Area Under the Curve
Differential Equation from Area Condition
nta_pyq_2024_jan
Grade 12
Question:
Let $Y=Y(X)$ be a curve lying in the first quadrant such that the area enclosed by the line $Y-y=Y'(X)(X-x)$ and the co-ordinate axes, where $(x,y)$ is any point on the curve, is always $\frac{-y^2}{2Y'(x)}+1$, $Y'(x)\ne0$. If $Y(1)=1$, then $12Y(2)$ equals
Step-by-Step Solution
Key Concept: The tangent at $(x,y)$ has intercepts: x-intercept $=x-y/Y'(x)$, y-intercept $=y-xY'(x)$. Area of triangle $=\frac{1}{2}|x-y/Y'||y-xY'|=\frac{-y^2}{2Y'}+1$. This gives the ODE $2xy-y^2y'=2Y'$. Solve with $Y(1)=1$.
ODE: $\frac{dy}{dx}-\frac{2}{x}y=-\frac{2}{x^2}$. IF $=e^{-2\ln x}=1/x^2$. $y/x^2=\frac{2}{3}x^{-3}+C$. At $(1,1)$: $1=2/3+C\Rightarrow C=1/3$. $Y(x)=\frac{2}{3x}+\frac{x^2}{3}$. $12Y(2)=12\cdot\frac{5}{3}=20$.
Correct Answer: 20