User:Egm6321.f10.team3.cook/HW2

=Problem 2: Verify (7) p. 7-1 is a Nonlinear ODE=

Given

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$$ 4 {x}^{7} + \sin\left(y\right) +  {x}^{2} {y}^{3} {y}^{'} = 0 $$ $$
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 * $$\displaystyle (7) p. 7-1
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Find
Verify that $$ 4 {x}^{7} + \sin\left(y\right) +  {x}^{2} {y}^{3} {y}^{'} $$ is a N1-ODE

Solution
For $$\displaystyle (7) p. 7-1$$ to be linear it must obey the following conditions for linearity given below:
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f(cx) = cf(x) $$ where $$ c \in \mathbb{R}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq 2.1)
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f(x+y) = f(x) + f(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq.2.2)
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To show that $$\displaystyle (7) p.7-1$$ is nonlinear we will use $$\displaystyle (Eq 2.1)$$ and show that this condition is not true


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$$
 * Let $$ \mathrm{L} \left( y\right):= 4 {x}^{7} + sin \left( y\right) + {x}^{2} {y}^{3} {y}^{'}$$
 * $$\displaystyle (Eq.2.3)
 * $$\displaystyle (Eq.2.3)
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$$
 * $$ \mathrm{L} \left( cy\right)= 4 {x}^{7} + sin \left( cy\right) + {x}^{2} {\left(cy\right)}^{3} \frac{\ {d} }{\ {d} x} \left( cy\right)$$
 * $$\displaystyle (Eq.2.4)
 * $$\displaystyle (Eq.2.4)
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$$ But
 * $$ \mathrm{L} \left( cy\right)= 4 {x}^{7} + sin \left( cy\right) + {c}^{4}{x}^{2} {y}^{3} \frac{\ {d} }{\ {d} x} \left( y\right)$$
 * $$\displaystyle (Eq.2.5)
 * $$\displaystyle (Eq.2.5)
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$$
 * $$ c\mathrm{L} \left( y\right)= c \left( 4 {x}^{7} + sin \left( y\right) + {x}^{2} {y}^{3} {y}^{'}\right)$$
 * $$\displaystyle (Eq.2.6)
 * $$\displaystyle (Eq.2.6)
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$$ \Rightarrow 4 {x}^{7} + sin \left( cy\right) + {c}^{4}{x}^{2} {y}^{3} {y}^{'}\neq c \left( 4 {x}^{7} + sin \left( y\right) + {x}^{2} {y}^{3} {y}^{'}\right)$$ By inspection, $$ \mathrm{L} \left( cy\right)\neq c\mathrm{L} \left( y\right)$$
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This verifies that $$ 4 {x}^{7} + \sin\left(y\right) +  {x}^{2} {y}^{3} {y}^{'} $$ is an N1-ODE =Problem 8=

Given
From $$\displaystyle p. 10-3$$ of the lecture notes there are two integration constants

$${k}_{1}$$ is contained in $$\displaystyle (1) p. 10-3 $$


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h \left( x\right)= \exp \left[ \int^{x} {a}_{0} \left( s\right)ds\right] $$ $$
 * $$\displaystyle (Eq 8.1)
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$$ {k}_{2}$$ is contained in $$\displaystyle (3) p. 10-3 $$


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\int^{x} h\left( s\right)b\left( s\right)ds $$ $$
 * $$\displaystyle (Eq 8.2)
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Find
Show that $$ {k}_{1}$$ is not necessary

Solution
To show this we will use $$\displaystyle (6) p. 10-3$$ and show that $$ {k}_{1}$$ cancels out of the equation, but first we must find the constants $$ {k}_{1}$$ and $$ {k}_{2}$$ from equations $$\displaystyle (Eq 8.1) $$ and $$\displaystyle (Eq 8.2) $$


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y \left( x\right)= \frac{1}{h \left( x\right)} \int^{x}h \left(s \right)b \left( s\right)ds $$ $$
 * $$\displaystyle (6) p. 10-3
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When we perform the integration in the RHS of $$\displaystyle (Eq 8.1) $$ it becomes


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\exp \left[ \int^{x} {a}_{0} \left( s\right)ds\right]= \exp \left[  \int {a}_{0} \left( x\right)dx + k \right] $$ $$
 * $$\displaystyle (Eq 8.3)
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Using the properties for exponential functions $$ \exp \left[ a+b\right] = \exp \left[ a\right]\exp \left[b\right]$$

