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.


closed as off-topic by 5xum, José Carlos Santos, Shailesh, Isaac Browne, Leucippus Jul 27 '18 at 0:05

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  • $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – 5xum Jul 26 '18 at 9:23
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  • $\begingroup$ i have edited the question. please check. $\endgroup$ – d.s Jul 26 '18 at 9:48

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 $K_1 , K_2$


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