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

R1.5A

Problem Statement

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

$y^{″}+4y^{\prime }+({\pi }^2+4)y=0$

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.

$\mathrm{\Delta }=a^2-4b$

(5.2)

$\mathrm{\Delta }=16-4({\pi }^2+4)$ $=16-4{\pi }^2-16$ $=-4{\pi }^2$

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)

$y_h(x)=c_1{e}^{a_{1}x}\cos b_{1}x+c_2{e}^{a_{2}x}\sin b_{2}x$

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

The final general solution is therefore

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

(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.

$y^{\prime }(x)=-2e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)+e^{-2x}(-\pi c_1\sin \pi x+\pi c_2\cos \pi x)$

(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

$y^{″}+4y^{\prime }+({\pi }^2+4)y=0$

(5.12)

$4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)-{\xcancel{4e^{-2x}(-\pi c_1\sin \pi x+\pi c_2\cos \pi x)}}^0+e^{-2x}(-{\pi }^2{c}_1\cos \pi x-{\pi }^2{c}_2\sin \pi x)-8e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)+{\xcancel{4e^{-2x}(-\pi c_1\sin \pi x+\pi c_2\cos \pi x)}}^0+({\pi }^2+4)(e^{-2x})(c_1\cos \pi x+c_2\sin \pi x)=0$

${\xcancel{-4e^{-2x}(c_1\cos \pi x+c_2\sin \pi x)}}^0+{\xcancel{e^{-2x}(-{\pi }^2{c}_1\cos \pi x-{\pi }^2{c}_2\sin \pi x)}}^0+{\xcancel{({\pi }^2+4)(e^{-2x})(c_1\cos \pi x+c_2\sin \pi x)}}^0=0$

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

Section 2 Lecture Notes