Question

The approximate solution of the initial value problem \( \frac{dy}{dx} = 1 + xy \), \( y(0) = 1 \) by Picard’s method will be

• A. \( y = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots \)
• B. \( y = 1 + x + \frac{x^2}{2} + \frac{x^3}{8} + \dots \)
• C. \( y = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!} + \dots \)
• D. \( y = 1 + \frac{x^2}{2} - \frac{x^4}{8} \dots \)

Hint:

In Picard’s method, use previous approximation inside the integral to construct the next.

Answer 1

The Correct Option is B

Given: \( \frac{dy}{dx} = 1 + xy \), with \( y(0) = 1 \)
Picard’s method constructs successive approximations: \[ y_0(x) = 1 \]
\[ y_1(x) = 1 + \int_0^x (1 + t \cdot y_0(t)) dt = 1 + \int_0^x (1 + t) dt = 1 + x + \frac{x^2}{2} \]
\[ y_2(x) = 1 + \int_0^x (1 + t \cdot y_1(t)) dt = 1 + \int_0^x \left(1 + t(1 + t + \frac{t^2}{2})\right) dt \]
\[ = 1 + \int_0^x \left(1 + t + t^2 + \frac{t^3}{2} \right) dt = 1 + x + \frac{x^2}{2} + \frac{x^3}{3} + \frac{x^4}{8} \]
Thus, the second iteration gives: \[ y \approx 1 + x + \frac{x^2}{2} + \frac{x^3}{8} + \dots \]