Now $$\displaystyle (Eq 8.3)$$ becomes


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\exp \left[ \int {a}_{0} \left( x\right)dx \right]\underbrace{\exp \left[k\right]}_{=: k_1} $$ $$
 * $$\displaystyle (Eq 8.4)
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Plugging this back into $$\displaystyle (Eq 8.1) $$ gives us
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$$ h \left( x\right)= k_{1}\exp \left[ \int {a}_{0} \left( x\right)dx\right] $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq 8.5)
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To find $$ {k}_{2}$$ we will evaluate the integral in $$\displaystyle (Eq 8.2) $$


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$$ \int^{x} h\left( s\right)b\left( s\right)ds= \int h\left( x\right)b\left( x\right)dx + k_{2} $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq 8.6)
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Finally plug equations $$\displaystyle (Eq 8.5) $$ and $$\displaystyle (Eq 8.6) $$ into $$\displaystyle (6) p. 10-3 $$


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y \left( x\right)= \frac{1}{\cancelto{}{k_{1}}\exp \left[  \int {a}_{0} \left( x\right)dx\right]}\left[ \int \cancelto{}{k_{1}}\exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.7)
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This shows $$k_1$$ is not necessary
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$$ y \left( x\right)= \frac{1}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]}\left[ \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\right] $$ $$
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 * <p style="text-align:right;">$$\displaystyle (Eq 8.8)
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Given
From the meeting 10 lecture notes


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y \left( x\right)= \frac{1}{h \left( x\right)} \int^{x}h \left(s \right)b \left( s\right)ds $$ $$
 * <p style="text-align:right;">$$\displaystyle (6) p. 10-3
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Similarly from King et al, p. 512


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y = A\exp \left\{  - \int ^{x}{P} \left( t\right)dt\right\}+\exp \left\{  - \int ^{x}{P} \left( t\right)dt\right\}\int ^{x}{Q} \left( s\right)\exp \left\{  \int ^{x}{P} \left(t\right)dt\right\}ds $$ $$
 * <p style="text-align:right;">$$\displaystyle K p. 512
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Find
Show that $$\displaystyle (6) p. 10-3 $$ agrees with $$\displaystyle K p. 512 $$, i.e.


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y = A{y}_{H}\left(x \right)+{y}_{P}\left(x \right) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.9)
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Solution
To show that thes two equations agree we need to get them in the form of $$\displaystyle (Eq 8.9)$$. Fortunately, $$\displaystyle K p. 512 $$ is already in this form where


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A{y}_{H}\left(x \right)= A\exp \left\{ - \int ^{x}{P} \left( t\right)dt\right\} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.10)
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{y}_{P}\left(x \right)= \exp \left\{ - \int ^{x}{P} \left( t\right)dt\right\}\int ^{x}{Q} \left( s\right)\exp \left\{  \int ^{x}{P} \left(t\right)dt\right\}ds $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.11)
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Now, using $$\displaystyle (Eq 8.8) $$ which is the expanded form of $$\displaystyle (6) p. 10-3 $$ we can further expand and rearrange to get


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y \left( x\right)= \frac{1}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]} \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +\frac{k_2}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.12)
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y \left( x\right)= \exp \left[  -\int {a}_{0} \left( x\right)dx\right] \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\exp \left[  -\int {a}_{0} \left( x\right)dx\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.13)
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Now we have $$\displaystyle (6) p. 10-3 $$ in the form of $$\displaystyle (Eq 8.9)$$ where


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A{y}_{H}\left(x \right)= k_2\exp \left[ -\int {a}_{0} \left( x\right)dx\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.14)
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{y}_{P}\left(x \right)= \exp \left[ -\int {a}_{0} \left( x\right)dx\right] \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.15)
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By comparing $$\displaystyle (Eq 8.10) $$ with $$\displaystyle (Eq 8.14)$$ and $$\displaystyle (Eq 8.11) $$ with $$\displaystyle (Eq 8.15) $$ we can see that these equations are in agreement

Given

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{y}^{'}+ {a}_{0}y=0 $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.16)
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Find
Find $$ y_H$$

Solution
Since $$\displaystyle (Eq 8.16)$$ is the homogeneous portion $$\displaystyle (2) p. 10-3$$ and is seperable, we can solve directly for $$ y_H$$


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\frac{1}{y}{y}^{'} = -{a}_{0} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.17)
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Integration of this equation gives


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\ln \left( y\right)= -\int^{x} {a}_{0} \left( s\right)ds + C $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.18)
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Solving for $$y$$ gives


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y= \exp\left[ -\int^{x} {a}_{0} \left( s\right)ds + C \right]= A\exp\left[ -\int^{x} {a}_{0} \left( s\right)ds\right]= y_H $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.19)
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