Solve $ y'' + \lambda y^2 = 0$

Question

Solve $ y'' + \lambda y^2 = 0$

Attempt 1 : if $\lambda =0 $ .then it's trivial to solve.

If $ \lambda <0 $ ,then $y '' \ge 0$

In particular when $y '' > 0 $ for some interval . Let $ y''= e^{u(x)}$

Then $y(x) =\frac {e^{\frac{u(x)}2}}{\sqrt{-\lambda}} $

Again differentiating this and substituting back leads to another non-linear differential equation .

Attempt 2 : Let $y' = u$

Then $\frac{du}{dx} = \frac{d^2y}{dx^2} = -\lambda y^2 $

or , $ \frac{du}{dy} \frac{dy}{dx} = -\lambda y^2 $

or , $ u du= -\lambda y^2 dy $

Integrating both sides,

$\frac{u^2}2 = -\lambda \frac{y^3}3 +c/2 $

or $u= \sqrt{-\frac 23 \lambda y^3 +c }$

or , $ \frac {dy}{ \sqrt{-\frac 23 \lambda y^3 +c} }=dx$ , (For appropriate values of $\lambda$ and c)

The left integral is definitely not simple looking . How may I proceed ? (I don't think we have to take the help of numerical methods)

Answer 1

Nonlinear differential equations rarely have closed form solutions in terms of "elementary" functions, depending on what you choose to call elementary of course. Even the logarithm was invented at one stage because someone needed a function to express the answer to the question "to what power should I raise 10 in order to get this number". Before the name logarithm was invented, that question was not solvable in terms of elementary function.

Your case is actually a lucky one. Because this type of integral occurs so often in practical problems, the resulting functions have received names.

Integrals of the square root of a cubic polynomial

Answer 2

$$2y'y''+2\lambda y^2y'=0$$ leads you directly to$$y'^2+\frac23\lambda y^3=C=(y'_0)^2+\frac23\lambda y_0^3$$ and

$$\frac{y'}{\sqrt{\frac23\lambda(y_0^3-y^3)+(y'_0)^2}}=\pm 1.$$

The three roots of the denominator are

$$y_k=\sqrt[3]{\frac3{2\lambda}(y'_0)^2+y_0^3}\,\omega^k$$ where $\omega$ is a complex cube root of unity. You will need an elliptic integral.