Definite Integration
Integral Equation — Differential Equation Approach
DAILY_CHALLENGE
Grade 12
Question:
Let $\int_0^x\sqrt{1-(y'(t))^2}\,dt=\int_0^x y(t)\,dt$, $0\leq x\leq3$, $y\geq0$, $y(0)=0$. Then at $x=2$, $y''+y+1$ is equal to:
Step-by-Step Solution
Key Concept: Differentiate both sides: $\sqrt{1-(y')^2}=y\Rightarrow1-(y')^2=y^2\Rightarrow(y')^2=1-y^2$. Solve: $y=\sin x$ (using $y(0)=0$, $y\geq0$).
$y=\sin x$. $y''+y+1=-\sin x+\sin x+1=1$.
Correct Answer: 1