If dydx=x2−2x−4, y(3)=−6 then 3y is equal to

dydx=x2−2x−4, y(3)=−6 ∫dy=∫x2-2x-4dx y=x33-x2-4x+c y(3)=−6 ⇒-6=9-9-12+c⇒c=6 ⇒y=x33-x2-4x+6 ⇒3y=x3−3x2−12x+18

If dydx=x2−2x−4, y(3)=−6 then 3y is equal to

dydx=x2−2x−4, y(3)=−6 ∫dy=∫x2-2x-4dx y=x33-x2-4x+c y(3)=−6 ⇒-6=9-9-12+c⇒c=6 ⇒y=x33-x2-4x+6 ⇒3y=x3−3x2−12x+18

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If $\frac{d y}{d x} = x^{2} - 2 x - 4, y (3) = - 6$ then 3y is equal to

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$x^{3} + 3 x^{2} + 12 x - 18$

$x^{3} - 3 x^{2} + 12 x + 18$

$x^{3} + 3 x^{2} + 12 x + 18$

$x^{3} - 3 x^{2} - 12 x + 18$

answer is D.

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Detailed Solution

$\frac{d y}{d x} = x^{2} - 2 x - 4, y (3) = - 6 \int d y = \int x^{2} - 2 x - 4 d x y = \frac{x^{3}}{3} - x^{2} - 4 x + c y (3) = - 6 \Rightarrow - 6 = 9 - 9 - 12 + c \Rightarrow c = 6 \Rightarrow y = \frac{x^{3}}{3} - x^{2} - 4 x + 6 \Rightarrow 3 y = x^{3} - 3 x^{2} - 12 x + 18$