User:Egm4313.s12.team20.chang/Report 1

R 1.5
Advanced Engineering Mechanics 10th Ed. Kreyszig 2011 p.59 problems.4-5

Problem 4
Find a general solution. Check your answer by substitution.

Given

 * {| style="width:100%" border="0"

$$\displaystyle y''+4y'+(\pi^2 + 4)y = 0$$ (5.0)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Solution
Rewrite as


 * {| style="width:100%" border="0"

$$\frac{d^2 y}{dx^2} + 4\frac{dy}{dx} + (\pi^2 + 4)y=0 $$ (5.1)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Set value for lambda
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$ (5.2)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Plug in the variable lambda to get the quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+4\lambda+(\pi^2+4)=0$$ (5.3)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The formula for the Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha \pm \beta i$$ (5.4)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Using the quadratic formula, the roots are determined to be
 * {| style="width:100%" border="0"

$$\lambda =-2\pm (\pi)i$$ (5.5)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The General Solution formula for Complex Conjugate roots is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1\cos \beta x+c_2\sin \beta x)$$ (5.6)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be

Problem 5
Find a general solution. Check your answer by substitution.

Given

 * {| style="width:100%" border="0"

$$\displaystyle y''+2\pi y'+\pi^2y = 0$$ (5.8)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Solution
Rewrite as


 * {| style="width:100%" border="0"

$$\frac{d^2 y}{dx^2} + 2\pi \frac{dy}{dx} + (\pi^2 y)=0 $$ (5.9)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Set value for lambda (refer to equation 5.2)
 * {| style="width:100%" border="0"

$$\lambda = \frac{dy}{dx}$$
 * style="width:95%" |
 * style="width:95%" |
 * }
 * }

Plug in the variable lambda to get the quadratic equation
 * {| style="width:100%" border="0"

$$\displaystyle \lambda^2+2\pi \lambda+\pi^2 y=0$$ (5.10)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The formula for the Real Double root is
 * {| style="width:100%" border="0"

$$\lambda=-\frac{1}{2}\alpha$$ (5.11)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Using the quadratic formula, the root is determined to be
 * {| style="width:100%" border="0"

$$\displaystyle\lambda=-\pi$$ (5.12)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

The General Solution formula for Real Double root is
 * {| style="width:100%" border="0"

$$\displaystyle y(x)=e^{-\frac{\alpha x}{2}}(c_1 + c_2 x)$$ (5.13)
 * style="width:95%" |
 * style="width:95%" |
 * 
 * }

Therefore, by plugging in the roots into the General Solution formula, the solution is determined to be