Clairaut's equation: Find the general solution of $x^2 (y-xy') =y(y') ^2$ if the singular solution doesn't...
Question:
Find the general solution of
$$x^2 (y-xy') =y(y') ^2$$
if the singular solution doesn't exist.
Now, I know that it has to be solved by Clairaut's equation.
However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.
I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.
Please let me know if you have any suggestions.
differential-equations
edited Nov 27 '18 at 2:38
Shaun
8,810113680
asked Nov 27 '18 at 2:11
MadCap
213
add a comment |
Question:
Find the general solution of
$$x^2 (y-xy') =y(y') ^2$$
if the singular solution doesn't exist.
Now, I know that it has to be solved by Clairaut's equation.
However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.
I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.
Please let me know if you have any suggestions.
differential-equations
edited Nov 27 '18 at 2:38
Shaun
8,810113680
asked Nov 27 '18 at 2:11
MadCap
213
add a comment |
Question:
Find the general solution of
$$x^2 (y-xy') =y(y') ^2$$
if the singular solution doesn't exist.
Now, I know that it has to be solved by Clairaut's equation.
However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.
I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.
Please let me know if you have any suggestions.
differential-equations
edited Nov 27 '18 at 2:38
Shaun
8,810113680
asked Nov 27 '18 at 2:11
MadCap
213
Question:
Find the general solution of
$$x^2 (y-xy') =y(y') ^2$$
if the singular solution doesn't exist.
Now, I know that it has to be solved by Clairaut's equation.
However, the given equation is not of the form $y=px+f(p)$ where $p=y'$ and cannot, as far as I know, be reduced to this form.
I even tried arranging the equation in the form of a quadratic in $y'$ but when I find the solution it yields a rather big equation that doesn't really lead anywhere.
Please let me know if you have any suggestions.
differential-equations
differential-equations
edited Nov 27 '18 at 2:38
Shaun
8,810113680
asked Nov 27 '18 at 2:11
MadCap
213
edited Nov 27 '18 at 2:38
Shaun
8,810113680
asked Nov 27 '18 at 2:11
MadCap
213
edited Nov 27 '18 at 2:38
Shaun
8,810113680
edited Nov 27 '18 at 2:38
Shaun
8,810113680
edited Nov 27 '18 at 2:38
Shaun
8,810113680
8,810113680
asked Nov 27 '18 at 2:11
MadCap
213
asked Nov 27 '18 at 2:11
MadCap
213
asked Nov 27 '18 at 2:11
MadCap
213
213
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
Hint
Rewrite the equation as
$$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
$$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
$$y^2 z'^2-z^2-4=0$$ that is to say
$$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
– MadCap
Nov 28 '18 at 0:27
@MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
– Claude Leibovici
Nov 28 '18 at 3:08
Oh now it's more clear.
– MadCap
Nov 28 '18 at 13:18
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Hint
Rewrite the equation as
$$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
$$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
$$y^2 z'^2-z^2-4=0$$ that is to say
$$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
– MadCap
Nov 28 '18 at 0:27
@MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
– Claude Leibovici
Nov 28 '18 at 3:08
Oh now it's more clear.
– MadCap
Nov 28 '18 at 13:18
add a comment |
Hint
Rewrite the equation as
$$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
$$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
$$y^2 z'^2-z^2-4=0$$ that is to say
$$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
– MadCap
Nov 28 '18 at 0:27
@MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
– Claude Leibovici
Nov 28 '18 at 3:08
Oh now it's more clear.
– MadCap
Nov 28 '18 at 13:18
add a comment |
Hint
Rewrite the equation as
$$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
$$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
$$y^2 z'^2-z^2-4=0$$ that is to say
$$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
Hint
Rewrite the equation as
$$x^2 left(y-frac{x}{x'}right)-frac{y}{x'^2}=0$$ Now, let $x=sqrt{y z}$ which makes
$$frac{y^2 z left(y^2 z'^2-z^2-4right)}{left(y z'+zright)^2}=0$$ Excluding the trivial solution, we are left with
$$y^2 z'^2-z^2-4=0$$ that is to say
$$frac{z'^2}{z^2+4}=frac 1 {y^2}implies frac{z'}{sqrt{z^2+4}}=pm frac 1y$$ which seems to be simple to integrate.
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
answered Nov 27 '18 at 3:49
Claude Leibovici
119k1157132
119k1157132
Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
– MadCap
Nov 28 '18 at 0:27
@MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
– Claude Leibovici
Nov 28 '18 at 3:08
Oh now it's more clear.
– MadCap
Nov 28 '18 at 13:18
add a comment |
Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
– MadCap
Nov 28 '18 at 0:27
@MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
– Claude Leibovici
Nov 28 '18 at 3:08
Oh now it's more clear.
– MadCap
Nov 28 '18 at 13:18
Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
– MadCap
Nov 28 '18 at 0:27
Shouldn't the denominator be yz'+zy' when you substitute for x or am I missing something here?
– MadCap
Nov 28 '18 at 0:27
@MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
– Claude Leibovici
Nov 28 '18 at 3:08
@MadCap. Since $y$ became the variable, $y'=1$. I double check and all of this seems to be correct.
– Claude Leibovici
Nov 28 '18 at 3:08
Oh now it's more clear.
– MadCap
Nov 28 '18 at 13:18
Oh now it's more clear.
– MadCap
Nov 28 '18 at 13:18
add a comment |
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