User:Egm6321.f10.team03/Hwk2

=Homework Two Summary=

= Problem 1 - Non-linearity of in general N1_ODE =

Given
Given is the in general form of a N1_ODE:
 * {| style="width:100%" border="0"

$$M(x,y)+N(x,y)\frac{dy}{dx}=0$$
 * $$\,(Eq. 1.1) $$
 * style= |
 * }

Find
It has to be shown, that IN GENERAL (Eq. 1.1) is non-linear

Solution
To show that (Eq. 1.1) is in general non-linear it needs to meet the non-linearity condition, which is:
 * {| style="width:100%" border="0"

$$F\left( \alpha u+\beta v \right)\ne F\left( \alpha u \right)+F\left( \beta v \right)=\alpha F\left( u \right)+\beta F\left( v \right)$$ Applying the condition above to (Eq. 1.1), we have to see if the following equation is true:
 * $$\,(Eq. 1.2) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$cM(x,y)+cN(x,y)\frac{dy}{dx}\overset{?}{\mathop{=}}\,M(x,cy)+N(x,cy)\frac{d(cy)}{dx}$$, where $$c$$ is constant
 * $$\,(Eq. 1.3) $$
 * style= |
 * }

Now, since N(x,y) can be ANY function, like a cos(2xy) which is non-linear, it turns out that (Eq. 1.3) is not true and is IN GENERAL not linear since:
 * {| style="width:100%" border="0"

$$cM(x,y)+cN(x,y)\frac{dy}{dx}\ne M(x,cy)+N(x,cy)\frac{dcy}{dx}=M(x,cy)+cN(x,cy)\frac{dy}{dx}$$ So it can be said that: :{| style="width:30%" border="0" $$M(x,y)+N(x,y)\frac{dy}{dx}=0\ is\ in\ general\ non\ linear!$$ For particular solutions it might be the case, that (Eq. 1.1) is linear, but a particular solution does not describe the most general case of an equation.
 * $$\,(Eq. 1.4) $$
 * style= |
 * }
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * }
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Solved and Proof read by Egm6321.f10.team3.Sudheesh 17:01, 20 September 2010 (UTC)
 * Solved and Proof read by Adam Franklin 20:23, 20 September 2010 (UTC)

= Problem 2 - Verification of an equation to be a N1-ODE=

Given

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

$$ 4 {x}^{7} + \sin\left(y\right) +  {x}^{2} {y}^{3} {y}^{'} = 0 $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * $$\displaystyle (7) p. 7-1
 * }

Find
Verify that $$ 4 {x}^{7} + \sin\left(y\right) +  {x}^{2} {y}^{3} {y}^{'} $$ is a N1-ODE

Solution
For $$\displaystyle (7) p. 7-1$$ (shown above in "Given") to be linear it must obey the following conditions for linearity given below:
 * {| style="width:100%" border="0"

f(cx) = cf(x) $$ where $$ c \in \mathbb{R}$$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq 2.1)
 * style= |
 * }


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

f(x+y) = f(x) + f(y) $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * $$\displaystyle (Eq.2.2)
 * style= |
 * }

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


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

$$
 * Let $$ \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)
 * style= |
 * }


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

$$
 * $$ \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)
 * style= |
 * }


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

$$ But
 * $$ \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)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$
 * $$ 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)
 * style= |
 * }


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

$$ \Rightarrow 4 {x}^{7} + sin \left( cy\right) + {c}^{4}{x}^{2} {y}^{3} {y}^{'}\neq c \left( 4 {x}^{7} + sin \left( y\right) + {x}^{2} {y}^{3} {y}^{'}\right)$$ By inspection, $$ \mathrm{L} \left( cy\right)\neq c\mathrm{L} \left( y\right)$$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style= |
 * }
 * }

Also by inspection, $$\displaystyle (7) p. 7-1$$ is first order because the equation contains the term $$\displaystyle y'$$ but not $$\displaystyle y''$$, which would make the equation 2nd order.

This verifies that $$ 4 {x}^{7} + \sin\left(y\right) +  {x}^{2} {y}^{3} {y}^{'} $$ is an N1-ODE

Contributing Members

 * Solved and posted by --Egm6321.f10.team3.cook 23:35, 19 September 2010 (UTC)
 * Proofed by Kurt Schulze
 * Proof read by : Egm6321.f10.team03.sigillo 01:04, 21 September 2010 (UTC)
 * Proof read by : Adam Franklin 20:04, 21 September 2010 (UTC)

=Problem 3: Linear independence of two homogeneous solutions=

Given
Given a linear second order differential equation with varying coefficient of the form:


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

$$\,{a}_{2} \left( x\right)\ddot{y}+{a}_{1} \left(x \right) \dot{y}+{a}_{0} \left( x\right)y=f\left(x \right)$$
 * $$\,(Eq.3.1)$$
 * }



Assume $$L_2=f \left( x \right)$$, a superposition of solution can be written as:


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

$$\,L_2\left( y_H \right)+L_2 \left( y_P\right)=f \left( x\right)$$
 * <p style="text-align:right">$$\,(Eq.3.2)$$
 * }

