Second derivative differential equations in terms of y?

[Saracen Rue]

Firstly I know how to do this with first derivatives in differential equations - for example say we had $\frac{dy}{dx}=4y^2-y$, and we're also told that $y=1$ when $x=0$.

$\frac{dy}{dx}=4y^2-y$
$\frac{dx}{dy}=\frac{1}{4y^2-y}=\frac{1}{y\left(4y-1\right)}=\frac{4}{4y-1}-\frac{1}{y}$
$\int _{ }^{ }\frac{dx}{dy}dy=\int _{ }^{ }\frac{4}{4y-1}dy-\int _{ }^{ }\frac{1}{y}dy$
$x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c$

Substituting in $y=1$ and $x=0$, we can find $c$:
$x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)+c$
$0=\ln \left(3\right)+c,\ c=-\ln \left(3\right)$
$x=\ln \left(\left|4y-1\right|\right)-\ln \left(\left|y\right|\right)-\ln \left(3\right)$
$x=\ln \left(\frac{\left|4y-1\right|}{3\left|y\right|}\right)$
$3e^x=\left|4-\frac{1}{y}\right|$
$4-\frac{1}{y}=-3e^x,4-\frac{1}{y}=3e^x$

Because we know that $y$ has to equal postive $1$ when $x$ equals $0$;
$4-1=-3e^0,\ 4-1=3e^0$
$3=-3,\ 3=3$
As the first statement is false, the second one must be true.
$∴4-\frac{1}{y}=3e^x$
$\frac{1}{y}=4-3e^x$
$y=\frac{1}{4-3e^x}$.

So as you can see it's relatively simple to solve a differential equation of the form $\frac{dy}{dx}=f\left(y\right)$, but I'm wondering if it would be possible to solve one in the form of $\frac{d^2y}{dx^2}=f\left(y\right)$. I've tried looking it up online but I haven't been able to find anything useful.

 

[fresh_42]

So as you can see it's relatively simple to solve a differential equation of the form $\frac{dy}{dx}=f\left(y\right)$, but I'm wondering if it would be possible to solve one in the form of $\frac{d^2y}{dx^2}=f\left(y\right)$. I've tried looking it up online but I haven't been able to find anything useful.

One can introduce $z:=\frac{dy}{dx}=f\left(x,y\right)$ and get a system of two differential equations of order one. The other equation is $\frac{dz}{dx}=f\left(x,y,z\right)$. This process can be iterated to higher orders. You can look for ODE (ordinary differential equations) to solve such systems.

 

[Mark44]

So as you can see it's relatively simple to solve a differential equation of the form $\frac{dy}{dx}=f\left(y\right)$,

It's not necessarily that simple -- you have to be able to integrate $\int \frac{dy}{f(y)}$.

 

[pasmith]

So as you can see it's relatively simple to solve a differential equation of the form $\frac{dy}{dx}=f\left(y\right)$, but I'm wondering if it would be possible to solve one in the form of $\frac{d^2y}{dx^2}=f\left(y\right)$. I've tried looking it up online but I haven't been able to find anything useful.

Multiply both sides by $y'$ and integrate once with respect to $x$ to obtain $$\left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.$$ Now just determine the constant of integration from the initial conditions, and work out whether you need the positive root or the negative root, and the problem is reduced to a first-order separable ODE.

 

[Saracen Rue]

Multiply both sides by $y'$ and integrate once with respect to $x$ to obtain $$\left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.$$ Now just determine the constant of integration from the initial conditions, and work out whether you need the positive root or the negative root, and the problem is reduced to a first-order separable ODE.

Thank you for you're help but I still find myself a little confused about how you arrived at $\left(\frac{dy}{dx}\right)^2 = 2\int f(y)\,dy.$

I tried multiplying both sides by $\frac{dy}{dx}$ and then integrating both sides with respect to $x$ and this is what I got;
$$\frac{d^2y}{dx^2}=f\left(y\right)$$
$$\frac{dy}{dx}\cdot \frac{d}{dx}\cdot \frac{dy}{dx}=f\left(y\right)\cdot \frac{dy}{dx}$$
$$\left(\frac{dy}{dx}\right)^2\cdot \frac{d}{dx}=f\left(y\right)\cdot \frac{dy}{dx}$$
$$\int _{ }^{ }\left(\frac{dy}{dx}\right)^2\cdot \frac{d}{dx}\cdot dx=\int _{ }^{ }f\left(y\right)\cdot \frac{dy}{dx}\cdot dx$$
$$\int _{ }^{ }\left(\frac{dy}{dx}\right)^2\cdot d=\int _{ }^{ }f\left(y\right)\cdot dy$$

I am unsure of what to do on the left hand side - how am I supposed to integrate without a variable for which to integrate with respect to. I'm also not sure why you have a two out the front of the integral on the right hand side, but I'm guessing that the answer to my first question will probably explain that as well.