User:Egm4313.s12.team14.pickett.report2

R2.8

2.8.8

Given:
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{y}'' + y' + 3.25y = 0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Problem: Find the general solution for this homogeneous differential equation.

Solution: Assume that the solution is of the form
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y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
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 * }


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\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
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 * }


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y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1 yields

$$\displaystyle \lambda^2 e^{\lambda x} + \lambda e^{\lambda x} + 3.25e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + \lambda + 3.25) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle \therefore (\lambda^2 + 4\lambda + 3.25) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{(1)^2 - 4(1)(3.25)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{1 - 13)}}{2}$$ $$\displaystyle \lambda = \frac{-1 \pm \sqrt{-12}}{2}$$ $$\displaystyle \lambda = -0.5 \pm j{\sqrt{3}}$$ Where $$\displaystyle j = \sqrt{-1}$$

$$\displaystyle y_h = {C_1}e^{(-0.5 + j\sqrt{3})x} + {C_2}e^{(-0.5 - j\sqrt{3})x} $$ In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation Let $$\displaystyle a = -0.5, b = \sqrt{3}$$ $$\displaystyle y_h = {C_1}e^{(a + jb)x} + {C_2}e^{(a - jb)x} $$ $$\displaystyle y_h = {C_1}e^{ax}e^{jbx} + {C_2}e^{ax}e^{-jbx} $$ $$\displaystyle {y_h}' = {C_1}[e^{ax}({jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle + {C_2}[e^{ax}(-{jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle {y_h}'' = {C_1}[e^{ax}((jb)^2e^{jbx}) + ({jb}e^{jbx})({a}e^{ax}) + e^{jbx}((a)^2e^{ax}) + ({a}e^{ax})({jb}e^{jbx})]$$ $$\displaystyle + {C_2}[e^{ax}((jb)^2e^{-jbx}) + (-{jb}e^{-jbx})({a}e^{ax}) + e^{-jbx}((a)^2e^{ax}) + ({a}e^{ax})(-{jb}e^{-jbx})] $$

Plugging $$\displaystyle y_h, {y_h}' , {y_h}'' $$ into Eq. 1 and collecting terms gives $$\displaystyle 0 = {C_1}e^{ax}e^{jbx}[-b^2 + jab +a^2 + jab + jb + a + 3.25] $$ $$\displaystyle + {C_2}e^{ax}e^{-jbx}[-b^2 - jab +a^2 - jab - jb + a + 3.25] $$ $$\displaystyle 0 = {C_1}e^{-0.5x}e^{j{\sqrt{3} x}}[-3 - j{\sqrt{3}}/2 + 1/4 - j{\sqrt{3}}/2 + j{\sqrt{3}} - 1/2 + 3.25] $$ $$\displaystyle + {C_2}e^{-0.5x}e^{-j{\sqrt{3}}x}[-3 + j{\sqrt{3}}/2 + 1/4 + j{\sqrt{3}}/2 - j{\sqrt{3}} - 1/2 + 3.25] $$ $$\displaystyle 0 = {C_1}e^{-0.5x}e^{j{\sqrt{3}}x}(0) + {C_2}e^{-0.5x}e^{-j{\sqrt{3}}x}(0) $$

Therefore the solution holds for any values of $$\displaystyle C_1, C_2 , x $$

2.8.15

Given:
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{y}'' + (0.54)y' + (0.0729 + \pi)y = 0, $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
 * }

Problem: Find the general solution for this homogeneous differential equation.

