Solving a non-linear differential equation of second degree
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How do I solve the differential equation $y^2= 2px+ p^2 $, $ $ $p$ $ $ being $frac{dy}{dx}$? I came up with this equation accidentally while trying to copy an exercise from the book. The actual problem is fine, but this equation is troubling me for a while. I tried reducing it to Clairaut's form to no avail. I tried solving for $y$ or $x$, but failed. Kindly help me out.
differential-equations derivatives
asked Nov 22 at 17:19
Subhasis Biswas
401211
|
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up vote
1
down vote
favorite
How do I solve the differential equation $y^2= 2px+ p^2 $, $ $ $p$ $ $ being $frac{dy}{dx}$? I came up with this equation accidentally while trying to copy an exercise from the book. The actual problem is fine, but this equation is troubling me for a while. I tried reducing it to Clairaut's form to no avail. I tried solving for $y$ or $x$, but failed. Kindly help me out.
differential-equations derivatives
asked Nov 22 at 17:19
Subhasis Biswas
401211
1
You can solve for $p$ as $$p = dfrac{dy}{dx} = -x~pm~ sqrt{x^2+y^2}$$ However, I am not sure you can solve the resulting differential equation.
– Moo
Nov 22 at 17:31
I am totally lost from here. No idea. In fact, I tried fiddling around this format before.
– Subhasis Biswas
Nov 22 at 17:33
You would to resort to numerical methods at this point.
– Moo
Nov 22 at 17:33
1
@Isham: Oops - just a typo - corrected - thanks.
– Moo
Nov 22 at 17:36
1
@Isham It should be a $-$
– Subhasis Biswas
Nov 22 at 17:36
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How do I solve the differential equation $y^2= 2px+ p^2 $, $ $ $p$ $ $ being $frac{dy}{dx}$? I came up with this equation accidentally while trying to copy an exercise from the book. The actual problem is fine, but this equation is troubling me for a while. I tried reducing it to Clairaut's form to no avail. I tried solving for $y$ or $x$, but failed. Kindly help me out.
differential-equations derivatives
asked Nov 22 at 17:19
Subhasis Biswas
401211
How do I solve the differential equation $y^2= 2px+ p^2 $, $ $ $p$ $ $ being $frac{dy}{dx}$? I came up with this equation accidentally while trying to copy an exercise from the book. The actual problem is fine, but this equation is troubling me for a while. I tried reducing it to Clairaut's form to no avail. I tried solving for $y$ or $x$, but failed. Kindly help me out.
differential-equations derivatives
differential-equations derivatives
asked Nov 22 at 17:19
Subhasis Biswas
401211
asked Nov 22 at 17:19
Subhasis Biswas
401211
edited Nov 22 at 17:27
asked Nov 22 at 17:19
Subhasis Biswas
401211
asked Nov 22 at 17:19
Subhasis Biswas
401211
asked Nov 22 at 17:19
Subhasis Biswas
401211
401211
1
You can solve for $p$ as $$p = dfrac{dy}{dx} = -x~pm~ sqrt{x^2+y^2}$$ However, I am not sure you can solve the resulting differential equation.
– Moo
Nov 22 at 17:31
I am totally lost from here. No idea. In fact, I tried fiddling around this format before.
– Subhasis Biswas
Nov 22 at 17:33
You would to resort to numerical methods at this point.
– Moo
Nov 22 at 17:33
1
@Isham: Oops - just a typo - corrected - thanks.
– Moo
Nov 22 at 17:36
1
@Isham It should be a $-$
– Subhasis Biswas
Nov 22 at 17:36
|
show 3 more comments
1
You can solve for $p$ as $$p = dfrac{dy}{dx} = -x~pm~ sqrt{x^2+y^2}$$ However, I am not sure you can solve the resulting differential equation.
– Moo
Nov 22 at 17:31
I am totally lost from here. No idea. In fact, I tried fiddling around this format before.
– Subhasis Biswas
Nov 22 at 17:33
You would to resort to numerical methods at this point.
– Moo
Nov 22 at 17:33
1
@Isham: Oops - just a typo - corrected - thanks.
– Moo
Nov 22 at 17:36
1
@Isham It should be a $-$
– Subhasis Biswas
Nov 22 at 17:36
1
1
You can solve for $p$ as $$p = dfrac{dy}{dx} = -x~pm~ sqrt{x^2+y^2}$$ However, I am not sure you can solve the resulting differential equation.
– Moo
Nov 22 at 17:31
You can solve for $p$ as $$p = dfrac{dy}{dx} = -x~pm~ sqrt{x^2+y^2}$$ However, I am not sure you can solve the resulting differential equation.
– Moo
Nov 22 at 17:31
I am totally lost from here. No idea. In fact, I tried fiddling around this format before.
– Subhasis Biswas
Nov 22 at 17:33
I am totally lost from here. No idea. In fact, I tried fiddling around this format before.
– Subhasis Biswas
Nov 22 at 17:33
You would to resort to numerical methods at this point.
– Moo
Nov 22 at 17:33
You would to resort to numerical methods at this point.
– Moo
Nov 22 at 17:33
1
1
@Isham: Oops - just a typo - corrected - thanks.
– Moo
Nov 22 at 17:36
@Isham: Oops - just a typo - corrected - thanks.
– Moo
Nov 22 at 17:36
1
1
@Isham It should be a $-$
– Subhasis Biswas
Nov 22 at 17:36
@Isham It should be a $-$
– Subhasis Biswas
Nov 22 at 17:36
|
show 3 more comments
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1
You can solve for $p$ as $$p = dfrac{dy}{dx} = -x~pm~ sqrt{x^2+y^2}$$ However, I am not sure you can solve the resulting differential equation.
– Moo
Nov 22 at 17:31
I am totally lost from here. No idea. In fact, I tried fiddling around this format before.
– Subhasis Biswas
Nov 22 at 17:33
You would to resort to numerical methods at this point.
– Moo
Nov 22 at 17:33
1
@Isham: Oops - just a typo - corrected - thanks.
– Moo
Nov 22 at 17:36
1
@Isham It should be a $-$
– Subhasis Biswas
Nov 22 at 17:36