User:Egm4313.s12.team13.steinberg.r2

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

$$ y'' + 6y' + 8.96y = 0\,\!$$

Solution
The homogeneous characteristic equation for linear 2nd order ODE's with constant coefficients is $$ \lambda^2 + a\lambda + b = 0\,\!$$(3.1a)

For the ODE in this problem, the characteristic equation becomes

$$ \lambda^2 + 6\lambda + 8.96 = 0\,\!$$

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

$$ \Delta = a^2 - 4b\,\!$$(3.2a)

$$ \Delta = 36 - 4(8.96) = 0.16\,\!$$

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

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$(3.3a)

$$ \lambda_{1,2} = \frac{-6 \pm \sqrt{0.16}}{2}\,\!$$ $$ = \frac{-6 \pm 0.4}{2}\,\!$$ $$ = -3 \pm 0.2\,\!$$

Therefore,

$$ \lambda_{1} = -2.8 \,\!$$(3.4a) $$ \lambda_{2} = -3.2 \,\!$$(3.5a)

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

$$ y_{h,1}(x) = e^{\lambda_1 x}\,\!$$(3.6a)

$$ y_{h,2}(x) = e^{\lambda_2 x}\,\!$$(3.7a)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$(3.8a)

$$ y_{h}(x) = c_1 e^{\lambda_1 x} + c_2 e^{\lambda_2 x}\,\!$$

The final general solution is therefore

$$ y_{h}(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$(3.9a)

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.

$$ y'(x) = -2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}\,\!$$(3.10a)

$$ y''(x) = 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x}\,\!$$(3.11a)

Substituting back into the original ODE gives

$$ y'' + 6y' + 8.96y = 0\,\!$$(3.12a)

$$ 7.84 c_1 e^{-2.8x} + 10.24 c_2 e^{-3.2x} + 6(-2.8 c_1 e^{-2.8x} - 3.2 c_2 e^{-3.2x}) + 8.96(c_1 e^{-2.8 x} + c_2 e^{-3.2 x}) = 0\,\!$$

$$ \cancelto{0}{(7.84 - 16.8 + 8.96)c_1 e^{-2.8x}} + \cancelto{0}{(10.24 - 19.2 + 8.96)c_2 e^{-3.2x}} = 0\,\!$$

Since all terms cancel to 0, $$ y_(x) = c_1 e^{-2.8 x} + c_2 e^{-3.2 x}\,\!$$ is a general solution to the original ODE.

Section 3 Lecture Notes

Section 5 Lecture Notes

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 $$ \lambda^2 + a\lambda + b = 0\,\!$$(3.1b)

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.

$$ \Delta = a^2 - 4b\,\!$$(3.2b)

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

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.

$$ \lambda_{1,2} = \frac{-a \pm \sqrt{a^2 - 4b}}{2}\,\!$$(3.3b)

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

Therefore,

$$ \lambda_{1} = -2 + \pi i \,\!$$(3.4b) $$ \lambda_{2} = -2 - \pi i \,\!$$<p style="text-align:right">(3.5b)

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

$$ y_{h,1}(x) = e^{a_1 x} \cos b_1 x\,\!$$<p style="text-align:right">(3.6b)

$$ y_{h,2}(x) = e^{a_2 x} \sin b_2 x\,\!$$<p style="text-align:right">(3.7b)

The homogeneous solution is

$$ y_{h} = c_1 y_{h,1} + c_2 y_{h,2}\,\!$$<p style="text-align:right">(3.8b)

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

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

The final general solution is therefore

$$ y_{h}(x) = e^{-2 x}(c_1 \cos \pi x + c_2 \sin \pi x)\,\!$$<p style="text-align:right">(3.9b)

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.

$$ y'(x) = -2e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) + e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x)\,\!$$<p style="text-align:right">(3.10b)

$$ y''(x) = 4e^{-2x}(c_1 \cos \pi x + c_2 \sin \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) - 2e^{-2x}(-\pi c_1 \sin \pi x + \pi c_2 \cos \pi x) + e^{-2x}(- \pi^{2} c_1 \cos \pi x - \pi^{2} c_2 \sin \pi x)\,\!$$<p style="text-align:right">(3.11b)

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

Substituting back into the original ODE gives

$$ y'' + 4y' + (\pi^2 + 4)y = 0\,\!$$<p style="text-align:right">(3.12b)

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

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

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 5 Lecture Notes

Section 6 Lecture Notes