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

$$\,y_{H}=c \left( x\right)+d \left( x \right)$$
 * <p style="text-align:right">$$\,(Eq.3.3)$$
 * }

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

$$\,{y^1_{H}} \left( x\right)=x$$ and
 * <p style="text-align:right">$$\,(Eq.3.4)$$
 * }
 * {|style="width:100%" border="0"

$$\,{y^2_{H}} \left( x\right)= \frac{x}{2} \log \left(  \frac{1+x}{1-x}\right)-1$$
 * <p style="text-align:right">$$\,(Eq.3.5)$$
 * }

Find
Plot functions $${y^1_{H}} \left( x\right)$$ and $${y^2_{H}} \left( x\right)$$ and show linear independence between the two functions.

Solution
In general, linear dependence among n number of vectors can be defined as $$a_1 \mathbf{v}_1 + \cdots + a_n \mathbf{v}_n=0 $$. In this case, linear independence can be defined as


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

$$\,{y^1_{H}} \left( x\right) \neq \alpha{y^2_{H}} \left( x\right)$$ or there exists no constant $$\alpha$$ that makes equations 3.4 and 3.5 equivalent for all $$\alpha$$ and for all $$x$$ in the real domain. To prove this, let us assume:
 * <p style="text-align:right">$$\,(Eq.3.6)$$
 * }


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

$$\,\alpha= \frac{{y^1_{H}}\left( x\right)}{{y^2_{H}} \left( x\right)}$$
 * <p style="text-align:right">$$\,(Eq.3.7)$$
 * }

Evaluating equation 3.7 at $$x=\frac{1}{2}$$ gives us $$\alpha=-0.6893$$.


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

$$\,{y^1_{H}} \left( x\right) =\alpha{y^2_{H}} \left( x\right)$$ Evaluating equation 3.8 with the assumed value of $$\alpha$$ at $$x=\frac{1}{4}$$ yields:
 * <p style="text-align:right">$$\,(Eq.3.8)$$
 * }


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

$$.25 \neq 0.6453$$
 * <p style="text-align:right">$$\,(Eq.3.9)$$
 * }
 * FunctionsY1H_and_Y2H.jpg

The values are not equivalent for all $$x$$ in the domain, therefore the two functions are linearly independent. Further more, figure 3.1 agrees with this conclusion.

Contributing Members

 * Solved and authored by James Roark 12:32, 18 September 2010 (UTC)
 * Proof read by: Egm6321.f10.team03.sigillo 01:00, 21 September 2010 (UTC)

= Problem 4 - Create Exact Nonlinear First Order Ordinary Differential Equation (N1-ODE) =

Given
<H4>Conditions of Exactness:</H4>


 * For function $$F\left( . \right)$$ to be exact:


 * $$\left( 1 \right)$$ $$F\left( . \right)$$ must be in the form:


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

$$
 * $$F\left( x,y,{{y}^{1}} \right)=M\left( x,y \right)+N\left( x,y \right){{y}^{'}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.1)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.1)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"


 * Where $$M\left( x,y \right)$$ and $$N\left( x,y \right)$$ are functions of $$\,x$$  and $$\,y$$ and where $$=\frac{dy}{dx}$$
 * }
 * }
 * }


 * $$\left( 2 \right)$$ $$M\left( x,y \right)$$, and $$N\left( x,y \right)$$ must satisfy:


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

$$
 * $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.2)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.2)
 * }
 * }

<H4>Given function:</H4>


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

$$
 * $$\phi \left( x,y \right)={{x}^{2}}{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\log \left( {{x}^{3}}{{y}^{2}} \right)=k$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.3)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.3)
 * }
 * }

Find

 * $$\left( 1 \right)$$ Find function $$F\left( . \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


 * $$\left( 2 \right)$$ Verify that $$F\left( x,y,{{y}^{1}} \right)$$ is an Exact N1-ODE


 * $$\left( 3 \right)$$ Invent three more Exact N1-ODE

Solution

 * $$\left( 1 \right)$$ Find function $$F\left( x,y,{{y}^{1}} \right)$$:


 * Derivative of $$\phi \left( x,y \right)$$ is given by
 * {| style="width:100%" border="0"

$$
 * $$\frac{d\phi }{dx}=\frac{\partial \phi }{\partial x}+\frac{\partial \phi }{\partial y}\cdot {{y}^{'}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.4)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.4)
 * }
 * }


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

$$
 * $$\frac{\partial \phi }{\partial x}=\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{1}\cdot 3{{x}^{2}}{{y}^{2}} \right]=2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x}\Rightarrow M(x,y)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.5)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.5)
 * }
 * }


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

$$
 * $$\frac{\partial \phi }{\partial y}=\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{1}\cdot 2{{x}^{3}}y \right]=\frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y}\Rightarrow N(x,y)$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.6)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.6)
 * }
 * Substituting Eq.4.5 & 4.6 into Eq.4.4 yields:
 * Substituting Eq.4.5 & 4.6 into Eq.4.4 yields:


