# Solve the initial value differential equation [closed]

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, LeucippusJul 27 '18 at 0:05

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – 5xum, José Carlos Santos, Shailesh, Isaac Browne, Leucippus
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• 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? – 5xum Jul 26 '18 at 9:23
• Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add an upvote. – 5xum Jul 26 '18 at 9:23
• Follow this link for MathJax tutorial math.meta.stackexchange.com/questions/5020/… – Indrajit Ghosh Jul 26 '18 at 9:32
• i have edited the question. please check. – 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 $K_1 , K_2$