English: So, second equation I can write as $(A - B) = - \frac{\alpha}{ik}D$ which is $\frac{i\alpha}{k}D$ and this immediately tells you that $D = \frac{2A}{1 + \frac{i\alpha}{k}}$ which is equal to $\frac{2kA}{k + i\alpha}\ $and B which is equal to D − A is going to be equal to $\frac{2kA}{k + i\alpha} - A\ $and that gives you So, what we found is that in this case when particles are coming and going back and there is some decaying wave function here I have found Ae^(ikx) which represent particles coming to right Be^(−ikx) represent particles going to left I have found that $\frac{B}{A}$ is equal to let see where it is $\left( \frac{k - i\alpha}{k + i\alpha}\ \right)$ and $\frac{D}{A} = \frac{2k}{k + i\alpha}$.
Kannada: ಆದ್ದರಿಂದ, ಎರಡನೇ ಸಮೀಕರಣವನ್ನು ನಾನು$(A - B) = - \frac{\alpha}{ik}D$ ಎಂದು ಬರೆಯಬಹುದು ಅದು $\frac{i\alpha}{k}D$ ಮತ್ತು ಇದು ತಕ್ಷಣವೇ ನಿಮಗೆ ಹೇಳುತ್ತದೆ $D = \frac{2A}{1 + \frac{i\alpha}{k}}$ ಇದು$\frac{2kA}{k + i\alpha}\ $ ಗೆ ಸಮನಾಗಿರುತ್ತದೆ ಮತ್ತು D − A ಗೆ ಸಮನಾಗಿರುವ B , $\frac{2kA}{k + i\alpha} - A\ $ ಗೆ ಸಮಾನವಾಗಿರುತ್ತದೆ.