We've updated our
Privacy Policy effective December 15.
Please read our updated Privacy Policy and tap
Graph
Hide Plot »
Frequently Asked Questions (FAQ)
What is the solution for y^{''}+10y^'+25y=t^{-2}e^{-5t} ?
- The solution for y^{''}+10y^'+25y=t^{-2}e^{-5t} is y=c_{1}e^{-5t}+c_{2}te^{-5t}-e^{-5t}ln(t)-e^{-5t}