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

$$
 * $$\frac{d\phi }{dx}=\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right]+\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right]\cdot {{y}^{'}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.7)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.7)
 * }
 * Therefore:
 * Therefore:


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

$$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=\underbrace{\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right]}_{M\left( x,y \right)}+\underbrace{\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right]}_{N\left( x,y \right)}\cdot {{y}^{'}}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.8)
 * }
 * }


 * $$\left( 2 \right)$$ Verify that $$F\left( x,y,{{y}^{1}} \right)$$ is an Exact N1-ODE:


 * The function $$F\left( x,y,{{y}^{1}} \right)$$ is a first order nonlinear differential equation as its highest order is 1 and does not satisfy the condtion of superposition (linearity).
 * The function $$F\left( x,y,{{y}^{1}} \right)$$ given in Eq. 4.8 has the structure of Eq.4.1 and hence first condition of Exactness is satisfied.
 * Definiton of the second condition of exactness yields to:
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}\left( 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right)=2x\cdot \frac{3}{2}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+0=3x{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.9)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.9)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial N}{\partial x}=\frac{\partial }{\partial x}\left( \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right)=\frac{3}{2}\cdot 2x{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+0=3x{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.10)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.10)
 * }
 * Eq.4.9 and 4.10 implies,
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

$$
 * $$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$$
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.11)
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.11)
 * }
 * Eq.4.11 satisfy the second condition of Exactness.
 * Eq.4.11 satisfy the second condition of Exactness.


 * Since both conditions of exactness are satisifed,
 * {| style="width:100%" border="0"


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=\underbrace{\left[ 2x{{y}^{{}^{3}\!\!\diagup\!\!{}_{2}\;}}+\frac{3}{x} \right]}_{M\left( x,y \right)}+\underbrace{\left[ \frac{3}{2}{{x}^{2}}{{y}^{{}^{1}\!\!\diagup\!\!{}_{2}\;}}+\frac{2}{y} \right]}_{N\left( x,y \right)}\cdot {{y}^{'}}$$ is an Exact N1-ODE
 * }
 * $$\left( 3 \right)$$ Invent three more Exact N1-ODE:
 * Example 1:
 * Consider, $$\phi \left( x,y \right)=3x\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}}= k $$
 * Example 1:
 * Consider, $$\phi \left( x,y \right)=3x\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}}= k $$


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


 * $$\frac{\partial \phi }{\partial x}=3\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}}\Rightarrow M(x,y)$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"


 * $$\frac{\partial \phi }{\partial y}=3x\cdot {{e}^{y}}+8y\cdot {{e}^{x}}\Rightarrow N\left( x,y \right)$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


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


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=M\left( x,y \right)+N\left( x,y \right)\cdot {{y}^{'}}=0$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"

$$F\left( x,y,{{y}^{1}} \right)=\left[ 3\cdot {{e}^{y}}+4{{y}^{2}}\cdot {{e}^{x}} \right]+\left[ 3x\cdot {{e}^{y}}+8y\cdot {{e}^{x}} \right]\cdot {{y}^{'}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.12)
 * }
 * This is an Exact N1-ODE
 * This is an Exact N1-ODE


 * Example 2:
 * Consider, $$\phi \left( x,y \right)=5y\log \left( {{x}^{2}} \right)+3{{x}^{2}}\sin \left( {{y}^{3}} \right)=k$$


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


 * $${{\phi }_{x}}\left( x,y \right)=5y\cdot \left( \frac{1}\cdot 2x \right)+6x\sin \left( {{y}^{3}} \right)=10\frac{y}{x}+6x\sin \left( {{y}^{3}} \right)\Rightarrow M(x,y)$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"


 * $${{\phi }_{y}}\left( x,y \right)=5\log \left( {{x}^{2}} \right)+3{{x}^{2}}\left( \cos \left( {{y}^{3}} \right)\cdot 3{{y}^{2}} \right)=5\log \left( {{x}^{2}} \right)+9{{x}^{2}}{{y}^{2}}\cos \left( {{y}^{3}} \right)\Rightarrow N\left( x,y \right)$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


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


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=M\left( x,y \right)+N\left( x,y \right)\cdot {{y}^{'}}=0$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"

$$F\left( x,y,{{y}^{'}} \right)=\left[ 10\frac{y}{x}+6x\sin \left( {{y}^{3}} \right) \right]+\left[ 5\log \left( {{x}^{2}} \right)+9{{x}^{2}}{{y}^{2}}\cos \left( {{y}^{3}} \right) \right]{{y}^{'}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.13)
 * }
 * This is an Exact N1-ODE
 * This is an Exact N1-ODE


 * Example 3:
 * Consider, $$\phi \left( x,y \right)=7{{x}^{3}}\cdot \cos \left( {{y}^{3}} \right)+2{{y}^{2}}{{e}^{{{x}^{2}}\cdot \sin y}}=k$$


