Question

The particular solution of the differential equation \((y - x^2) dy = (1 - x^3) dx\) with \(y(0) = 1\), is:

• A. \(y^2 = x^2 + 2 \log_e |1 + x| + 1\)
• B. \(y^2 = 1 + x^2 + 2 \log_e \left| \frac{1 + x}{2} \right|\)
• C. \(y^2 = x^2 + 2x - 3\)
• D. \(y^2 = x^2 + 2x + 1\)

Answer 1

The Correct Option is A

Rewrite the given differential equation as:

\[ \frac{dy}{dx} = \frac{1 - x^3}{y - x^2 y}. \]

Factor \( y \) in the denominator:

\[ \frac{dy}{dx} = \frac{1 - x^3}{y(1 - x^2)}. \]

Separate variables:

\[ y(1 - x^2) dy = (1 - x^3) dx. \]

Integrate both sides:

\[ \int y \, dy = \int \frac{1 - x^3}{1 - x^2} dx. \]

The left-hand side integrates to:

\[ \int y \, dy = \frac{y^2}{2}. \]

Simplify the right-hand side using partial fractions:

\[ \frac{1 - x^3}{1 - x^2} = 1 + x. \]

Thus:

\[ \int (1 + x) dx = x + \frac{x^2}{2}. \]

Equating both sides:

\[ \frac{y^2}{2} = x + \frac{x^2}{2} + C. \]

Multiply through by 2:

\[ y^2 = x^2 + 2x + 2C. \]

Using the initial condition \( y(0) = 1 \):

\[ 1^2 = 0 + 0 + 2C \implies C = \frac{1}{2}. \]

Thus, the solution is:

\[ y^2 = x^2 + 2x + 1. \]