Q.48 What is the solution of the differential equation dy/dx = x/y, with the initial condition, at x = 0, y = 1?
(A) x2 = y2 + 1
(B) y2 = x2 + 1
(C) y2 = 2x2 + 1
(D) x2 − y2 = 0
Solution of Differential Equation dy/dx = x/y with y(0)=1 Explained
dy/dx = x/y with initial condition y(0) = 1 is y2 = x2 + 1. This separable differential equation yields a hyperbola satisfying the given condition. Option (B) is correct.
Solving the Equation
Separate variables: y dy = x dx. Integrate both sides: ∫y dy = ∫x dx gives (1/2)y2 = (1/2)x2 + C, or y2 = x2 + K where K = 2C.
Apply y(0) = 1: 1 = 0 + K, so K = 1, yielding y2 = x2 + 1.
Option Analysis
- (A)
x2 = y2 + 1 - rearranges to
y2 = x2 - 1. Atx=0,y2 = -1(impossible for real y), fails initial condition. - (B)
y2 = x2 + 1 - satisfies
y(0)=1anddy/dx = x/y: differentiate implicitly(2y y' = 2x), soy' = x/y, correct. - (C)
y2 = 2x2 + 1 - gives
y(0)=1✓, buty' = (2x)/y≠x/yunlessx=0, incorrect coefficient. - (D)
x2 − y2 = 0 - or
y = ±x.y(0)=0≠ 1, fails initial condition.
Step-by-Step Derivation
dy/dx = x/y
y dy = x dx
∫y dy = ∫x dx
(1/2)y² = (1/2)x² + C
y² = x² + 2C
y(0) = 1 → 1 = 2C → y² = x² + 1 ✓Option Verification Table
| Option | Equation | y(0) Check | Derivative Check | Status |
|---|---|---|---|---|
| (A) | x² = y² + 1 |
y² = -1 ✗ |
— | Invalid |
| (B) | y² = x² + 1 |
1=1 ✓ |
y' = x/y ✓ |
Correct |
| (C) | y² = 2x² + 1 |
1=1 ✓ |
y' = 2x/y ✗ |
Wrong |
| (D) | x² = y² |
0=1 ✗ |
— | Invalid |
dy/dx = x/y with initial condition y(0) = 1 appears frequently in entrance exams testing separable equations. The solution y² = x² + 1 represents a hyperbola passing through origin’s upper branch.
This hyperbola solution demonstrates perfect separation of variables technique for first-order ODEs in mathematical biology and physics modeling.
