Simplification of nonlinear differential equation
$begingroup$
Given this nonlinear differential equation
$$
[(1+epsilon y) y']'+ 2xy'= 0, 0 < x < infty,
$$
why we can write it like the following?
$$
[(1+epsilon y) (1+ epsilon y)' ]' + 2x(1+epsilon y)'= 0 , 0 < x < infty
$$
calculus ordinary-differential-equations
edited Dec 1 '18 at 17:57
rafa11111
1,119417
asked Dec 1 '18 at 17:48
U22U22
12
$endgroup$
add a comment |
$begingroup$
Given this nonlinear differential equation
$$
[(1+epsilon y) y']'+ 2xy'= 0, 0 < x < infty,
$$
why we can write it like the following?
$$
[(1+epsilon y) (1+ epsilon y)' ]' + 2x(1+epsilon y)'= 0 , 0 < x < infty
$$
calculus ordinary-differential-equations
edited Dec 1 '18 at 17:57
rafa11111
1,119417
asked Dec 1 '18 at 17:48
U22U22
12
$endgroup$
add a comment |
$begingroup$
Given this nonlinear differential equation
$$
[(1+epsilon y) y']'+ 2xy'= 0, 0 < x < infty,
$$
why we can write it like the following?
$$
[(1+epsilon y) (1+ epsilon y)' ]' + 2x(1+epsilon y)'= 0 , 0 < x < infty
$$
calculus ordinary-differential-equations
edited Dec 1 '18 at 17:57
rafa11111
1,119417
asked Dec 1 '18 at 17:48
U22U22
12
$endgroup$
Given this nonlinear differential equation
$$
[(1+epsilon y) y']'+ 2xy'= 0, 0 < x < infty,
$$
why we can write it like the following?
$$
[(1+epsilon y) (1+ epsilon y)' ]' + 2x(1+epsilon y)'= 0 , 0 < x < infty
$$
calculus ordinary-differential-equations
calculus ordinary-differential-equations
edited Dec 1 '18 at 17:57
rafa11111
1,119417
asked Dec 1 '18 at 17:48
U22U22
12
edited Dec 1 '18 at 17:57
rafa11111
1,119417
asked Dec 1 '18 at 17:48
U22U22
12
edited Dec 1 '18 at 17:57
rafa11111
1,119417
edited Dec 1 '18 at 17:57
rafa11111
1,119417
edited Dec 1 '18 at 17:57
rafa11111
1,119417
1,119417
asked Dec 1 '18 at 17:48
U22U22
12
asked Dec 1 '18 at 17:48
U22U22
12
asked Dec 1 '18 at 17:48
U22U22
12
12
add a comment |
add a comment |
1 Answer
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votes
$begingroup$
Note that $(1+epsilon y)'=epsilon y'$. So if $epsilon neq 0$, we can cancel out $epsilon$ from second equation to get back the first one.
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
add a comment |
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$begingroup$
Note that $(1+epsilon y)'=epsilon y'$. So if $epsilon neq 0$, we can cancel out $epsilon$ from second equation to get back the first one.
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
add a comment |
$begingroup$
Note that $(1+epsilon y)'=epsilon y'$. So if $epsilon neq 0$, we can cancel out $epsilon$ from second equation to get back the first one.
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
add a comment |
$begingroup$
Note that $(1+epsilon y)'=epsilon y'$. So if $epsilon neq 0$, we can cancel out $epsilon$ from second equation to get back the first one.
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
$endgroup$
Note that $(1+epsilon y)'=epsilon y'$. So if $epsilon neq 0$, we can cancel out $epsilon$ from second equation to get back the first one.
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
answered Dec 1 '18 at 17:57
Thomas ShelbyThomas Shelby
2,119220
2,119220
add a comment |
add a comment |
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