Solve the following initial value differential equations $20y''+4y'+y=0, y(0)=3.2, y'(0)=0$.
To solve this I substituted $D= \frac{\mathrm d}{\mathrm dx}$. and solved the auxiliary equation to get roots of $D$. and then the solution was $y= \exp\left(-\dfrac{x}{10}\right)(A \cos(x/5)+ B \sin(x/5))$ and tried to get the constants through given question. But I am not sure if the answer is correct.
Your solution is correct $$ 20y′′+4y′+y=0$$ $$\implies 20r^2+4r+1=0$$ $$\Delta_r=16-4.20.1=-64=(8i)^2$$ $$r=\frac {-1\pm 2i}{10}$$ Therefore $$y(x)=e^{-x/10}(K_1\cos(x/5)+K_2\sin(x/5))$$ Apply the initial conditions to find the constant $K_1 , K_2$