Step 1 : Concept:
This question tests initial value problems (IVPs) for first-order and second-order ordinary differential equations. Step 2 : Key Formulas and Approach:
Solve each differential equation by direct integration, separation of variables, or characteristic equations, and apply the given initial conditions. Step 3 : Step-by-step Explanation:
• Item A: $y'' = 1$ with $y(0) = 1, y'(0) = 2$.
Integrating once: $y' = x + C_1$. Since $y'(0) = 2 \implies C_1 = 2$.
Integrating again: $y = \frac{x^2}{2} + 2x + C_2$. Since $y(0) = 1 \implies C_2 = 1$.
Thus, $y = \frac{x^2}{2} + 2x + 1$. Matches with II.
• Item B: $\sin\left(\frac{dy}{dx}\right) = 1 \implies \frac{dy}{dx} = \sin^{-1}(1) = \frac{\pi}{2}$.
Integrating: $y = \frac{\pi}{2} x + C$.
Using $y(0) = 1 \implies C = 1$, so $y - 1 = \frac{\pi}{2} x \implies \frac{y-1}{x} = \frac{\pi}{2}$.
Taking sine on both sides: $\sin\left(\frac{y-1}{x}\right) = \sin\left(\frac{\pi}{2}\right) = 1$. Matches with IV.
• Item C: $\frac{dy}{dx} = e^{x+y} = e^x e^y \implies e^{-y} dy = e^x dx$.
Integrating: $-e^{-y} = e^x + C$.
Using $y(0) = 1 \implies -e^{-1} = e^0 + C = 1 + C \implies C = -1 - e^{-1}$.
Substituting $C$: $-e^{-y} = e^x - 1 - e^{-1} \implies e^x + e^{-y} = 1 + e^{-1}$.
Multiplying the entire equation by $e$: $e^{x+1} + e^{-(y-1)} = e + 1$. Matches with I.
• Item D: $\frac{d^2 y}{dx^2} + y = 0$ with $y(0) = -1, y'(0) = 1$.
General solution: $y(x) = c_1 \cos x + c_2 \sin x$.
Using $y(0) = -1 \implies c_1 = -1$.
$y'(x) = -c_1 \sin x + c_2 \cos x \implies y'(0) = c_2 = 1$.
Thus, $y(x) = \sin x - \cos x$. Matches with III. Step 4 : Final Answer:
The correct matching is A-II, B-IV, C-I, D-III, which corresponds to option (B).
