2} + \frac{x^{4}}{6}$
C

$y = 1 + x + x^{2} + x^{3}$
D

none of these

Solution

we have $\frac{d^{3} y}{d x^{3}} = 4 x - 3$
Integrating, we get    $\frac{d^{2} y}{d x^{2}} = 2 x^{2} - 3 x + c$
Given slope of tangent to the curve is maximum at x = 1,
$s l o p e i s \frac{d y}{d x}, t o f i n d s l o p e m a x e q u a t e \frac{d^{2} y}{d x^{2}} t o z e r o$
Then ${(\frac{d^{2} y}{d x^{2}})}_{x = 1} = 0 \Rightarrow 2 - 3 + c = 0 \Rightarrow c = 1$
$\frac{d^{2} y}{d x^{2}} = 2 x^{2} - 3 x + 1$
Integrating we get
$\frac{d y}{d x} = \frac{2 x^{3}}{3} - \frac{3 x^{2}}{2} + x + D$
$∴$ curve has a minimum value 1 at x = 0
$t o f i n d y = f (x) m i n i m u m e q u a t e f^{'} (x) = 0 o r \frac{d y}{d x} = 0$
$∴ {(\frac{d y}{d x})}_{x = 0} = 0 \Rightarrow D = 0$
$∴ \frac{d y}{d x} = \frac{2 x^{3}}{3} - \frac{3 x^{2}}{2} + x$
Again integrating, we get    $y = \frac{x^{4}}{6} - \frac{x^{3}}{2} + \frac{x^{2}}{2} + E$
Given minimum value of curve is 1 at x = 0
Then 1 = 0 + E $\Rightarrow$ E = 1
Hence the curve is    $y = 1 + \frac{x^{2}}{2} - \frac{x^{3}}{2} + \frac{x^{4}}{6}$

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