It is instructive to see how picard’s method works with a choice of the initial approximation other than the constant function y0(x) = y0. apply the method to the initial value …

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asked Feb 28, 2024 226k views

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Patrick Yoder · Answer 1 · 2024-03-06T05:14:33+0000

The initial value problem of equivalent integral equation is:

1) Given initial value problem is:

(dy)/(dx) =x+y

y(x) =
e^x , y = 1

The equivalent integral equation is,

$y = y_o + \int\limits^x_0 {(s+e^s)} \, dx \\$

Then by pieard's method,

$y = 1 + \int\limits^x_0 {(s+e^s)} \, dx \\= 1+\int\limits^0_xsds+\int\limits^x_0 {e^s} \, dx$

= 1+(x^2)/(2) +e^x

y(x) =
e^(x^2) -1

y(x) = x²

2) The given initial value problem is,

(dy)/(dx) =x+y

y(x) = 1+x

The equivalent integral equation is,

$y = y_o + \int\limits^x_0 {(s+e^s)} \, dx \\$

Then by pieard's method,

$y = 1 + \int\limits^x_0 {(s+1+s)} \, dx \\\\= 1+\int\limits^0_x {(1+2s)} \, dx \\= 1+[s]^x_0+2[(s^2)/(2) ]^x_0\\=1+x+x^2$

y(x) = 1+x+2[
e^x-x-1 ]

3) The given initial value problem is,

(dy)/(dx) =cosx

y(x) = cosx

The equivalent integral equation is,

$y = y_o + \int\limits^x_0 {(s+e^s)} \, dx \\$

Then by pieard's method,

$y = 1 + \int\limits^x_0 {(s+cos s)} \, dx \\\\= 1+(x^2)/(2)+sinx \\ = (sinx-x)+1+x+(x^2)/(2)$

y = -sinx - x +
(x^3)/(3!) + 1 +x + x² + x³/3! + x⁴/3!

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