Solution: Assume that the solution is of the form
 * {| style="width:100%" border="0" align="left"

y_h = e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
 * }
 * }


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

\therefore y'_h = \lambda e^{\lambda x} $$ $$ And
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
 * }
 * }


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

y''_h = \lambda^2 e^{\lambda x} $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 4)
 * }
 * }

Substituting equations 2, 3, and 4 into equation 1 yields

$$\displaystyle \lambda^2 e^{\lambda x} + (0.54)\lambda e^{\lambda x} + (0.0729 + \pi)e^{\lambda x} = 0$$ $$\displaystyle e^{\lambda x}(\lambda^2 + (0.54)\lambda + (0.0729 + \pi)) = 0$$ Since $$\displaystyle e^{\lambda x} \neq 0$$ $$\displaystyle \therefore (\lambda^2 + (0.54)\lambda + (0.0729 + \pi)) = 0$$

We must use the quadratic formula to obtain values for lambda $$\displaystyle \lambda = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$ $$\displaystyle \lambda = \frac{-(0.54) \pm \sqrt{(0.54)^2 - 4(1)(0.0729 + \pi)}}{2(1)}$$ $$\displaystyle \lambda = \frac{-0.54 \pm j{3.545}}{2}$$ $$\displaystyle \lambda = -0.27 \pm j(1.772)$$ Where $$\displaystyle j = \sqrt{-1}$$

$$\displaystyle y_h = {C_1}e^{(-0.27 + j(1.772))x} + {C_2}e^{(-0.27 - j(1.772))x} $$ In order to verify solution, we must evaluate it's first and second derivative, then plug them into the equation Let $$\displaystyle a = -0.27, b = 1.772$$ $$\displaystyle y_h = {C_1}e^{(a + jb)x} + {C_2}e^{(a - jb)x} $$ $$\displaystyle y_h = {C_1}e^{ax}e^{jbx} + {C_2}e^{ax}e^{-jbx} $$ $$\displaystyle {y_h}' = {C_1}[e^{ax}({jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle + {C_2}[e^{ax}(-{jb}e^{jbx}) + e^{jbx}({a}e^{ax})] $$ $$\displaystyle {y_h}'' = {C_1}[e^{ax}((jb)^2e^{jbx}) + ({jb}e^{jbx})({a}e^{ax}) + e^{jbx}((a)^2e^{ax}) + ({a}e^{ax})({jb}e^{jbx})]$$ $$\displaystyle + {C_2}[e^{ax}((jb)^2e^{-jbx}) + (-{jb}e^{-jbx})({a}e^{ax}) + e^{-jbx}((a)^2e^{ax}) + ({a}e^{ax})(-{jb}e^{-jbx})] $$

Plugging $$\displaystyle y_h, {y_h}' , {y_h}'' $$ into Eq. 1 and collecting terms gives $$\displaystyle 0 = {C_1}e^{ax}e^{jbx}[-b^2 + jab +a^2 + jab + (0.54)(jb + a) + 0.0729 + \pi]$$ $$\displaystyle + {C_2}e^{ax}e^{-jbx}[-b^2 - jab +a^2 - jab + (0.54)(-jb + a) + 0.0729 + \pi] $$ $$\displaystyle 0 = {C_1}e^{-0.27x}e^{j{1.772x}}[-(1.772)^2 + j(-0.27)(1.772) + (0.27)^2 $$ $$\displaystyle + j(-0.27)(1.772) + (0.54)(j(1.772) - 0.27) + 0.0729 + \pi] $$ $$\displaystyle + {C_2}e^{-0.27x}e^{-j1.772x}[-(1.772)^2 - j(-0.27)(1.772) + (0.27)^2 $$ $$\displaystyle - j(-0.27)(1.772) + (0.54)(-j(1.772) - 0.27) + 0.0729 + \pi] $$ $$\displaystyle 0 = {C_1}e^{-0.27x}e^{j1.772x}(0) + {C_2}e^{-0.27x}e^{-j1.772x}(0) $$

Therefore the solution holds for any values of $$\displaystyle C_1, C_2 , x $$

$$\displaystyle y = {C_1}e^{5x} + {C_2}{x}e^{5x} $$ $$\displaystyle y(x) = e^{5x} - {5x}e^{5x} $$

Given: $$\displaystyle {y_p}'' + p(x)y' +q(x)y = r(x) $$ (1) Where p(x) and q(x) are any functions of "x".

