The solution of the differential equation d^ 2 y dx^ 2 = 3x + e^ x + x^ 2 when y _ 1 (0) = 1 and y(0) = 0 , is

Correct answer: A. - ( 3x)/(9) + e^ x + (x4)/(12) + (1)/(3)x - 1 Integrating the given differential equation, we have (dy)/(dx) = (- 3x)/(3) + e^ x + (x3)/(3) + C _ 1 But y _ 1 (0) = 1 So, 1 = (- (1)/(3)) + 1 + C _ 1 ⇒ C _ 1 = (1)/(3) ∴ (dy)/(dx) = (- 3x)/(3) + e^ x + (x3)/(3) + (1)/(3) Again integrating, we get y = - ( 3x)/(9) + e^ x + (x4)/(12) + (1)/(3)x + C _ 2 But y(0) = 0, so 0 = 0 + 1 + C _ 2 ⇒ C _ 2 = - 1 Thus, y = - ( 3x)/(9) + e^ x + (x4)/(12) + (1)/(3)x - 1 Hence, option (a) correct.

BITSAT 2018

Exam: BITSAT
Year: 2018
Subject: Mathematics
Chapter: Differential Equations

BITSAT 2018 · Mathematics · Differential Equations

The solution of the differential equation d^ 2 y dx^ 2 = 3x + e^ x...

The solution of the differential equation

\dfrac{d^{2}y}{dx^{2}} = \sin 3x + e^{x} + x^{2}

when

{y}_{1}\left(0\right) = 1

and

y\left(0\right) = 0

, is

Options

A $- \dfrac{\sin 3x}{9} + e^{x} + \dfrac{x4}{12} + \dfrac{1}{3}x - 1$
B $- \dfrac{\sin 3x}{9} + e^{x} + \dfrac{x4}{12} + \dfrac{1}{3}x$
C $- \dfrac{\cos 3x}{9} + e^{x} + \dfrac{x4}{12} + \dfrac{1}{3}x + 1$
DNone of the above

Correct answer

A. $- \dfrac{\sin 3x}{9} + e^{x} + \dfrac{x4}{12} + \dfrac{1}{3}x - 1$

Step-by-step solution

Integrating the given differential equation, we have

$\dfrac{dy}{dx} = \dfrac{- \cos 3x}{3} + e^{x} + \dfrac{x3}{3} + {C}_{1}$

But ${y}_{1}\left(0\right) = 1$

So, $1 = \left(- \dfrac{1}{3}\right) + 1 + {C}_{1}$

$⇒ {C}_{1} = \dfrac{1}{3}$

$∴ \dfrac{dy}{dx} = \dfrac{- \cos 3x}{3} + e^{x} + \dfrac{x3}{3} + \dfrac{1}{3}$

Again integrating, we get

$y = - \dfrac{\sin 3x}{9} + e^{x} + \dfrac{x4}{12} + \dfrac{1}{3}x + {C}_{2}$

But $y\left(0\right) = 0,$ so $0 = 0 + 1 + {C}_{2}$

$⇒ {C}_{2} = - 1$

Thus, $y = - \dfrac{\sin 3x}{9} + e^{x} + \dfrac{x4}{12} + \dfrac{1}{3}x - 1$

Hence, option (a) correct.

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