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

& {{\phi }_{x}}\left( x,y \right)=7\times 3{{x}^{2}}\cdot \cos \left( {{y}^{3}} \right)+2{{y}^{2}}\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\cdot \left( 2x\cdot \sin y \right)=21{{x}^{2}}\cdot \cos \left( {{y}^{3}} \right)+4x{{y}^{2}}\cdot \sin y\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\Rightarrow M(x,y) \\ & \\ \end{align}$$
 * $$\begin{align}
 * $$\begin{align}


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


 * $${{\phi }_{y}}\left( x,y \right)=7{{x}^{3}}\cdot \left[ -\sin \left( {{y}^{3}} \right)\cdot 3{{y}^{2}} \right]+2\left[ 2y\cdot {{e}^{{{x}^{2}}\cdot \sin y}}+{{y}^{2}}\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\cdot {{x}^{2}}\cdot \cos y \right]=-21{{x}^{3}}{{y}^{2}}\sin \left( {{y}^{3}} \right)+4y{{e}^{{{x}^{2}}\cdot \sin y}}+2{{x}^{2}}{{y}^{2}}\cos y\cdot {{e}^{{{x}^{2}}\cdot \sin y}}\Rightarrow N\left( x,y \right)$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$
 * }
 * Now consider a function $$F\left( x,y,{{y}^{1}} \right)$$ such that $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}$$


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


 * $$F\left( x,y,{{y}^{1}} \right)=\frac{d\phi }{dx}=M\left( x,y \right)+N\left( x,y \right)\cdot {{y}^{'}}=0$$
 * }
 * {| style="width:100%" border="0"
 * }
 * {| style="width:100%" border="0"

$$\left( x,y,{{y}^{'}} \right)=\left[ 21{{x}^{2}}\cdot \cos \left( {{y}^{3}} \right)+4x{{y}^{2}}\cdot \sin y\cdot {{e}^{{{x}^{2}}\cdot \sin y}} \right]+\left[ -21{{x}^{3}}{{y}^{2}}\sin \left( {{y}^{3}} \right)+4y{{e}^{{{x}^{2}}\cdot \sin y}}+2{{x}^{2}}{{y}^{2}}\cos y\cdot {{e}^{{{x}^{2}}\cdot \sin y}} \right]{{y}^{'}}=0$$ $$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq. 4.14)
 * }
 * This is an Exact N1-ODE
 * This is an Exact N1-ODE

Contributing Members

 * Solved and posted by --Egm6321.f10.team3.Sudheesh 15:20, 20 September 2010 (UTC)
 * Proofed by Kurt Schulze
 * Proof read by: Egm6321.f10.team03.sigillo 01:39, 21 September 2010 (UTC)

= Problem 5 - Non exact N1_ODE=

Given
The general form of a N1_ODE is given as being:
 * {| style="width:100%" border="0"

$$F(x,y,{y}')=M(x,y)+N(x,y)f\left( {{y}'} \right)=0$$
 * <p style="text-align:right;">$$\,(Eq. 5.1) $$
 * style= |
 * }

Find
To be found is a function $$f\left( {{y}'} \right)$$ such that there is no analytical solution to $$f\left( {{y}'} \right)=\frac{-M}{N}$$ , hence a non exact N1_ODE has to be found.

Solution
Let's have a look at the chosen non-linear equation :
 * {| style="width:100%" border="0"

$$2x{{y}^{2}}+\cos \left( 3y \right)x\cdot {y}'\exp \left( {{y}'} \right)=0$$ where the three terms of the general equation (Eq. 5.1) in this application case can be found as:
 * <p style="text-align:right;">$$\,(Eq. 5.2) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$M\left( x,y \right)=2x{{y}^{2}}\quad \quad N\left( x,y \right)=\cos \left( 3y \right)x\quad \quad f({y}')={y}'\cdot \exp \left( {{y}'} \right)$$ Since the first conditoin for an exact N1_ODE is defined to be:
 * style= |
 * }
 * }
 * {| style="width:100%" border="0"

$$\,M(x,y)+N(x,y){y}'=0$$ It's needed to put (Eq. 5.2) in the form of (Eq. 5.3). Thus the next step is to solve (Eq. 5.2) for y' :
 * <p style="text-align:right;">$$\,(Eq. 5.3) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'\cdot \exp \left( {{y}'} \right)=\frac{-2x{{y}^{2}}}{\cos \left( 3y \right)x}=\frac{-2{{y}^{2}}}{\cos \left( 3y \right)}$$ trying to get y' out of the exp-function we apply the logarithm-funtion:
 * <p style="text-align:right;">$$\,(Eq. 5.4) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\ln \left( {y}'\cdot \exp \left[ {{y}'} \right] \right)=\ln \left( {{y}'} \right)+\ln \left( \exp \left[ {{y}'} \right] \right)=\ln \left( \frac{-2{{y}^{2}}}{\cos \left( 3y \right)} \right)$$
 * <p style="text-align:right;">$$\,(Eq. 5.5) $$
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\ln \left( {{y}'} \right)+{y}'=\ln \left( \frac{-2{{y}^{2}}}{\cos \left( 3y \right)} \right)$$ :{| style="width:90%" border="0" As it can be seen, there is on analytical solution for y' in (Eq. 5.6), this means that (Eq. 5.2) can not be put in the form of (Eq. 5.3), hence it is NOT an exact N1_ODE !
 * <p style="text-align:right;">$$\,(Eq. 5.6) $$
 * style= |
 * }
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * }
 * }

