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

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

ajprincess Group TitleBest 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

phi Group TitleBest 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

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

phi Group TitleBest 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

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

phi Group TitleBest ResponseYou've already chosen the best response.0
x= 0.3 y will be 0.862
 one year ago

ajprincess Group TitleBest 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

phi Group TitleBest 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

phi Group TitleBest 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

phi Group TitleBest 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_Goutham Group TitleBest 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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