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How to solve a differential equation with a term to a power?
The Next CEO of Stack OverflowHow do I solve this differential equation?Solving differential equation $x^2y''-xy'+y=0, x>0$ with non-constant coefficients using characteristic equation?Solution to differential equationCan I solve an Euler differential equation by using the Frobenius method?How to solve a differential equationHelp needed with differential equationHow to solve differential equations that look like theseHow to solve differential equation with one differential termsolve differential equation $fracdPdt = kPcos^2(rt-Theta)$How to solve an exponential differential equation?
$begingroup$
How would I solve an equation where one of the differential terms is to a power? For example:
$fracd^2ydx^2+k(fracdydx)^2=0$?
I've been given advice to use the $D$ operator which apparently means $fracddx()$ 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:
$fracd^2ydx^2+k(fracdydx)^2=0$?
I've been given advice to use the $D$ operator which apparently means $fracddx()$ 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:
$fracd^2ydx^2+k(fracdydx)^2=0$?
I've been given advice to use the $D$ operator which apparently means $fracddx()$ 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:
$fracd^2ydx^2+k(fracdydx)^2=0$?
I've been given advice to use the $D$ operator which apparently means $fracddx()$ 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 $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.
answered 4 hours ago
mickepmickep
18.7k12250
$endgroup$
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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 $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.
answered 4 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 $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.
answered 4 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 $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.
answered 4 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 $v=fracdydx$. Your differential equation will turn into $fracdvdx+kv^2=0$, and I guess you will manage to take it from here.
answered 4 hours ago
mickepmickep
18.7k12250
answered 4 hours ago
mickepmickep
18.7k12250
answered 4 hours ago
mickepmickep
18.7k12250
answered 4 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.
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