Contributing Members

 * Solved by Michele Sigilló
 * Posted by Michele Sigilló
 * Proof read by:Egm6321.f10.team3.Sudheesh 15:29, 20 September 2010 (UTC)

= Problem 6 - Verification of a particular solution y(x)=

Given
Suppose a function $$\phi $$ can be found to be a reduced order solution and its derivative is equal to:
 * {| style="width:100%" border="0"

$$\frac{d\phi (x,y,{y}')}{dx}=F(x,y,{y}')=75{{x}^{4}}+\cos \left( y \right){y}'=0$$ $$ and the explicit solution for $$\,y(x)$$ is given as:
 * <p style="text-align:right;">$$\,(Eq. 6.1)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$y(x)={{\sin }^{-1}}\left( k-15{{x}^{5}} \right)$$. $$
 * <p style="text-align:right;">$$\,(Eq. 6.2)
 * style= |
 * }

Find
It has to be proven that (Eq. 6.2) satisfies (Eq. 6.1)

Solution
In general $$F(x,y,{y}')$$ can be written as:
 * {| style="width:100%" border="0"

$$\,F(x,y,{y}')=M(x,y)+N(x,y){y}'=0$$ $$ where from (Eq. 6.1) it is known that
 * <p style="text-align:right;">$$\,(Eq. 6.3)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,M(x,y)=75{{x}^{4}}$$ $$ and
 * <p style="text-align:right;">$$\,(Eq. 6.4)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\,N(x,y)=\cos \left( y \right)$$ $$ The chosen approach is to derivate $$y(x)$$ ,insert the result into (Eq. 6.2) and check if it satisfies the condition $$F(x,y,{y}')=0$$.In order to have a better overview during deferentiating, let's define
 * <p style="text-align:right;">$$\,(Eq. 6.5)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$u(x)=\left( k-15{{x}^{5}} \right)$$ $$ The derivative is defined to be:
 * <p style="text-align:right;">$$\,(Eq. 6.6)
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{dy}{dx}=\frac{dy}{du}\frac{du}{dx}$$ $$ which in the particular case it is:
 * <p style="text-align:right;">$$\,(Eq. 6.7)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{dy}{du}=\frac{1}{\sqrt{1-{{u}^{2}}}}$$ $$ and
 * <p style="text-align:right;">$$\,(Eq. 6.8)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\frac{du}{dx}=-75{{x}^{4}}$$ $$ Inserting (Eq. 6.8) and (Eq. 6.9) into (Eq. 6.7) while substituiting $$u$$ back, (Eq. 6.7) can be rewritten as:
 * <p style="text-align:right;">$$\,(Eq. 6.9)
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{-75{{x}^{4}}}{\sqrt{1-{{\left( k-15{{x}^{5}} \right)}^{2}}}}$$ $$ The key-point of this Problem in terms of quickness, lays in the fact that is has to be recognized that $${{\left( k-15{{x}^{5}} \right)}^{2}}$$ is the same like $${{sin }^{2}}\left( y \right)$$, hence (Eq. 6.10) becomes:
 * <p style="text-align:right;">$$\,(Eq. 6.10)
 * style= |
 * }
 * {| style="width:100%" border="0"

$${y}'=\frac{-75{{x}^{4}}}{\sqrt{1-{{\sin }^{2}}\left( y \right)}}=\frac{-75{{x}^{4}}}{\sqrt{{{\cos }^{2}}\left( y \right)}}=\frac{-75{{x}^{4}}}{\cos \left( y \right)}$$ $$ Now substituting y' in (Eq. 6.1) with (Eq. 6.11) we get:
 * <p style="text-align:right;">$$\,(Eq. 6.11)
 * style= |
 * }
 * {| style="width:100%" border="0"

$$75{{x}^{4}}+\cos \left( y \right)\frac{-75{{x}^{4}}}{\cos \left( y \right)}=75{{x}^{4}}-75{{x}^{4}}=0$$ $$ :{| style="width:70%" border="0" Left-Hand-Side and Right-Hand-Side of the equation are the same, hence (Eq. 6.2) satisfies (Eq. 6.1) !
 * <p style="text-align:right;">$$\,(Eq. 6.12)
 * style= |
 * }
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * }
 * }

Contributing Members

 * Solved by Kurt Schulze
 * Posted by Michele Sigilló
 * Proof read by: Egm6321.f10.team03.sigillo 00:39, 21 September 2010 (UTC)

= Problem 7 - Solving a L1-ODE using Euler integrationfactor method=

Part 1
Solution to linear, first order, ordinary differential equation with varying coefficients using Euler integrating factor method

