Differential equation $y''=y^2$

Question

I must solve the differential equation $y''=y^2$. Clearly, the function $y=0$ is a solution. So, assume that $y$ is not identically zero. Unless I'm not mistaken, there is a trick to solve equations of this kind. If I multiply for $2y'$, the left-hand member is $2y'y''$, that is the derivative of $(y')^2$, while the right-hand member is $2y^2 y'$. At this point, how can I continue this exercise?

Answer 1

$$y''=y^2$$ $$2y''y'=2y^2y' \quad\to\quad (y')^2=\frac23 y^3+c_1 \quad\to\quad y'=\pm\sqrt{\frac23 y^3+c_1}$$ $$\pm\frac{dy}{\sqrt{\frac23 y^3+c_1}}=dt \quad\to\quad t+c_2=\pm\int\frac{dy}{\sqrt{\frac23 y^3+c_1}}$$ This integral is of the elliptic kind and can be expressed on closed form thanks to elliptic function : http://mathworld.wolfram.com/EllipticIntegral.html

The function $t(y)$ is rather ugly and moreover have to be inverted in order to obtain $y(t)$. This is far to be the simplest way.

As a matter of fact, the ODE : $$(y')^2=\frac23 y^3+c_1$$ is a case of Weierstrass equation, which solutions are the Weierstrass functions $\wp$. http://mathworld.wolfram.com/WeierstrassEllipticFunction.html

The general form of Weierstass equation is : $$(\wp'(x))^2=4(\wp(x))^3-g_2\wp(x)-g_3$$ In the present case, with $\quad y=\sqrt[3]6\:\wp \quad\;\quad x=\frac{t}{\sqrt[3]6}\quad\;\quad g_2=0\quad\;\quad g_3=-c_1 \quad$ , leading to : $$y(t)=\sqrt[3]6\: \wp\left(\frac{t+C_2}{\sqrt[3]6} \:;\: 0 \:,\:C_1\right)$$ where $C_1$ and $C_2$ are arbitrary constant.

Answer 2

HINT. $$2y'y''=2y'y^2$$ $$(y')^2=\frac23(y^3)'$$ now let $y^3=z$.

Answer 3

Write

$$y'y''=y^2y'$$ and integrate:

$$\frac{y'^2}2=\frac{y^3}3+C,$$ which is separable. Then

$$\frac{y'}{\sqrt{y^3+C}}=\pm\sqrt{\frac23}.$$

Now, unless $C=0$, the left integral is terrible.

Answer 4

Hint

Use $p=\frac {dy}{dx}$ and $p'=\frac {dp}{dy}$ for this kind of ode...

Then solve $$pp'=y^2$$ $$\int pdp=\int y^2dy$$ Or just consider: $$\frac {dy'}{dx}=y^2$$ $$\frac {dy'}{dx}\frac {dy}{dy}=y^2$$ $$\int y' {dy'}=\int y^2dy=\frac {y^3} 3+K$$

Answer 5

Another hint could be:

$y''=y^2$

assume exists (at least one) solution $y$ and then set $y=z+bx+c$ since $bx+c$ dies with double differentiation $$z''=(z+bx+c)^2 \\z''= z^2+2z(bx+c)+(bx+c)^2$$

Now the $z''=z^2$ we recognize but what about the rest?

Answer 6

We must solve $y'' = y^2$. Assuming the independent variable is $x$, let us write this as:

$\frac{d^2y}{dx^2} = y^2$. Then, using the chain rule, we can write the left hand side as:

$\frac{d^2y}{dx^2} = \frac{d}{dx} \left(\frac{dy}{dx}\right) = \frac{d}{dy} \left(\frac{dy}{dx}\right) \frac{dy}{dx}$.

But, this is equivalent to:

$\frac{d}{dy} \left[ \left(\frac{dx}{dy}\right)^{-1}\right] \left( \frac{dx}{dy}\right)^{-1}$.

Computing this derivative, we get:

$-\left(\frac{dx}{dy}\right)^{-2} \frac{d^2 x}{dy^2} \left( \frac{dx}{dy}\right)^{-1} = -\left(\frac{dx}{dy}\right)^{-3} \frac{d^2x}{dy^2}$.

But, this is simply:

$\frac{d}{dy} \left[ \frac{1}{2} \left( \frac{dx}{dy}\right)^{-2}\right]$.

Now, we go back to your original ODE:

$\frac{d^2 y}{dx^2} = y^2$ we can write as:

$\frac{d}{dy} \left[ \frac{1}{2} \left( \frac{dx}{dy}\right)^{-2}\right] = y^2$

Integrating both sides wrto $y$, we get:

$\frac{1}{2} \left( \frac{dx}{dy}\right)^{-2} = 2 \int y^2 dy + A$, where $A$ is just a constant.

Solving for $dx/dy$, we get that:

$\boxed{x + B = \pm \int \frac{dy}{\sqrt{2 \int y^2 dy + A}}}$.

The right side is a very messy integral, you get some elliptic functions, in fact, Mathematica gives that the right-hand-side is: $\frac{\sqrt[6]{-1} 2^{2/3} \sqrt[12]{3} \sqrt[3]{A} \sqrt{(-1)^{5/6} \left(\frac{\sqrt[3]{-\frac{2}{3}} y}{\sqrt[3]{A}}-1\right)} \sqrt{\frac{\left(-\frac{2}{3}\right)^{2/3} y^2}{A^{2/3}}+\frac{\sqrt[3]{-\frac{2}{3}} y}{\sqrt[3]{A}}+1} F\left(\sin ^{-1}\left(\frac{\sqrt{-\frac{i y \sqrt[3]{-\frac{2}{3}}}{\sqrt[3]{A}}-(-1)^{5/6}}}{\sqrt[4]{3}}\right)|\sqrt[3]{-1}\right)}{\sqrt{A+\frac{2 y^3}{3}}}$,

where $F$ denotes the elliptic integral of the first kind.

In general, the boxed equation is the general implicit solution to this problem. These types of ODEs occur all the time in classical mechanics, and most often are solved numerically or using dynamical systems techniques.

Hope this is helpful!