User:Egm4313.s12.team8.tclamb/R2

=R2.8 - Two ODEs with Complex Roots=

Problem Statement
Find a general solution. Check your answer by substitution.

Part A:
 * $$\displaystyle y'' + y' + 3.25y = 0$$

Part B:
 * $$y'' + 0.54 y' + \left(0.0729 + \pi\right) y = 0$$

Finding the General Solution
Assume solutions are proportional to $$e^{rt}$$:
 * $$e^{rt}\left(r^2 + r + 3.25\right) = 0$$

Observe $$e^{rt} \neq 0$$ for any $$t$$:
 * $$\displaystyle r^2 + r + 3.25 = 0$$

Solve for $$r$$:
 * $$r = -\frac{1}{2} \pm \sqrt{3} i$$

Substitute in $$r$$ for initial assumption:
 * $$y(t) = c_1 e^{\left(-\frac{1}{2} + \sqrt{3} i\right) t} + c_2 e^{\left(-\frac{1}{2} - \sqrt{3} i\right) t}$$

Expand the solution using Euler's identity ($$e^{a + bi}=e^{a}\left(\cos{b} + i\sin{b}\right)$$):
 * $$\begin{align}

y(t) &= c_1 e^{-\frac{1}{2} t} \left(\cos{\sqrt{3} t} + i\sin{\sqrt{3} t}\right) \\ &+ c_2 e^{-\frac{1}{2} t} \left(\cos{\sqrt{3} t} - i\sin{\sqrt{3} t}\right) \end{align}$$ Collect like terms:
 * $$\begin{align}

y(t) &= \left(c_1 + c_2\right) e^{-\frac{1}{2} t} \cos{\sqrt{3} t} \\ &+ \left(c_1 - c_2\right) i e^{-\frac{1}{2} t} \sin{\sqrt{3} t} \end{align}$$ Define new constants $$c'_1 = c_1 + c_2$$ and $$c'_2 = \left(c_1 - c_2\right) i$$: $$y(t) = c'_1 e^{-\frac{1}{2} t} \cos{\sqrt{3} t} + c'_2 e^{-\frac{1}{2} t} \sin{\sqrt{3} t}$$

Checking by Substitution
Determine $$3.25 y$$, $$y'$$, and $$y''$$ from $$y(t)$$:
 * $$\begin{array}{rllrlrllcrrlcll}

3.25 & y  & = e^{-\frac{1}{2} t} (   &  \frac{13}{4} & c'_1 &               &        & \cos{\sqrt{3} t} & + &   &           &      &   \frac{13}{4} & c'_2   & \sin{\sqrt{3} t} ) \\ & y' & = e^{-\frac{1}{2} t} ( ( & -\frac{1}{2}  & c'_1 & + \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} & + & ( & -\sqrt{3} & c'_1 & - \frac{1}{2}  & c'_2 ) & \sin{\sqrt{3} t} ) \\ & y'' & = e^{-\frac{1}{2} t} ( ( & -\frac{11}{4} & c'_1 & - \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} & + & ( & \sqrt{3} & c'_1 & - \frac{11}{4} & c'_2 ) & \sin{\sqrt{3} t} ) \end{array}$$ Substitute into original ODE:
 * $$\begin{array}{rrrcrllrlrlll}

( e^{-\frac{1}{2} t} & ( ( & -\frac{11}{4} & c'_1 & - \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} + ( & \sqrt{3} & c'_1 & - \frac{11}{4} & c'_2 ) & \sin{\sqrt{3} t} ) ) & \\ + \; ( e^{-\frac{1}{2} t} & ( ( & -\frac{1}{2} & c'_1 & + \; \sqrt{3} & c'_2 ) & \cos{\sqrt{3} t} + ( & -\sqrt{3} & c'_1 & - \frac{1}{2}  & c'_2 ) & \sin{\sqrt{3} t} ) ) & \\ + \; ( e^{-\frac{1}{2} t} & (  &  \frac{13}{4} & c'_1 &               &        & \cos{\sqrt{3} t} +   &           &      &   \frac{13}{4} & c'_2   & \sin{\sqrt{3} t} ) ) & = 0 \end{array}$$ Reduce: $$\displaystyle 0 = 0$$

Finding the General Solution
Assume solutions are proportional to $$e^{rt}$$:
 * $$e^{rt}\left(r^2 + 0.54 r + \left(0.0729 + \pi\right)\right) = 0$$

