Electronics and Communication Engineering (ECE) Exam > Electronics and Communication Engineering (ECE) Questions > Givendy/dx = x2 − y2, y(0)The value of y(0/1...
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Given
dy/dx = x2 − y2, y(0)
The value of y(0/1) using Runge-kutta Second order method is (Take h = 0.1)
- a)0.89
- b)0.93
Correct answer is between '0.89,0.93'. Can you explain this answer?
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Most Upvoted Answer
Givendy/dx = x2 − y2, y(0)The value of y(0/1) using Runge-kutta Secon...
**Solution:**
Given differential equation is given as:
$\frac{dy}{dx} = x^2 - y^2$
Initial condition is given as $y(0) = 1$.
Using Runge-Kutta Second Order method, we can find the value of $y$ at $x=0.1$.
Let us first find the values of $k_1$ and $k_2$:
$k_1 = hf(x_n,y_n) = 0.1(0^2 - 1^2) = -0.1$
$k_2 = hf(x_n + h/2, y_n + k_1/2) = 0.1(0.05^2 - 0.955^2) = -0.045$
Now, we can find the value of $y_{n+1}$ as:
$y_{n+1} = y_n + \frac{k_1+k_2}{2} = 1 + \frac{-0.1-0.045}{2} = 0.9275$
Therefore, the value of $y$ at $x=0.1$ using Runge-Kutta Second Order method is approximately 0.9275.
Since the correct answer is between 0.89 and 0.93, our answer is plausible.
Given differential equation is given as:
$\frac{dy}{dx} = x^2 - y^2$
Initial condition is given as $y(0) = 1$.
Using Runge-Kutta Second Order method, we can find the value of $y$ at $x=0.1$.
Let us first find the values of $k_1$ and $k_2$:
$k_1 = hf(x_n,y_n) = 0.1(0^2 - 1^2) = -0.1$
$k_2 = hf(x_n + h/2, y_n + k_1/2) = 0.1(0.05^2 - 0.955^2) = -0.045$
Now, we can find the value of $y_{n+1}$ as:
$y_{n+1} = y_n + \frac{k_1+k_2}{2} = 1 + \frac{-0.1-0.045}{2} = 0.9275$
Therefore, the value of $y$ at $x=0.1$ using Runge-Kutta Second Order method is approximately 0.9275.
Since the correct answer is between 0.89 and 0.93, our answer is plausible.
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Community Answer
Givendy/dx = x2 − y2, y(0)The value of y(0/1) using Runge-kutta Secon...
Y’ = x2 - y2
y(0) = 1
y1p = y0 + h.f (xo,yo)
=1 + 0.1(-1) = 0.9
Modified value of y1 is
y1c= y0+ h/2 [f(x0,y0) + f(x1,y1p)]
=1+ 0.1/2[-1 + (0.01 - 0.81)]
=1 - 0.09
= 0.91
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Givendy/dx = x2 − y2, y(0)The value of y(0/1) using Runge-kutta Second order method is (Take h = 0.1)a)0.89b)0.93Correct answer is between '0.89,0.93'. Can you explain this answer?
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