University of Florida/Egm4313/s12.team11.gooding/R2/2.8

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

Problem 8
$$y''+ y'+3.25y=0\!$$

Let: $$ \lambda=d/dx\!$$

Characteristic Equation
$$ \lambda^2+\lambda+3.25=0\!$$ Using the quadratic equation to find roots we get: $$\lambda_1=\frac{-1+i\sqrt(12)}{2}\!$$ $$\lambda_2=\frac{-1-i\sqrt(12)}{2}\!$$ Therefore: $$y_h(x)=e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})\!$$

Check By Substitution
$$y'(x)=-\frac{1}{2}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})+e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})\!$$ $$y''(x)=\frac{1}{4}e^{-\frac{1}{2}x}(c_1\cos(x\sqrt{3})+c_2\sin(x\sqrt{3})-\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})-\!$$ $$\frac{1}{2}e^{-\frac{1}{2}x}(-\sqrt{3}c_1\sin{\sqrt{3}x}+\sqrt{3}c_1\cos{\sqrt{3}x})e^{-\frac{1}{2}x}(-3c_1\cos(x\sqrt{3})-3c_2\sin(x\sqrt{3})\!$$ Substituting $$y,y',y\!$$ into the original equation, the result is  $$y+ y'+3.25y=0\!$$

Problem 15
$$y''+0.54y'+(0.0729+\pi)y=0\!$$ Let: $$\frac{d}{dx}=\lambda\!$$

Characteristic Equation
$$ \lambda^2+0.54\lambda+(0.0729+\pi)=0\!$$ Using the quadratic equation to find roots we get: $$\lambda_1=\frac{-0.27+i\sqrt(\pi)}{2}\!$$ $$\lambda_2=\frac{-0.27-i\sqrt(\pi)}{2}\!$$ Therefore: $$y_h(x)=e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})\!$$

Check By Substitution
$$y'(x)=-0.27e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})+e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})\!$$

$$y''(x)=0.0729e^{-0.27x}(c_1\cos(x\sqrt{\pi})+c_2\sin(x\sqrt{\pi})-0.27e^{-0.27x}(-\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})-\!$$

$$0.27e^{-0.27x}(\sqrt{\pi}c_1\sin{\sqrt{\pi}x}+\sqrt{\pi}c_1\cos{\sqrt{\pi}x})+e^{-0.27x}(-\pi(c_1\cos(x\sqrt{\pi}))-\pi(c_2\sin(x\sqrt{\pi})))\!$$

Substituting $$y,y',y\!$$ into the original equation, the result is  $$y+0.54y'+(0.0729+\pi)y=0\!$$ Egm4313.s12.team11.gooding 03:41, 7 February 2012 (UTC)