Find: Prove that $$\displaystyle y_p(x) = \sum_{i=1}^n y_{p,i}(x) $$ is the overall solution to (1) when $$\displaystyle r(x) = \sum_{i=1}^n r_{i}(x) $$ Also, discuss the need to have both sine and cosine functions in the particular solution "guess" when you only have a sine or cosine function as the input, r(x)

Solution: Let the indices "i" for the particular solutions correspond to the respective indices for the forcing functions such that $$\displaystyle {y_1}'' + p(x){y_1}' +q(x){y_1} = {r_1}(x) $$ (2) and $$\displaystyle {y_2}'' + p(x){y_2}' +q(x){y_2} = {r_2}(x) $$ (3)

Adding both equations together yields: $$\displaystyle {y_1} + p(x){y_1}' +q(x){y_1} + {y_2} + p(x){y_2}' +q(x){y_2} = {r_1}(x) + {r_2}(x) $$ By collecting terms and using summation notation we obtain: $$\displaystyle \sum_{i=1}^2 y_{p,i}''(x) + p(x){\sum_{i=1}^2 y_{p,i}'(x)} +q(x){\sum_{i=1}^2 y_{p,i}(x)} = \sum_{i=1}^2 r_{i}(x) $$ This concept is easily extended to an "n" number of excitation functions with a corresponding "n" number of particular solutions such that: $$\displaystyle \sum_{i=1}^n y_{p,i}''(x) + p(x){\sum_{i=1}^n y_{p,i}'(x)} +q(x){\sum_{i=1}^n y_{p,i}(x)} = \sum_{i=1}^n r_{i}(x) $$ This proves the concept of superposition for excitation functions and their particular solutions for second order differential equations with varying coefficients.

The reason for having both sine and cosine in the particular solution when there is only either sine or cosine as the forcing function is due to the fact that the derivative of sine is cosine and the derivative of cosine is negative sine. Thus, without including both in the initial "guess" for the particular solution, we would be missing half of the solution. Or in other words: $$\displaystyle {y_p} = M\cos(wx) + N\sin(wx) $$ $$\displaystyle {y_p}' = -M\sin(wx) + N\cos(wx) $$ $$\displaystyle {y_p}'' = -M\cos(wx) - N\sin(wx) $$ So plugging these into an arbitrary second order differential equation would yield values for "N" and "M", when there is a forcing function of sine or cosine.

= Problem 8 =

Given

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x^{n}log(1+x)dx $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 1)
 * }
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n=0, n=1 $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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Find
Find the integral of EQ 1 using integration by parts for n=0 and n=1

Solution
For n=0

For n=1


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\int xlog(1+x)dx$$ $$ Take u=(x+1) and du=dx
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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\int (u-1)log(u)du$$
 * $$\displaystyle
 * $$\displaystyle

$$ This gives two separate integrals:
 * $$\displaystyle
 * }
 * }


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\int u log(u)du-\int log(u)du$$ $$ Integrate the first integral (left hand) by parts, taking:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 2)
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 * }
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f=log(u), df=\frac{1}{u}du, dg=udu, g=\frac{u^{2}}{2}$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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 * }
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\frac{1}{2}u^{2}log(u)-\frac{1}{2}\int udu$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle
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 * }


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\frac{1}{2}u^{2}log(u) - \frac{u^2}{4}$$ $$ Integrate the second integral (right hand) by parts, taking: :
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq. 3)
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 * }
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f=log(u), df=\frac{1}{u}du, dg=du, g=u$$ $$ Giving:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
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ulog(u) - \int du $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
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ulog(u)- u $$ $$ Substituting EQ.3 and EQ.4 into EQ.2 gives:
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle (Eq. 4)
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(\frac{1}{2}u^{2}log(u) - \frac{u^2}{4}) - (ulog(u)- u) + Constant (K) $$ $$ Since u=x+1
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
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 * }
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\frac{1}{2}(x + 1)^{2}log(x + 1) - \frac{(x + 1)^2}{4} - (x + 1)log(x + 1) + (x + 1) + K$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
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 * }