How to integral $z^{\prime\prime}+ 2 \eta z^\prime=0$?

Question

From the given ODE, $$z^{\prime\prime} + 2 \eta z^\prime = 0 $$ The rest procedure is as below $$\int \frac{z^{\prime\prime}}{z^\prime} d \eta = \int -2 \eta \, d \eta$$ $$ \ln z^\prime = - \eta ^2 + c_0 $$ $$ z^\prime = \frac{d z}{d \eta} = C_1 \exp (- \eta ^2 )$$ $$ z(\eta) = c_1 \int _0 ^{\eta} {\exp (-x^2) } \, {d x } + c_2$$

I'd like to know how the fraction of derivatives ($z^{\prime\prime}/z^\prime$) is integrated as $\ln z^\prime$.

There are similar equations but there is not exactly the same one.

Answer 1

You have to take into account that $z'=\frac{dz}{d\eta}$ and $\frac{d\log z'}{d\eta}=\frac{1}{z'}\frac{dz'}{d\eta}=\frac{z''}{z'}$. The rest is just an application of the fundamental theorem of calculus.

Answer 2

$$I=\int \dfrac {z''}{z'}d\eta$$ Since $z''=\dfrac {dz'}{d \eta}$: $$I=\int \dfrac {dz'}{d \eta} \dfrac {d\eta}{z'}$$ Therefore: $$I=\int \dfrac {dz'}{z'} =C+\ln |z'| $$

---

You can also substitute $u=z'$ then the DE is a first linear DE: $$z'' + 2 \eta z'=0$$ $$u' + 2 \eta u=0$$ This is separable. It leads to the same result. $$\int \dfrac {du}u =- 2 \int \eta d\eta$$

Answer 3

It's a funny thing. You have to treat the second derivative of $z$ as a fraction.

$\int \frac{z''}{z'} d\eta =\int \frac{1}{z'} \cdot \frac{dz'}{d\eta } d\eta $

As you can see, the differential terms $d\eta$ cancel each other out.

Use the identity

$ \begin{array}{l} \int \frac{1}{x} dx=ln( x) +C\\ \\ \int \frac{z''}{z'} d\eta =\int \frac{1}{z'} \cdot \frac{dz'}{d\eta } d\eta =\int \frac{1}{z'} dz'=ln( z') +C \end{array}$

Originally, the derivative was meant to be a fraction. It still works as a fraction, but its meaning has evolved to be more of a notation. It's a similar story for the integral notation.