University of Florida/Egm4313/s12.team13.steinberg.r1

Problem Statement
4.) Find the general solution to the following ODE and check the result by substitution.

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is $$ \lambda^2 + a\lambda + b = 0\,\!$$(5.1)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 4\lambda + (\pi^2 + 4) = 0\,\!$$

To solve for the solutions, the discriminant to the quadratic equation must first be calculated.

$$ \Delta = a^2 - 4b\,\!$$(5.2)

$$ \Delta = 16 - 4(\pi^2 + 4)\,\!$$ $$ = 16 - 4\pi^2 - 16\,\!$$ $$ = -4\pi^2\,\!$$

Since the discriminant is negative, there will be two imaginary solutions to the ODE. The solutions can be obtained by solving the remainder of the quadratic formula.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$(5.3)

$$ \lambda_{1,2} = \frac{-4 \pm \sqrt{-4\pi^2}}{2}\,\!$$ $$ = \frac{-4 \pm 2\pi i}{2}\,\!$$ $$ = -2 \pm \pi i\,\!$$

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$(5.4) $$ \lambda_{2} = -2 - \pi i \,\!$$(5.5)

The two distinct, linearly independent, homogeneous solutions for the case with two imaginary roots are

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$(5.6)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$(5.7)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$(5.8)

$$ y_{h}(x) = c_1 e^{a_1 x} \cos b_1 x + c_2 e^{a_2 x} \sin b_2 x\,\!$$

$$ y_{h}(x) = c_1 e^{-2 x} \cos \pi x + c_2 e^{-2 x} \sin \pi x\,\!$$

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$(5.9)

To check this result using substitution, we must take the first and second derivatives of the general solution to substitute back into the original ODE.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$(5.10)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$(5.11)

$$ = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$(5.12)

$$ 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x) - 8e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + \cancelto{0}{4e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)} + (\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x) = 0\,\!$$

$$ \cancelto{0}{-4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x)} + \cancelto{0}{e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)} + \cancelto{0}{(\pi^{2} + 4)(e^{-2x})(c_1 \cos \pi x + c_2 \sin \pi x)} = 0\,\!$$

Since all terms cancel to 0, $$ y_(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$ is a general solution to the original ODE.

Section 2 Lecture Notes