Given
Given


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

$$a_{1}\left( x \right){y}'+a_{0}\left( x \right)y=b\left( x \right)$$


 * <p style="text-align:right">$$\,(Eq.7.1)$$
 * }
 * where
 * {|style="width:100%" border="0"

$$a_{1}\left( x \right)=1$$
 * <p style="text-align:right">$$\,(Eq.7.2)$$
 * }


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

$$a_{0}\left( x \right)=x$$
 * <p style="text-align:right">$$\,(Eq.7.3)$$
 * }


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

$$b\left( x \right)=2x+3$$
 * <p style="text-align:right">$$\,(Eq.7.4)$$
 * }

Find
Find $$y$$ using Euler Integration Factor

Solution
Substituting equations 7.2, 7.3, and 7.4 into 7.1:


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

$$\,{y}'+xy=2x+3$$ Equation 7.5 is of the form $${y}'+P\left( x \right)y=Q\left( x \right)$$ The integrating factor for this form is $$I\left( x \right)=\exp \left[{P}\left( x \right)dx \right]$$. For equation 7.5, the integrating factor is
 * <p style="text-align:right">$$\,(Eq.7.5)$$
 * }


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

$$I\left( x \right)=\exp \left[ \int{x}dx \right]=\exp \left[ \frac{x^{2}}{2} \right]$$
 * <p style="text-align:right">$$\,(Eq.7.6)$$
 * }

Applying the integrating factor for equation 7.5:


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

$$\exp \left[ \frac{2} \right]\frac{dy}{dx}+\exp \left[ \frac{2} \right]xy=2x\exp \left[ \frac{2} \right]+3\exp \left[ \frac{2} \right]$$
 * <p style="text-align:right">$$\,(Eq.7.7)$$
 * }

(Eq. 7.7) reduces to
 * {|style="width:100%" border="0"

$$\frac{d}{dx}\left( y\exp \left( \frac{2} \right) \right)=2x\exp \left[ \frac{2} \right]+3\exp \left[ \frac{2} \right]$$
 * <p style="text-align:right">$$\,(Eq.7.8)$$
 * }

Integrating


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

$$y\exp \left( \frac{2} \right)=\int{2x\exp \left[ \frac{2} \right]}\,dx+\int{3\exp \left[ \frac{2} \right]}\,dx$$
 * <p style="text-align:right">$$\,(Eq.7.9)$$
 * }

results in the following equation:
 * {|style="width:100%" border="0"

$$y\exp \left( \frac{2} \right)=2\exp \left[ \frac{2} \right]\,+\int{3\exp \left[ \frac{2} \right]}\,dx$$
 * <p style="text-align:right">$$\,(Eq.7.10)$$
 * }

There is no known anti-derivative of the last term, so using mathematical analysis tools the last therm results in the error-function. So that rearranging the equation we get for y(x):

$$y=2+3\exp \left[ \frac{x^{2}}{2} \right]\left[ \sqrt{\frac{\pi }{2}} erfi\left( \frac{x}{\sqrt{2}} \right) \right]$$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right">$$\,(Eq.7.11)$$
 * }

Part 2
A general solution from the integrating factor method

Given
Assume $$a\left( x \right)\ne 0$$ for all $$x$$, rearranging equation 7.1 gives


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

$${y}'+\frac{a_{0}\left( x \right)}{a_{1}\left( x \right)}y=\frac{b\left( x \right)}{a_{1}\left( x \right)}$$
 * <p style="text-align:right">$$\,(Eq.7.12)$$
 * }

Find
Find an expression for $$y\left( x \right)$$ in terms of $$a_{0}$$, $$a_{1}$$ and $$b$$

Solution
Define $${P}\left( x \right)=\frac{a_{0}\left( x \right)}{a_{1}\left( x \right)}$$ and $$Q\left( x \right)=\frac{b\left( x \right)}{a_{1}\left( x \right)}$$ Equation 7.12 is then in the form $${y}'+{P}\left( x \right)y=Q\left( x \right)$$ The integrating factor for this form is $$I\left( x \right)=\exp \left[ \int{ {P}\left( x \right)}dx \right]$$ Introducing a dummy factor $$t$$ and substituting in for $$P\left( x \right)$$


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

$$\exp \left[ \int\limits_ – ^{x}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]\frac{dy}{dx}+\exp \left[ \int\limits_ – ^{x}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]\frac{a_{0}\left( x \right)}{a_{1}\left( x \right)}y=\frac{b\left( x \right)}{a_{1}\left( t \right)}\exp \left[ \int\limits_ – ^{x}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]$$ Reducing
 * <p style="text-align:right">$$\,(Eq.7.13)$$
 * }
 * {|style="width:100%" border="0"

$$\frac{d}{dx}\left[ \exp \left[ \int\limits_ – ^{x}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]y \right]=\frac{b\left( x \right)}{a_{1}\left( x \right)}\exp \left[ \int\limits_ – ^{x}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]$$ Integrating
 * <p style="text-align:right">$$\,(Eq.7.14)$$
 * }

$$y=A\exp \left[ -\int\limits_ – ^{x}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]+\exp \left[ -\int\limits_ – ^{x}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]\int\limits_ – ^{x}{\frac{b\left( s \right)}{a_{1}\left( s \right)}ds\exp }\left[ -\int\limits_ – ^{s}{\frac{a_{0}\left( t \right)}{a_{1}\left( t \right)}dt} \right]ds$$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right">$$\,(Eq.7.15)$$
 * }

Part 3
Solution to linear, first order, ordinary differential equation with varying coefficients using Euler integrating factor method.

