The elimination of the arbitrary constants A, | vedclass

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Question

(C) Given equation is $y = A + Bx + C{e^{ - x}} \dots (i)$Differentiating with respect to $x$:$\frac{dy}{dx} = B - C{e^{ - x}} \dots (ii)$Differentiat

Vedclass · Accepted Answer

$y''' + y'' = 0$

Vedclass · Answer

(C) Given equation is $y = A + Bx + C{e^{ - x}} \dots (i)$Differentiating with respect to $x$:$\frac{dy}{dx} = B - C{e^{ - x}} \dots (ii)$Differentiating again with respect to $x$:$\frac{d^2y}{dx^2} = C{e^{ - x}} \dots (iii)$Differentiating a third time with respect to $x$:$\frac{d^3y}{dx^3} = -C{e^{ - x}} \dots (iv)$From equation $(iii)$,we have $C{e^{ - x}} = \frac{d^2y}{dx^2}$.Substituting this into equation $(iv)$:$\frac{d^3y}{dx^3} = -\frac{d^2y}{dx^2}$Rearranging the terms,we get:$\frac{d^3y}{dx^3} + \frac{d^2y}{dx^2} = 0$,which is $y''' + y'' = 0$.

The elimination of the arbitrary constants $A, B$ and $C$ from $y = A + Bx + C{e^{ - x}}$ leads to the differential equation:

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