Let f(x) be a positive function such that the area bounded by y=f(x), y=0 from x=0 to x=a…

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 $e^{-a}+4 a^2+a-1$. Then the differential equation, whose general solution is $y=c_1 f(x)+c_2$, where $c_1$ and $c_2$ are...

Accepted Answer

$\int _{0}^{a}f(x)dx={e}^{-a}+4{a}^{2}+a-1$

Differentiate equation w.r.t. 'a'

$\begin{gathered} f(a)=-{e}^{-a}+8a+1 \ \Rightarrow f(x)=-{e}^{-x}+8x+1 \end{gathered}$

And $y={c}_{1}f(x)+{c}_{2}$

$\begin{gathered} y={c}_{1}(-{e}^{-x}+8x+1)+{c}_{2} \ {y}^{'}={c}_{1}({e}^{-x}+8)\Rightarrow {c}_{1}=\frac{{y}^{'}}{{e}^{-x}+8} \end{gathered}$

${y}^{''}=-{c}_{1}{e}^{-x}$

put value of ${c}_{1}$

$\begin{gathered} \frac{{d}^{2}y}{d{x}^{2}}=\frac{-\frac{dy}{dx}\cdot {e}^{-x}}{({e}^{-x}+8)}=\frac{\frac{dy}{dx}}{(1+8{e}^{x})} \ \Rightarrow (1+8{e}^{x})\frac{{d}^{2}y}{d{x}^{2}}+\frac{dy}{dx}=1 \end{gathered}$

Solutions · 1