JEE Mains · Maths · STD 12 - 9. differential equations
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 \(\mathrm{e}^{-\mathrm{a}}+4 \mathrm{a}^2+\mathrm{a}-1\). 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 :
- A \(\left(8 e^x-1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0\)
- B \(\left(8 e^x+1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0\)
- C \(\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0\)
- D \(\left(8 e^x-1\right) \frac{d^2 y}{d x^2}-\frac{d y}{d x}=0\)
Answer & Solution
Correct Answer
(C) \(\left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0\)
Step-by-step Solution
Detailed explanation
\( \int_0^a f(x) d x=e^{-a}+4 a^2+a-1 \) \( f(a)=-e^{-a}+8 a+1 \) \( f(x)=-e^{-x}+8 x+1\) Now \(y=\mathrm{C}_1 \mathrm{f}(\mathrm{x})+\mathrm{C}_2\) \( \frac{d y}{d x}=C_1 f^{\prime}(x)=C_1\left(e^{-x}+8\right) \) ..................(\(1\))…
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