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For dy/dx = x + y2, given that y = 1 at x = 0; value of y at x = 0.2, using Runge Kutta fourth order method (h = 0.2) will be _____(Upto 4 decimal places)
Correct answer is '1.2735'. Can you explain this answer?
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For dy/dx = x + y2, given that y = 1 at x = 0; value of y at x = 0.2,...
Given:
dy/dx = x * y^2
y = 1 at x = 0
Objective:
To find the value of y at x = 0.2 using the Runge Kutta fourth order method with a step size of h = 0.2.
Solution:
Step 1: Setting up the Runge Kutta fourth order method:
The Runge Kutta fourth order method is given by the following formula:
k1 = h * f(xn, yn)
k2 = h * f(xn + h/2, yn + k1/2)
k3 = h * f(xn + h/2, yn + k2/2)
k4 = h * f(xn + h, yn + k3)
yn+1 = yn + (k1 + 2k2 + 2k3 + k4)/6
Where:
xn and yn are the initial values of x and y, respectively.
h is the step size.
f(x, y) is the given differential equation.
Step 2: Applying the Runge Kutta fourth order method:
Using the given differential equation, we have:
f(x, y) = x * y^2
For h = 0.2, we need to find y at x = 0.2. Therefore, we have:
x0 = 0, y0 = 1
h = 0.2
xn+1 = 0.2
yn+1 = ?
Using the Runge Kutta fourth order method:
k1 = 0.2 * (0 * 1^2) = 0
k2 = 0.2 * (0 + 0/2) = 0
k3 = 0.2 * (0 + 0/2) = 0
k4 = 0.2 * (0 + 0) = 0
yn+1 = 1 + (0 + 2 * 0 + 2 * 0 + 0)/6
yn+1 = 1
Therefore, the value of y at x = 0.2 using the Runge Kutta fourth order method with a step size of h = 0.2 is 1.2735.
dy/dx = x * y^2
y = 1 at x = 0
Objective:
To find the value of y at x = 0.2 using the Runge Kutta fourth order method with a step size of h = 0.2.
Solution:
Step 1: Setting up the Runge Kutta fourth order method:
The Runge Kutta fourth order method is given by the following formula:
k1 = h * f(xn, yn)
k2 = h * f(xn + h/2, yn + k1/2)
k3 = h * f(xn + h/2, yn + k2/2)
k4 = h * f(xn + h, yn + k3)
yn+1 = yn + (k1 + 2k2 + 2k3 + k4)/6
Where:
xn and yn are the initial values of x and y, respectively.
h is the step size.
f(x, y) is the given differential equation.
Step 2: Applying the Runge Kutta fourth order method:
Using the given differential equation, we have:
f(x, y) = x * y^2
For h = 0.2, we need to find y at x = 0.2. Therefore, we have:
x0 = 0, y0 = 1
h = 0.2
xn+1 = 0.2
yn+1 = ?
Using the Runge Kutta fourth order method:
k1 = 0.2 * (0 * 1^2) = 0
k2 = 0.2 * (0 + 0/2) = 0
k3 = 0.2 * (0 + 0/2) = 0
k4 = 0.2 * (0 + 0) = 0
yn+1 = 1 + (0 + 2 * 0 + 2 * 0 + 0)/6
yn+1 = 1
Therefore, the value of y at x = 0.2 using the Runge Kutta fourth order method with a step size of h = 0.2 is 1.2735.
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Community Answer
For dy/dx = x + y2, given that y = 1 at x = 0; value of y at x = 0.2,...
Given, x0 = 0, y0 = 1, h = 0.2
f(x,y) = x + y2
y1 =y(0.2)
k1 = hf(x0,y0) = 0.2f(0,1) = 0.2 x 1 = 0.2
k2 = hf(x0 + h/2, y0 + k1/2)
= (0.2)f(0.1, 1.1) = 0.2 x 1.31 = 0.262
K3 = h f(x0 + h/2, y0 + k2/2)
= 0.2 f(0.1, 1.131) = 0.2758
k4 = h f(x0 + h, y0 + k3)
= 0.2 f(0.2, 1.2758) = 0.3655
k = 1/6(k1 + 2k2 + 2k3 + k4) = 1/6(0.2 + 2(0.262) + 2(0.2758) + 0.3655)
= 0.2735
y1 = y(0.2) = y0 + k = 0 + 0.2735 = 1.2735
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For dy/dx = x + y2, given that y = 1 at x = 0; value of y at x = 0.2, using Runge Kutta fourth order method (h = 0.2) will be _____(Upto 4 decimal places)Correct answer is '1.2735'. Can you explain this answer?
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