Author Topic: Quiz3 TUT0402  (Read 373 times)


  • Jr. Member
  • **
  • Posts: 7
  • Karma: 0
    • View Profile
Quiz3 TUT0402
« on: October 11, 2019, 01:59:59 PM »
Question: Find the solution of the given initial value problem.
$$y''+5y'+3y=0, y(0)=1, y'(0)=0$$

Find roots of characteristic equation:

$$r = \frac{-5\pm\sqrt{5^2-4(1)(3)}}{2(1)}
= \frac{-5\pm\sqrt{13}}{2}$$

So the general solution is:

    y &= c_1e^{\frac{-5+\sqrt{13}}{2}}+
    \implies y' &= \frac{-5+\sqrt{13}}{2}c_1e^{\frac{-5+\sqrt{13}}{2}}
- \frac{5+\sqrt{13}}{2}c_2e^{\frac{-5-\sqrt{13}}{2}}\notag

Plug in the given initial value:
    y(0)=1 &\implies c_1+c_2=1\notag\\
    y'(0)=0 &\implies \frac{-5+\sqrt{13}}{2}c_1- \frac{5+\sqrt{13}}{2}c_2=0\notag

Solving this system for $c_1$, $c_2$, we get:
    c_1 &= \frac{5+\sqrt{13}}{2\sqrt{13}}\notag\\
    c_2 &= 1-\frac{5+\sqrt{13}}{2\sqrt{13}}\notag

$$y= \frac{5+\sqrt{13}}{2\sqrt{13}}e^{\frac{-5+\sqrt{13}}{2}}+
    \left(\frac{\sqrt{13}-5}{2\sqrt{13}} \right) e^{\frac{-5-\sqrt{13}}{2}}$$