Given
Given an equation in the form of 7.1, where $$a_{1}\left( x \right)=x^{2}+1$$, $$a_{0}\left( x \right)=x$$, and $$b\left( x \right)=2x$$, equation 7.1 becomes
 * {|style="width:100%" border="0"

$$\left( x^{2}+1 \right){y}'+xy=2x$$ Rearranging
 * <p style="text-align:right">$$\,(Eq.7.16)$$
 * }
 * {|style="width:100%" border="0"

$${y}'+\frac{x}{x^{2}+1}y=\frac{2x}{x^{2}+1}$$ Equation 7.17 is in the form $${y}'+{P}\left( x \right)y=Q\left( x \right)$$ The integrating factor for this form is $$I\left( x \right)=\exp \left[ \int{ {P}\left( x \right)}dx \right]$$ or
 * <p style="text-align:right">$$\,(Eq.7.17)$$
 * }


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

$$I\left( x \right)=\exp \left[ \int{\frac{x}{x^{2}+1}}dx \right]=\exp \left[ \frac{1}{2}\ln \left( x^{2}+1 \right) \right]=\sqrt{x^{2}+1}$$||<p style="text-align:right">$$\,(Eq.7.18)$$ Multiply 7.17 by 7.18 and simplify
 * }
 * {|style="width:100%" border="0"



$$\frac{d}{dx}\left( y\sqrt{x^{2}+1} \right)=\frac{2y}{\sqrt{x^{2}+1}}$$ Integrating
 * <p style="text-align:right">$$\,(Eq.7.19)$$
 * }
 * {|style="width:100%" border="0"

$$\int\limits_ – ^{x}{\frac{d}{dx}\left( y\sqrt{x^{2}+1} \right)}=\int\limits_ – ^{x}{\frac{2y}{\sqrt{x^{2}+1}}dy}$$ Gives
 * <p style="text-align:right">$$\,(Eq.7.20)$$
 * }



$$\,y=2$$
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * style="width:2%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right">$$\,(Eq.7.21)$$
 * }

Contributing Members

 * Solved by Kurt Schulze
 * Posted by James Roark 12:32, 18 September 2010 (UTC)
 * Proof read by Adam Franklin 20:23, 20 September 2010 (UTC)
 * Proof read by Egm6321.f10.team03.sigillo 20:45, 21 September 2010 (UTC)

=Problem 8=

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

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


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

h \left( x\right)= \exp \left[ \int^{x} {a}_{0} \left( s\right)ds\right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.1)
 * style= |
 * }

$$ {k}_{2}$$ is contained in $$\displaystyle (3) p. 10-3 $$


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

\int^{x} h\left( s\right)b\left( s\right)ds $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.2)
 * style= |
 * }

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 $$\displaystyle (Eq 8.2) $$


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

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
 * style= |
 * }

When we perform the integration in the RHS of $$\displaystyle (Eq 8.1) $$ it becomes


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

\exp \left[ \int^{x} {a}_{0} \left( s\right)ds\right]= \exp \left[  \int {a}_{0} \left( x\right)dx + k \right] $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.3)
 * style= |
 * }

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


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

\exp \left[ \int {a}_{0} \left( x\right)dx \right]\underbrace{\exp \left[k\right]}_{=: k_1} $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.4)
 * style= |
 * }

Plugging this back into $$\displaystyle (Eq 8.1) $$ gives us
 * {| style="width:100%" border="0"

$$ h \left( x\right)= k_{1}\exp \left[ \int {a}_{0} \left( x\right)dx\right] $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq 8.5)
 * style= |
 * }

To find $$ {k}_{2}$$ we will evaluate the integral in $$\displaystyle (Eq 8.2) $$


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

$$ \int^{x} h\left( s\right)b\left( s\right)ds= \int h\left( x\right)b\left( x\right)dx + k_{2} $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq 8.6)
 * style= |
 * }

Finally plug equations $$\displaystyle (Eq 8.5) $$ and $$\displaystyle (Eq 8.6) $$ into $$\displaystyle (6) p. 10-3 $$


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

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)
 * style= |
 * }

This shows $$k_1$$ is not necessary
 * {| style="width:100%" border="0"

$$ y \left( x\right)= \frac{1}{\exp \left[  \int {a}_{0} \left( x\right)dx\right]}\left[ \int \exp \left[  \int {a}_{0} \left( x\right)dx\right]b \left( x\right)dx +k_2\right] $$ $$
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * style="width:25%; padding:1px; border:2px solid #0000FF" |
 * <p style="text-align:right;">$$\displaystyle (Eq 8.8)
 * style= |
 * }

Given
From the meeting 10 lecture notes


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

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
 * style= |
 * }

Similarly from King et al, p. 512


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

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
 * style= |
 * }

Find
Show that $$\displaystyle (6) p. 10-3 $$ agrees with $$\displaystyle K p. 512 $$, i.e.


