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 positive $1$ 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&#039; and integrate once with respect to x to obtain <br /> \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&#039; and integrate once with respect to x to obtain <br /> \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.