Question

Differential equation \(y^3\frac{dy}{dx}\;+\;x^3=0\), y(0) = 1 has a solution given by y. The value of y(-1) is:
1. -1
2. 0
3. 1
4. 2

Answer 1

Correct Answer - Option 2 : 0

Concept:

Variable separable method

∫f(y)dy = ∫f(x)dx

Calculation:

Given:

\(y^3\frac{dy}{dx}\;+\;x^3=0\)

y³dy = -x³dx

Integrating both sides

\(\frac{y^4}{4}=-\frac{x^4}{4}+C\)

At y(0) = 1

\(C=\frac{1}{4}\)

\(\frac{y^4}{4}=-\frac{x^4}{4}+\frac{1}{4}\)

y⁴ = -x⁴ + 1

At y(-1)

y⁴ = -(-1)⁴ + 1 = 0