User:Egm6321.f10.team5.elliott/hw2

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Homework 2, Problem 8

Find: $$\left(\alpha \right)$$ show that k1 is not necessary.

If we integrate left side of(6) pg 10-3(lecture) it will reveal how k1 cancels during integration.


 * $$ \int_{}^{x} \frac{ {h}_{x}}{h}dx= \int_{}^{s}n \left(s \right)ds$$


 * $$ \left[ \ln \left(h \right)+ {k}_{1}\right]- \left[ \ln \left(h \left(0 \right) \right)+ {k}_{1} \right]= \int_{}^{s}n \left(s \right)ds$$


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 * $$\ln \left(h \right)+ \ln \left(h \left(0 \right) \right)+\left[{k}_{1}- {k}_{1} \right]= \int_{}^{s}n \left(s \right)ds$$
 * $$\ln \left(h \right)+ \ln \left(h \left(0 \right) \right)+\left[{k}_{1}- {k}_{1} \right]= \int_{}^{s}n \left(s \right)ds$$

k1 term cancels on left side
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$$\left( \beta \right)$$ show that equation (6) pg 10-3(lecture) agrees with King pg 512.

The linear, 1st order, ODE in King is written in the form:


 * $$ \frac{dy}{dx}+P \left(x \right)y=Q \left(x \right)$$

The structure of the solution in King is written as:

$$y= {y}_{h}+ {y}_{p}$$

where the particular solution is in the form:

$$ {y}_{p}= \exp \left\{- \int_{}^{x}P \left(x \right)dt \right\} \int_{}^{x}Q \left(s \right) \exp \left\{ \int_{}^{s}P \left(t \right)dt \right\}ds$$

However, equation 6 pg10-3 (lecture) is a particular solution written in the form:


 * $${y}_{p} \left(x \right)= \frac{1}{h \left(x \right)} \int_{0}^{x}h \left(s \right)b \left(s \right)ds$$

Where,from (1) and (5) pg 10-2 (lecture) we get:


 * $$h \left(x\right)= \exp \left[ \int_{}^{x}n \left(s \right)ds\right]$$


 * $$b \left(x\right)= \frac{R \left(x \right)}{P \left(x \right)}=Q \left(x \right)$$


 * $$n \left(x\right)=P \left(x \right)$$

Finally,plugging h(x),b(x), n(x) into the particular solution form given by King we get:


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 * $$ {y}_{p}= \exp \left\{- \int_{}^{x}h \left(s \right)ds\right\} \int_{}^{x}n \left(s \right) \exp \left\{ \int_{}^{x}b \left(s \right)ds \right\}ds$$
 * $$ {y}_{p}= \exp \left\{- \int_{}^{x}h \left(s \right)ds\right\} \int_{}^{x}n \left(s \right) \exp \left\{ \int_{}^{x}b \left(s \right)ds \right\}ds$$


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'''$$ \left( \gamma \right)$$ solve the following equation:

$$\frac{dy}{dx}+ {a}_{o}y=0$$

$$ {a}_{o}=x $$

solve by separation of variables:

$$ \int_{}^{y} \frac{dy}{y}+ \int_{}^{x}xdx=0$$

$$ \ln \left(y \right)+ \frac{ {x}^{2}}{2}=C$$

$$\ln \left(y \right)=-\frac{ {x}^{2}}{2}+C$$


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 * $$y= \exp \left[- \frac{ {x}^{2}}{2}+C \right]= {y}_{h}$$
 * $$y= \exp \left[- \frac{ {x}^{2}}{2}+C \right]= {y}_{h}$$


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