Area Under the Curve
DE from Area Function
nta_pyq_2024_apr
Grade 12
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 $e^{-a}+4a^2+a-1$. Then the differential equation whose general solution is $y=c_1f(x)+c_2$, where $c_1,c_2$ are arbitrary constants, is:
$(8e^x-1)\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0$
$(8e^x-1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0$
$(8e^x+1)\dfrac{d^2y}{dx^2}-\dfrac{dy}{dx}=0$
$(8e^x+1)\dfrac{d^2y}{dx^2}+\dfrac{dy}{dx}=0$
Step-by-Step Solution
Key Concept: $f(x)=-e^{-x}+8x+1$. $y'=c_1(e^{-x}+8)$, $y''=-c_1e^{-x}$. Eliminate $c_1$.
$(8e^x+1)\frac{d^2y}{dx^2}+\frac{dy}{dx}=0$.
Correct Answer: 4