A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 years 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.
 2 years 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.

This Question is Closed

phi
 2 years ago
Best ResponseYou've already chosen the best response.0this sounds like you need a computer program. what language?

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0No. I am learning this under maths, numerical analysis.

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0I am k with the first part of the question bt am stuck in the comparision part.

phi
 2 years ago
Best ResponseYou've already chosen the best response.0You 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

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0ya. bt hw abt the comparision part.

phi
 2 years ago
Best 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

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0bt y specifically 0.3

ajprincess
 2 years ago
Best ResponseYou've already chosen the best response.0ok 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

phi
 2 years ago
Best ResponseYou've already chosen the best response.0Here 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...

phi
 2 years ago
Best ResponseYou've already chosen the best response.0The matlab gives y= 0.861834234021667 the exact answer e^(0.15)0.3 = 0.861834242728283

phi
 2 years ago
Best ResponseYou've already chosen the best response.0Here 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

A.Avinash_Goutham
 2 years ago
Best ResponseYou've already chosen the best response.0wat's the problem here? i mean with the second part? i think it's straight forward.....did i miss smthin?
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.