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 +4 a^2+a-1 . Then the differential equation, whose general solution is y=c₁ f(x)+c₂ , where c₁ and c₂ are arbitrary constants, is

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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=c₁ f(x)+c₂ , where c₁ and c₂ are arbitrary constants, is

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Correct answer: D. (8 e^x+1 ) d^2 y d x^2 + d y d x =0

aligned & ₀^a f(x) d x=e^ -a +4 a^2+a-1 & f(a)=-e^ -a +8 a+1 & f(x)=-e^ -x +8 x+1 aligned Now y=C₁ f ( x )+ C ₂ dy dx = C ₁ f ^ ( x )= C ₁ ( e ^ - x +8 ) ...(1) d^2 y d x^2 =-C₁ e^ -x -e^x d^2 y d x^2 Put in equation (1) aligned & d y d x =-e^x d^2 y d x^2 (e^ -x +8 ) & (8 e^x+1 ) d^2 y d x^2 + d y d x =0 aligned

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=c₁ f(x)+c₂ , where c₁ and c₂ are arbitrary constants, is

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