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Solve the initial value problem \(dy/dx=\frac{1}{2](x+y)1\), \(0\le x\le0.3\), \(y(0)=1\), using fourthorder RungeKutta method with h=0.1. Compare your results with the analytical solution \(y=e^{\frac{x}{2}}x\) to this differential equation.
 one year ago
 one year ago
Please help:) Solve the initial value problem \(dy/dx=\frac{1}{2](x+y)1\), \(0\le x\le0.3\), \(y(0)=1\), using fourthorder RungeKutta method with h=0.1. Compare your results with the analytical solution \(y=e^{\frac{x}{2}}x\) to this differential equation.
 one year ago
 one year ago

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phiBest ResponseYou've already chosen the best response.0
this sounds like you need a computer program. what language?
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
No. I am learning this under maths, numerical analysis.
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
I am k with the first part of the question bt am stuck in the comparision part.
 one year ago

phiBest ResponseYou've already chosen the best response.0
You have to evaluate a number of expressions at every step. It looks like they want 3 steps 0 to 0.3 in steps of 0.1 so you could do it with a calculator
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
ya. bt hw abt the comparision part.
 one year ago

phiBest ResponseYou've already chosen the best response.0
\(y=e^{\frac{x}{2}}x\) plug in x=0.3 and evaluate. compare to your answer using rungekutta
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
bt y specifically 0.3
 one year ago

ajprincessBest ResponseYou've already chosen the best response.0
ok so I shoud check if I get the same when I solve dy/dx=1/2(x+3)1 using rangekutta method for x=0.3
 one year ago

phiBest ResponseYou've already chosen the best response.0
Here is a bit of matlab y=1; h=0.1; x=0; for ii=1:3 k1= h*fff(x,y); k2= h*fff(x+h/2,y+k1/2); k3= h*fff( x+h/2, y+ k2/2); k4= h*fff( x+h,y+k3); y= y+ (k1+2*k2+2*k3+k4)/6 x= x+h end where fff(x,y) is defined as function z=fff(x,y) z= 0.5*(x+y)1; I still have to check if I did this correctly...
 one year ago

phiBest ResponseYou've already chosen the best response.0
The matlab gives y= 0.861834234021667 the exact answer e^(0.15)0.3 = 0.861834242728283
 one year ago

phiBest ResponseYou've already chosen the best response.0
Here are the partial results Step 1: x0= 0.000000 y0= 1.000000 k1 = 0.050000000000000 k2 = 0.048750000000000 k3 = 0.048718750000000 k4 = 0.047435937500000 y = 0.951271093750000 Step 2: x1= 0.100000 y1= 0.951271 k1 = 0.047436445312500 k2 = 0.046122356445313 k3 = 0.046089504223633 k4 = 0.044740920523682 y = 0.905170912554321 Step 3: x2= 0.200000 y2= 0.905171 k1 = 0.044741454372284 k2 = 0.043359990731591 k3 = 0.043325454140574 k4 = 0.041907727079313 y = 0.861834234021667 y =0.861834234021667 x =0.300000000000000
 one year ago

A.Avinash_GouthamBest ResponseYou've already chosen the best response.0
wat's the problem here? i mean with the second part? i think it's straight forward.....did i miss smthin?
 one year ago
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