Let f (x ) be a positive function such that the area bounded by y=f (x ) ,y=0 from x=0 to x=a0 is e^-a+4a^2+a-1. Then the differential equation, whose general solution is y=c_1f (x )+c_2, where c_1…

Question

Let f (x ) be a positive function such that the area bounded by y=f (x ) ,y=0 from x=0 to x=a>0 is e^-a+4a^2+a-1. Then the differential equation, whose general solution is y=c_1f (x )+c_2, where c_1 and c_2 are arbitrary constants, is

Accepted Answer

Correct answer: D. (d) 0.05em 0.05emf (x ) dx=e^-a+4a^2+a-1f (a )=-e^-a+8a+1f (x )=-e^-x+8x+1 Now y=C_1f (x )+C_2 dy dx=C_1f^ ' (x )=C_1(e^-x+8) … d^2ydx^2=-C_1e^-x ⇒-e^x d^2ydx^2 Put in equation (1) dy dx=-e^x d^2ydx^2(e^-x+8)(8e^x+1) d^2ydx^2+ dy dx=0

$Let f (x) be a positive function such that the area bounded by y = f (x), y = 0 from x = 0 to x = a > 0 is e^{- a} + 4 a^{2} + a - 1 . Then the differential equation, whose general solution is y = c_{1} f (x) + c_{2}, where c_{1} and c_{2} are arbitrary constants, is$

Solution

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