Observe $$e^{rt} \neq 0$$ for any $$t$$:
 * $$r^2 + 0.54 r + \left(0.0729 + \pi\right) = 0$$

Rearrange by completing the square:
 * $$\left(r + 0.27\right)^2 + \pi = 0$$

Solve for $$r$$:
 * $$r = -0.27 \pm \sqrt{\pi} i$$

Substitute in $$r$$ for initial assumption:
 * $$y(t) = c_1 e^{\left(-0.27 + \sqrt{\pi} i\right) t} + c_2 e^{\left(-0.27 - \sqrt{\pi} i\right) t}$$

Expand the solution using Euler's identity ($$e^{a + bi}=e^{a}\left(\cos{b} + i\sin{b}\right)$$):
 * $$\begin{align}

y(t) &= c_1 e^{-0.27 t} \left(\cos{\sqrt{\pi} t} + i\sin{\sqrt{\pi} t}\right) \\ &+ c_2 e^{-0.27 t} \left(\cos{\sqrt{\pi} t} - i\sin{\sqrt{\pi} t}\right) \end{align}$$ Collect like terms:
 * $$\begin{align}

y(t) &= \left(c_1 + c_2\right) e^{-0.27 t} \cos{\sqrt{\pi} t} \\ &+ \left(c_1 - c_2\right) i e^{-0.27 t} \sin{\sqrt{\pi} t} \end{align}$$ Define new constants $$c'_1 = c_1 + c_2$$ and $$c'_2 = \left(c_1 - c_2\right) i$$: $$\displaystyle y(t) = c'_1 e^{-0.27 t} \cos{\sqrt{\pi} t} + c'_2 e^{-0.27 t} \sin{\sqrt{\pi} t}$$

Checking by Substitution
Determine $$( 0.0729 + \pi ) y$$, $$0.54 y'$$, and $$y''$$ from $$y(t)$$:
 * $$\begin{array}{rllrlrllcrrlcll}

( 0.0729 + \pi ) & y  & = e^{-0.27 t} (   &  ( 0.0729 + \pi )  & c'_1 &                        &        & \cos{\sqrt{\pi} t} & + &   &                     &      & ( 0.0729 + \pi )  & c'_2   & \sin{\sqrt{\pi} t} ) \\ & y' & = e^{-0.27 t} ( ( &            -0.27   & c'_1 & + \;               \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( &                -\sqrt{\pi} & c'_1 &       - \; 0.27   & c'_2 ) & \sin{\sqrt{\pi} t} ) \\ 0.54 & y' & = e^{-0.27 t} ( ( & -2 \times  0.0729  & c'_1 & + \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( & - 2 \times 0.27 \sqrt{\pi} & c'_1 & - \; 2 \times 0.0729 & c'_2 ) & \sin{\sqrt{\pi} t} ) \\ & y'' & = e^{-0.27 t} ( ( & ( 0.0729 - \pi ) & c'_1 & - \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + & ( &  2 \times 0.27 \sqrt{\pi} & c'_1 & + \; (0.0729 - \pi) & c'_2 ) & \sin{\sqrt{\pi} t} ) \end{array}$$ Substitute into original ODE:
 * $$\begin{array}{rrrlrlllrrllll}

( e^{-0.27 t} & ( ( & ( -\pi + 0.0729 )  & c'_1 & - \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + \; ( &   2 \times 0.27 \sqrt{\pi} & c'_1 & + \; ( 0.0729 - \pi) & c'_2 ) & \sin{\sqrt{\pi} t} ) ) & \\ + \; ( e^{-0.27 t} & ( ( & -2 \times 0.0729   & c'_1 & + \; 2 \times 0.27 \sqrt{\pi} & c'_2 ) & \cos{\sqrt{\pi} t} & + \; ( & - 2 \times 0.27 \sqrt{\pi} & c'_1 & - \; 2 \times 0.0729 & c'_2 ) & \sin{\sqrt{\pi} t} ) ) & \\ + \; ( e^{-0.27 t} & (  &  ( 0.0729 + \pi  )  & c'_1 &                               &        & \cos{\sqrt{\pi} t} & +   &                            &      & ( 0.0729 + \pi )  & c'_2   & \sin{\sqrt{\pi} t} ) ) & = 0 \end{array}$$ Reduce: $$\displaystyle 0 = 0$$