How to solve a differential equation with a term to a power? The Next CEO of Stack OverflowHow...
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How to solve a differential equation with a term to a power?
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$begingroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$frac{d^2y}{dx^2}+k(frac{dy}{dx})^2=0$?
I've been given advice to use the $D$ operator which apparently means $frac{d}{dx}()$ but I'm not sure how that's applicable to this scenario. Any alternative suggestions or explanations would be appreciated!
calculus ordinary-differential-equations
calculus ordinary-differential-equations
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
asked 4 hours ago
Ammar TarajiaAmmar Tarajia
111
111
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ammar Tarajia is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
1 Answer
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$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered 3 hours ago
mickepmickep
18.7k12250
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1 Answer
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1 Answer
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$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered 3 hours ago
mickepmickep
18.7k12250
$endgroup$
add a comment |
$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered 3 hours ago
mickepmickep
18.7k12250
$endgroup$
add a comment |
$begingroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered 3 hours ago
mickepmickep
18.7k12250
$endgroup$
Since you only have second and first derivatives of $y$ and no
un-differentiated $y$, you could try to introduce the new function $v=frac{dy}{dx}$. Your differential equation will turn into $frac{dv}{dx}+kv^2=0$, and I guess you will manage to take it from here.
answered 3 hours ago
mickepmickep
18.7k12250
answered 3 hours ago
mickepmickep
18.7k12250
answered 3 hours ago
mickepmickep
18.7k12250
answered 3 hours ago
mickepmickep
18.7k12250
18.7k12250
add a comment |
add a comment |
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
Ammar Tarajia is a new contributor. Be nice, and check out our Code of Conduct.
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