Question:
Using Runge-Kutta method of fourth order, an approximate value of \(y\) at \(x=0.2\), given that \(\frac{dy}{dx}=\frac{y^2-x^2}{y^2+x^2}\) and \(y(0)=1\), is:
Using Runge-Kutta method of fourth order, an approximate value of \(y\) at \(x=0.2\), given that \(\frac{dy}{dx}=\frac{y^2-x^2}{y^2+x^2}\) and \(y(0)=1\), is:
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Always compute \(k_1, k_2, k_3, k_4\) carefully—small arithmetic mistakes lead to large final errors.
Updated On: May 22, 2026
- 1.196
- 1.926
- 1.862
- 2.124
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The Correct Option is A
Solution and Explanation
Concept:
Runge-Kutta 4th order method uses weighted averages of slopes:
\[
y_{n+1} = y_n + \frac{h}{6}(k_1 + 2k_2 + 2k_3 + k_4)
\]
Step 1: Given values.
\[ x_0=0,\quad y_0=1,\quad h=0.2 \]
Step 2: Compute slopes.
\[ k_1 = f(0,1) = \frac{1-0}{1+0} = 1 \] \[ k_2 = f(0.1,1+0.1)=f(0.1,1.1) \] \[ k_3 = f(0.1,1+0.1k_2) \] \[ k_4 = f(0.2,1+0.2k_3) \] After full substitution and simplification: \[ y(0.2) \approx 1.196 \]
Step 1: Given values.
\[ x_0=0,\quad y_0=1,\quad h=0.2 \]
Step 2: Compute slopes.
\[ k_1 = f(0,1) = \frac{1-0}{1+0} = 1 \] \[ k_2 = f(0.1,1+0.1)=f(0.1,1.1) \] \[ k_3 = f(0.1,1+0.1k_2) \] \[ k_4 = f(0.2,1+0.2k_3) \] After full substitution and simplification: \[ y(0.2) \approx 1.196 \]
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