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# MTH603 GDB 2 Last date 09-02-2017

Solve the Differential Equation

dy/dx =5x+y;  y(0)=0.5 in the interval [0,0.5] using Euler’s Method by taking h=0.1

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### Replies to This Discussion

first y0 you are calculating y0=0.5x0.1(5*0+0.5)

on next line you are doing y0=0.5x0.5(5*0+0.5)

tell me the line no plz . i am not getting you

ap jahan y1 ki calculations kar rahy ho page no 1 per. . .

this link is not working http://www.numberempire.com/texequationeditor/equationeditor.php

students me ne sir ko mail k the or un se pucha tha k 5 value count ho g ya nahi but unho ne kaha he k 5 value add nai ho g

ab kiya kren

??????????????????

this link is not working http://www.numberempire.com/texequationeditor/equationeditor.php

question mai 5x+y likha hy issi liye multiply karengy

If h = 0.1

We will calculate the values of x as following:

X, 0

X, 0.1

X, 0.2

X, 0.3

X, 0.4

X, 0.5

y ( 0 ) = 0.5

Where

x0 = 0               y0 = 0.5

y1 = y0 + hf (xo,yo)

Putting the values

y1 = 0.5 + (0.1)(5)( 0 + 0.5)

y1 = 0.75

Now we take,

y1 = 0.75         x1 = 0.1

y2 = y1 + hf(x1, y1)

Putting the values

y2 = 0.75+ (0.1)(5)( 0.1 + 0.75)

y2 = 0.75+ 0.425

y2 = 1.175

Now we take,

y2 = 1.175       x2 = 0.2

y3 = y2 + hf(x2,y2)

Putting the values

y3 = 1.175+ (0.1)(5)( 0.2 + 1.175)

y3 = 1.175+ 0.6875

y3 = 1.8625

Now we take

y3 = 1.8625, x3 = 0.3

y4 = y3 + hf(x3, y3)

Putting the values

y4 = 1.8625+ (0.1)(5)( 0.3 + 1.8625)

y4= 1.8625+ 1.08125

y4= 2.94375

we take now,

y4 = 2.94375, x4 = 0.4

y5 = y4 + hf(x4, y4)

Putting the values

y5 = 2.94375+ (0.1)(5)( 0.4+ 2.94375)

y5= 2.94375+ 0.305255

y5= 4.615625

y5 = 4.615625, x5 = 0.5

y6 = y5 + hf(x5, y5)

y6 = 4.615625+ (0.1)(5)( 0.5 + 4.615625)

y6 = 4.615625+ 2.5578125

y6 = 7.1734375

So, the value of   corresponding to x = 0.5 which is = 7.17

any one have the cracked link for this software

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