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 arbitrary constants, is

Accepted Answer

$$\int_\limits0^a f(x) d x=e^{-a}+4 a^2+a-1$$

Differentiate equation w.r.t. 'a'

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

And $y=c_1 f(x)+c_2$

$$\begin{aligned}
& y=c_1\left(-e^{-x}+8 x+1\right)+c_2 \
& y^{\prime}=c_1\left(e^{-x}+8\right) \Rightarrow c_1=\frac{y^{\prime}}{e^{-x}+8}
\end{aligned}$$

$$y^{\prime \prime}=-c_1 e^{-x}$$

put value of $c_1$

$$\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{-\frac{d y}{d x} \cdot e^{-x}}{\left(e^{-x}+8\right)}=\frac{\frac{d y}{d x}}{\left(1+8 e^x\right)} \
& \Rightarrow\left(1+8 e^x\right) \frac{d^2 y}{d x^2}+\frac{d y}{d x}=1
\end{aligned}$$

Solutions · 1