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…

$\displaystyle \begin{aligned} & \int_0^a f(x) d x=e^{-a}+4 a^2+a-1 \\ & f(a)=-e^{-a}+8 a+1 \\ & f(x)=-e^{-x}+8 x+1\end{aligned}$​∫0a​f(x)dx=e−a+4a2+a−1f(a)=−e−a+8a+1f(x)=−e−x+8x+1​ Now $y=C_1 \mathrm{f}(\mathrm{x})+\mathrm{C}_2$y=C1​f(x)+C2​ $\frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{C}_1 \mathrm{f}^{\prime}(\mathrm{x})=\mathrm{C}_1\left(\mathrm{e}^{-\mathrm{x}}+8\right)$dxdy​=C1​f′(x)=C1​(e−x+8) ...(1) $\frac{d^2 y}{d x^2}=-C_1 e^{-x} \Rightarrow-e^x \frac{d^2 y}{d x^2}$dx2d2y​=−C1​e−x⇒−exdx2d2y​ Put in equation (1) $\begin{aligned} & \frac{d y}{d x}=-e^x \frac{d^2 y}{d x^2}\left(e^{-x}+8\right) \\ & \left(8 e^x+1\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=0\end{aligned}$​dxdy​=−exdx2d2y​(e−x+8)(8ex+1)dx2d2y​+dxdy​=0​