How can i solve this Cauchy-Euler equation?
My problem is this given Cauchy-Euler equation:
$$x^{3}y^{prime prime prime} +xy^{prime}-y=0$$
My approach: this is an differential equation, so i was looking for a solution with the method of undetermined coefficients. but honestly, i failed.
differential-equations integration
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
add a comment |
My problem is this given Cauchy-Euler equation:
$$x^{3}y^{prime prime prime} +xy^{prime}-y=0$$
My approach: this is an differential equation, so i was looking for a solution with the method of undetermined coefficients. but honestly, i failed.
differential-equations integration
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
add a comment |
My problem is this given Cauchy-Euler equation:
$$x^{3}y^{prime prime prime} +xy^{prime}-y=0$$
My approach: this is an differential equation, so i was looking for a solution with the method of undetermined coefficients. but honestly, i failed.
differential-equations integration
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
My problem is this given Cauchy-Euler equation:
$$x^{3}y^{prime prime prime} +xy^{prime}-y=0$$
My approach: this is an differential equation, so i was looking for a solution with the method of undetermined coefficients. but honestly, i failed.
differential-equations integration
differential-equations integration
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
edited Nov 10 '17 at 20:12
Jsevillamol
3,03511124
3,03511124
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
asked Jun 15 '13 at 15:39
Toralf Westström
44711020
44711020
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
Besides to another answer, you may substitute $y=x^m$ wherein $m$ is a number and then form the certain $y_c(x)$. Let's do that: $$y=x^mto y'=mx^{m-1},~~y'''=m(m-1)(m-2)x^{m-3}$$ So $$x^3y'''+xy'-y=0Longrightarrow x^m(m(m-1)(m-2)+m-1)=0$$ If $xneq 0$ then $$(m-1)^3=0$$ This means that $$y_c(x)=x^1(1+ln x+ln^2 x), ~x>0$$
answered Jun 15 '13 at 16:30
mrs
1
1
$quad langle +rangle_+^+quad bf ddotsmile;$
– amWhy
Jun 16 '13 at 1:07
add a comment |
Hint: Use substitute $t=log x$ for $x>0$ and $t=log (-x)$ for $x<0$
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
Besides to another answer, you may substitute $y=x^m$ wherein $m$ is a number and then form the certain $y_c(x)$. Let's do that: $$y=x^mto y'=mx^{m-1},~~y'''=m(m-1)(m-2)x^{m-3}$$ So $$x^3y'''+xy'-y=0Longrightarrow x^m(m(m-1)(m-2)+m-1)=0$$ If $xneq 0$ then $$(m-1)^3=0$$ This means that $$y_c(x)=x^1(1+ln x+ln^2 x), ~x>0$$
answered Jun 15 '13 at 16:30
mrs
1
1
$quad langle +rangle_+^+quad bf ddotsmile;$
– amWhy
Jun 16 '13 at 1:07
add a comment |
Besides to another answer, you may substitute $y=x^m$ wherein $m$ is a number and then form the certain $y_c(x)$. Let's do that: $$y=x^mto y'=mx^{m-1},~~y'''=m(m-1)(m-2)x^{m-3}$$ So $$x^3y'''+xy'-y=0Longrightarrow x^m(m(m-1)(m-2)+m-1)=0$$ If $xneq 0$ then $$(m-1)^3=0$$ This means that $$y_c(x)=x^1(1+ln x+ln^2 x), ~x>0$$
answered Jun 15 '13 at 16:30
mrs
1
1
$quad langle +rangle_+^+quad bf ddotsmile;$
– amWhy
Jun 16 '13 at 1:07
add a comment |
Besides to another answer, you may substitute $y=x^m$ wherein $m$ is a number and then form the certain $y_c(x)$. Let's do that: $$y=x^mto y'=mx^{m-1},~~y'''=m(m-1)(m-2)x^{m-3}$$ So $$x^3y'''+xy'-y=0Longrightarrow x^m(m(m-1)(m-2)+m-1)=0$$ If $xneq 0$ then $$(m-1)^3=0$$ This means that $$y_c(x)=x^1(1+ln x+ln^2 x), ~x>0$$
answered Jun 15 '13 at 16:30
mrs
1
Besides to another answer, you may substitute $y=x^m$ wherein $m$ is a number and then form the certain $y_c(x)$. Let's do that: $$y=x^mto y'=mx^{m-1},~~y'''=m(m-1)(m-2)x^{m-3}$$ So $$x^3y'''+xy'-y=0Longrightarrow x^m(m(m-1)(m-2)+m-1)=0$$ If $xneq 0$ then $$(m-1)^3=0$$ This means that $$y_c(x)=x^1(1+ln x+ln^2 x), ~x>0$$
answered Jun 15 '13 at 16:30
mrs
1
edited Jun 15 '13 at 17:49
answered Jun 15 '13 at 16:30
mrs
1
answered Jun 15 '13 at 16:30
mrs
1
answered Jun 15 '13 at 16:30
mrs
1
1
1
$quad langle +rangle_+^+quad bf ddotsmile;$
– amWhy
Jun 16 '13 at 1:07
add a comment |
1
$quad langle +rangle_+^+quad bf ddotsmile;$
– amWhy
Jun 16 '13 at 1:07
1
1
$quad langle +rangle_+^+quad bf ddotsmile;$
– amWhy
Jun 16 '13 at 1:07
$quad langle +rangle_+^+quad bf ddotsmile;$
– amWhy
Jun 16 '13 at 1:07
add a comment |
Hint: Use substitute $t=log x$ for $x>0$ and $t=log (-x)$ for $x<0$
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
add a comment |
Hint: Use substitute $t=log x$ for $x>0$ and $t=log (-x)$ for $x<0$
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
add a comment |
Hint: Use substitute $t=log x$ for $x>0$ and $t=log (-x)$ for $x<0$
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
Hint: Use substitute $t=log x$ for $x>0$ and $t=log (-x)$ for $x<0$
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
answered Jun 15 '13 at 16:02
Cortizol
2,6911235
2,6911235
add a comment |
add a comment |
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