What is $\frac{d}{dx}$ in the differential equation $\left(\frac{\ d}{dx}+2x\right)·\left(\frac{\ dy}{dx}+2xy\right)= 2e^{-x^2}$

Question

$$\left(\frac{\ d}{dx}+2x\right)·\left(\frac{\ dy}{dx}+2xy\right)= 2e^{-x^2}$$

When ı solve this differential equation, ı multiply both sides with $y$ then $\frac{d}{dx}(y)=0$. But then ı realised it is wrong usage mathematically. It does not mean multiplication. At the end ı thought that $\frac{d}{dx}$ is already equals to $0$ because if the operator $\frac{d}{dx}$ is alone, it means a derivative of constant function $f(x)=1$. $\frac{d}{dx}(1)=0$. Then the differential equation solved.

But I'm still not sure. Am I right or not?

Answer 1

The equation is given compactly. You can't consider $\frac{d}{dx}$ as $\frac{d}{dx}(1) = 0$.

Expand it and use $\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d^2y}{dx^2}$ to get,

$\begin{align}&\left[\frac{d^2y}{dx^2} + \frac{d}{dx}(2xy)\right] +\left[2x\frac{dy}{dx}+4x^2y\right] = 2e^{-x^2}\\\\ \Rightarrow&\left[\frac{d^2y}{dx^2} + 2x\frac{dy}{dx} +2y\right] +\left[2x\frac{dy}{dx}+4x^2y\right] = 2e^{-x^2}\\\\ \Rightarrow&\frac{d^2y}{dx^2} + 4x\frac{dy}{dx} +2y +4x^2y = 2e^{-x^2}\end{align}$

Answer 2

This mixed use is rightfully confusing. If one does operator algebra, then please in all terms, $$ (D+2x)(D+2x)y=2e^{-x^2}, ~~~ D=\frac{d}{dx}. $$ Then use that $e^{x^2}$ is the integrating factor for $(D+2x)$, that is, $e^{x^2}(D+2x)y=D(e^{x^2}y)$. This results in $$ D^2(e^{x^2}y)=2 $$ which can now be nicely integrated.

Answer 3

$$\left(\frac{\ d}{dx}+2x\right)·\color {red}{\left(\frac{\ d}{dx}+2x\right)y}= 2e^{-x^2}$$ If you only know how to solve first order DE then rewrite the DE as: $$\left(\frac{\ d}{dx}+2x\right)·\color{red}w= 2e^{-x^2}$$ Where $w$ is the unknown function then use integrating factor method to solve the DE: $$w'+2xw= 2e^{-x^2}$$ $$e^{x^2}(w'+2xw)= 2$$ $$(e^{x^2}w)'= 2$$ Integrate: $$e^{x^2}w= 2x+C_1$$ Substitute back $w$: $$e^{x^2}\color{red}{(y'+2xy)}= 2x+C_1$$ Solve this forst order DE. $$(e^{x^2}y)'= 2x+C_1$$ Integrate again: $$e^{x^2}y= x^2+C_1x+C_2$$ $$y(x)=e^{-x^2}(x^2+C_1x+C_2)$$