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 $y=c_1 f(x)+c_2$, where $c_1$ and $c_2$ 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 $y=c_1 f(x)+c_2$, where $c_1$ and $c_2$ are arbitrary constants, is

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$\begin{aligned} & \int_0^a f(x) d x=e^{-a}+4 a^2+a-1 \ & f(a)=-e^{-a}+8 a+1 \ & f(x)=-e^{-x}+8 x+1\end{aligned}$ Now $y=C_1 \mathrm{f}(\mathrm{x})+\mathrm{C}_2$ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{C}_1 \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{C}_1\left(\mathrm{e}^{-\mathrm{x}}+8ight)$ ...(1) $\frac{d^2 y}{d x^2}=-C_1 e^{-x} \Rightarrow-e^x \frac{d^2 y}{d x^2}$ Put in equation (1) $\begin{aligned} & \frac{d y}{d x}=-e^x \frac{d^2 y}{d x^2}\left(e^{-x}+8ight) \ & \left(8 e^x+1ight) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0\end{aligned}$

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…

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