User:Egm4313.s12.team14.pickett.report1

R1.5

Given:
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{y}'' + 4{y}' + ({\pi}^2 + 4)y = 0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
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Problem: Find the general solution for this homogeneous differential equation.

Solution: Assume that the solution is of the form
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y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
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\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
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y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
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Substituting equations 2, 3, and 4 into equation 1 yields

$$\displaystyle \lambda^2 e^{\lambda x} + 4\lambda e^{\lambda x} + (\pi^2 + 4)e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + 4\lambda +(\pi^2 + 4)) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle (\lambda^2 + 4\lambda +(\pi^2 + 4)) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-4 \pm \sqrt{(4)^2 - 4(1)(\pi^2 + 4)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-4 \pm \sqrt{16 - 4\pi^2 - 16)}}{2}$$ $$\displaystyle \lambda = \frac{-4 \pm \sqrt{-4\pi^2)}}{2}$$ $$\displaystyle \lambda = -2 \pm i{\pi}$$

A solution of the form $$\displaystyle y_h = e^{\lambda z}$$ where z is a complex number of the form $$\displaystyle a + ib$$ requires the use of Euler's Formula, where $$\displaystyle e^{\lambda z} = e^{a}(\cos{b} + i\sin{b})$$

$$\displaystyle \therefore y_{h1} = e^{-2x}(\cos{\pi x} + i\sin{\pi x})$$ $$\displaystyle y_{h2} = e^{-2x}(\cos{-\pi x} + i\sin{-\pi x})$$ Finally $$\displaystyle y_h = C_1[e^{-2x}(\cos{\pi x} + i\sin{\pi x})] + C_2[e^{-2x}(\cos{-\pi x} + i\sin{-\pi x})]$$

R1.6

(A) Falling Stone (y represents height above the ground) $$\displaystyle y'' = g $$ This governing equation is SECOND order and LINEAR

To prove whether the principle of superposition applies to the differential equation, we will suppose that there exists two independent solutions. One solution for the homogeneous case would have any forcing function set to zero, while the particular solution would take the forcing function into account. $$\displaystyle y_{h} $$ $$\displaystyle y_{p} $$ Plugging these two solutions into the equation, we obtain $$\displaystyle {y_h}'' = 0 $$ $$\displaystyle {y_p}'' = g $$ Adding these equations together, we obtain $$\displaystyle {y_h} + {y_p} = g $$

We can see that $$\displaystyle {y_h} + {y_p} = (y_h + y_p)'' $$, which proves that superposition is applicable to this linear differential equation.

(H) Pendulum (theta represents the angle with respect to vertical) $$\displaystyle L{\theta}'' + g{\sin(\theta)} = 0 $$ This governing equation is SECOND order and NON-LINEAR

$$\displaystyle y_{h} $$ $$\displaystyle y_{p} $$ Plugging these two solutions into the equation, we obtain $$\displaystyle L{{\theta}_h}'' + g{\sin({\theta}_h)} = 0 $$ $$\displaystyle L{{\theta}_p}'' + g{\sin({\theta}_p)} = 0 $$ Adding these equations together, we obtain $$\displaystyle L{{\theta}_p} + L{{\theta}_h} + g{\sin({\theta}_p)} + g{\sin({\theta}_h)} = 0 $$

If we plug in a linear combination of the particular and homogeneous solutions into the differential equation, we obtain $$\displaystyle L({\theta}_p +{\theta}_h)'' + g{\sin({\theta}_p + {\theta}_h)} = 0 $$ We can see that $$\displaystyle L({\theta}_p +{\theta}_h) + g{\sin({\theta}_p + {\theta}_h)} \neq L{{\theta}_p} + L{{\theta}_h}'' + g{\sin({\theta}_p)} + g{\sin({\theta}_h)} $$

Therefore the principle of superposition does not hold for this non-linear differential equation.