diff ekv

$$\begin{gathered} Bestäm kons\tan ten k så att y=e^{-k{x}^2} blir en lösning till y\text{'}\text{'}+2kxy\text{'}+y=0: \\ y=e^{-k{x}^2} \\ g\left(x\right)=x^2 \\ y\text{'}=y\text{'}\left(g\right(x\left)\right)\times g\text{'}\left(x\right) \\ g\text{'}\left(x\right)=2x \\ y=e^{kz} \\ y\text{'}\left(z\right)=-k{e}^{kz} \\ y\text{'}=-2xk{e}^{k{x}^2} \\ y\text{'}\text{'}=-2k{e}^{k{x}^2}-2x{k}^2{e}^{k{x}^2} \\ y\text{'}\text{'}+2kxy\text{'}+y=0 \\ -2k{e}^{-k{x}^2}-2x{k}^2{e}^{-k{x}^2}-4x^2{k}^2{e}^{-k{x}^2}+e^{-k{x}^2}=0 \\ {e}^{-k{x}^2}(-2k-2x{k}^2-4x^2{k}^2+1)=0 // e^{-k{x}^2} \\ -2k-2x{k}^2-4x^2{k}^2+1=0 \\ 2x^2{k}^2+2x{k}^2+k-\frac{1}{2}=0 \\ {x}^2+x+\frac{k-1}{2k^2}=0 \end{gathered}$$

Jag är nog ute o cyklar