First Order De

Mediummathematics

The solution of dy/dx + y = 0 with initial condition y(0) = 1 is

Select the correct option:

$y = e^{(} - x)$

$y = e^{x}$

y = 1 - x

y = 1 + x

🔒 Solution Hidden from View

Submit your answer to unlock the detailed step-by-step solution.

About This Question

Subject: mathematics
Chapter: differential equations
Topic: first order de
Difficulty: Medium
Year: 2025

Solution

Correct Answer:

$y = e^{(} - x)$

Solve General Equation: $\frac{d y}{d x} = - y$ .
- $\frac{d y}{y} = - d x ⟹ ln y = - x + C$
- $y = e^{- x + C} = K e^{- x}$ .
Apply Initial Condition: Given $y = 1$ when $x = 0$ .
- $1 = K e^{0} = K \cdot 1$
- $K = 1$ .
Particular Solution: Substituting $K = 1$ back, we get $y = e^{- x}$ .

This medium difficulty mathematics question is from the chapter differential equations, covering the topic of first order de. It appeared in the 2025 exam.

Looking for more practice? Explore all mathematics questions or browse differential equations questions on RankGuru.

Rank Guru

First Order De

Select the correct option:

Solution

🔒 Solution Hidden from View

More first order de Practice Questions

The differential equation dy/dx = x/y represents

The degree of the differential equation (dy/dx)³ + 2(dy/dx)² + y = 0 is

The solution of dy/dx = y is

The order of the differential equation d²y/dx² + 3(dy/dx) + 2y = 0 is

About This Question

Solution