Second order ODE, separation of variables ??












0












$begingroup$


I have an ODE of the form:



$frac{d^2y}{dx^2}=frac{f(y)}{g(x)}$



I understand how to separate variables and integrate if its first order, but it looks trickier if its second order, is there a general way to solve it ?










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  • 1




    $begingroup$
    $frac{d^2y}{dx^2}=frac{dy'}{dx}$, so if you have a function of $y'$ instead of $y$ on the $RHS$, you can solve using separation of variables
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:44












  • $begingroup$
    Yeah I know that, doesn't solve the problem and not the question I asked, but thank you anyway
    $endgroup$
    – ODE
    Dec 23 '18 at 18:55










  • $begingroup$
    Are you asking about how to solve $2^{nd}$ order $ODE$s using separation of variables or otherwise?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:58












  • $begingroup$
    I'm asking how to solve a second order ODE of that form, where we can separate the variables into two independent functions,
    $endgroup$
    – ODE
    Dec 23 '18 at 19:03
















0












$begingroup$


I have an ODE of the form:



$frac{d^2y}{dx^2}=frac{f(y)}{g(x)}$



I understand how to separate variables and integrate if its first order, but it looks trickier if its second order, is there a general way to solve it ?










share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    $frac{d^2y}{dx^2}=frac{dy'}{dx}$, so if you have a function of $y'$ instead of $y$ on the $RHS$, you can solve using separation of variables
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:44












  • $begingroup$
    Yeah I know that, doesn't solve the problem and not the question I asked, but thank you anyway
    $endgroup$
    – ODE
    Dec 23 '18 at 18:55










  • $begingroup$
    Are you asking about how to solve $2^{nd}$ order $ODE$s using separation of variables or otherwise?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:58












  • $begingroup$
    I'm asking how to solve a second order ODE of that form, where we can separate the variables into two independent functions,
    $endgroup$
    – ODE
    Dec 23 '18 at 19:03














0












0








0


0



$begingroup$


I have an ODE of the form:



$frac{d^2y}{dx^2}=frac{f(y)}{g(x)}$



I understand how to separate variables and integrate if its first order, but it looks trickier if its second order, is there a general way to solve it ?










share|cite|improve this question









$endgroup$




I have an ODE of the form:



$frac{d^2y}{dx^2}=frac{f(y)}{g(x)}$



I understand how to separate variables and integrate if its first order, but it looks trickier if its second order, is there a general way to solve it ?







differential






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 18:40









ODEODE

141




141








  • 1




    $begingroup$
    $frac{d^2y}{dx^2}=frac{dy'}{dx}$, so if you have a function of $y'$ instead of $y$ on the $RHS$, you can solve using separation of variables
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:44












  • $begingroup$
    Yeah I know that, doesn't solve the problem and not the question I asked, but thank you anyway
    $endgroup$
    – ODE
    Dec 23 '18 at 18:55










  • $begingroup$
    Are you asking about how to solve $2^{nd}$ order $ODE$s using separation of variables or otherwise?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:58












  • $begingroup$
    I'm asking how to solve a second order ODE of that form, where we can separate the variables into two independent functions,
    $endgroup$
    – ODE
    Dec 23 '18 at 19:03














  • 1




    $begingroup$
    $frac{d^2y}{dx^2}=frac{dy'}{dx}$, so if you have a function of $y'$ instead of $y$ on the $RHS$, you can solve using separation of variables
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:44












  • $begingroup$
    Yeah I know that, doesn't solve the problem and not the question I asked, but thank you anyway
    $endgroup$
    – ODE
    Dec 23 '18 at 18:55










  • $begingroup$
    Are you asking about how to solve $2^{nd}$ order $ODE$s using separation of variables or otherwise?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 18:58












  • $begingroup$
    I'm asking how to solve a second order ODE of that form, where we can separate the variables into two independent functions,
    $endgroup$
    – ODE
    Dec 23 '18 at 19:03








1




1




$begingroup$
$frac{d^2y}{dx^2}=frac{dy'}{dx}$, so if you have a function of $y'$ instead of $y$ on the $RHS$, you can solve using separation of variables
$endgroup$
– Shubham Johri
Dec 23 '18 at 18:44






$begingroup$
$frac{d^2y}{dx^2}=frac{dy'}{dx}$, so if you have a function of $y'$ instead of $y$ on the $RHS$, you can solve using separation of variables
$endgroup$
– Shubham Johri
Dec 23 '18 at 18:44














$begingroup$
Yeah I know that, doesn't solve the problem and not the question I asked, but thank you anyway
$endgroup$
– ODE
Dec 23 '18 at 18:55




$begingroup$
Yeah I know that, doesn't solve the problem and not the question I asked, but thank you anyway
$endgroup$
– ODE
Dec 23 '18 at 18:55












$begingroup$
Are you asking about how to solve $2^{nd}$ order $ODE$s using separation of variables or otherwise?
$endgroup$
– Shubham Johri
Dec 23 '18 at 18:58






$begingroup$
Are you asking about how to solve $2^{nd}$ order $ODE$s using separation of variables or otherwise?
$endgroup$
– Shubham Johri
Dec 23 '18 at 18:58














$begingroup$
I'm asking how to solve a second order ODE of that form, where we can separate the variables into two independent functions,
$endgroup$
– ODE
Dec 23 '18 at 19:03




$begingroup$
I'm asking how to solve a second order ODE of that form, where we can separate the variables into two independent functions,
$endgroup$
– ODE
Dec 23 '18 at 19:03










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