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

y = A{y}_{H}\left(x \right)+{y}_{P}\left(x \right) $$ $$
 * <p style="text-align:right;">$$\displaystyle (Eq 8.9)
 * style= |
 * }

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


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

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)
 * style= |
 * }


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

{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)
 * style= |
 * }

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


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

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)
 * style= |
 * }


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

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)
 * style= |
 * }

Now we have $$\displaystyle (6) p. 10-3 $$ in the form of $$\displaystyle (Eq 8.9)$$ where


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

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)
 * style= |
 * }


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

{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)
 * style= |
 * }

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

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

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

Find

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


 * Find $$\displaystyle y_H$$
 * style= |
 * }
 * }

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


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

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

Integration of this equation gives


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

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

Solving for $$y$$ gives


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

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)
 * style= |
 * }

Contributing Members
Solved and Posted by:--Egm6321.f10.team3.cook 23:47, 19 September 2010 (UTC)
 * Solved and Proof read by Adam Franklin 20:23, 20 September 2010 (UTC)

=Problem 9- Find N1-ODE that is either exact or can be made exact=

Given
Class of N1-ODE is given by (eq. (1),(2)& (3), page 12-1 of the lecture note):


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

$$ \underbrace{\bar{b} \left( x\right)  c\left( y\right)}_{N\left( x,y \right)}{y'} +   \underbrace{a\left( x\right) \bar{c}\left( x\right)}_{M\left( x,y \right)} = 0 $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\,(Eq.9.1)$$
 * }

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

$$ \bar{b} \left( x\right) = \int\limits_ – ^{x}{{b} \left( s\right)}ds $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\,(Eq.9.2)$$
 * }


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

$$ \bar{c} \left( y\right) = \int\limits_ – ^{y}{{c} \left( s\right)}ds $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\,(Eq.9.3)$$
 * }

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

$$ a\left( x\right) = sin(x^3) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }


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

$$ b\left( x\right) = cos(x) $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }


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

$$ c\left( y\right) = e^{2y} $$ $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right;">$$\displaystyle
 * }

Find

 * (1) Find a N1-ODE that is either exact or can be made exact


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

\phi(x,y)=k $$ $$
 * (2) Find, $$\displaystyle
 * (2) Find, $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }

Solution
Part 1
 * {| style="width:100%" border="0"

$$ \bar{b} \left( x\right) = \int\limits_ – ^{x}{{b} \left( s\right)}ds = \int\limits_ – ^{x}{cos(s)}ds = sin(x) $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\,(Eq.9.4)$$


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

$$ \bar{c} \left( y\right) = \int\limits_ – ^{y}{{c} \left( s\right)}ds = \int\limits_ – ^{y}{e^{2s}} ds = \frac{1}{2}e^{2y} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">$$\,(Eq.9.5)$$
 * }

Now Eq.1 becomes:
 * {|style="width:100%" border="0"

$$ sin(x) e^{2y} y' + \frac{1}{2}sin(x^{3}) e^{(2y)} = 0 $$
 * <p style="text-align:right">$$\,(Eq.9.6)$$
 * }

Eq. 9.6 is the form of
 * {| style="width:100%" border="0"

$$\,M(x,y)+N(x,y)y'=0$$ and first condition of Exactness is satisfied. $$
 * <p style="text-align:right;">$$\,(Eq.9.7)
 * style= |
 * }

The condition below is not only necessary but also sufficient for Eq.9.6 to be an exact differencial equation.


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

$$ $$
 * $$\frac{\partial M }{\partial y} = \frac{\partial N }{\partial x}
 * $$\frac{\partial M }{\partial y} = \frac{\partial N }{\partial x}
 * <p style="text-align:right;">$$\displaystyle (Eq.9.8)
 * }
 * }


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


 * $$\frac{\partial M }{\partial y} = \frac{\partial (\frac{1}{2}sin(x^{3})e^{2y})}{\partial y} =sin(x^{3})e^{2y} $$
 * <p style="text-align:right">$$\,(Eq.9.9)$$
 * }
 * }
 * {| style="width:100%" border="0"


 * $$\frac{\partial N }{\partial x} = \frac{\partial (sin(x)e^{2y})}{\partial x} = cos(x)e^{2y}$$
 * <p style="text-align:right">$$\,(Eq.9.10)$$
 * }
 * }

Eq.9.9 & 9.10 reveals that


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

$$
 * $$\frac{\partial M }{\partial y}  \neq  \frac{\partial N }{\partial x} $$
 * <p style="text-align:right;">$$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * }
 * }

Hence the Eq.9.6 is not exact.

But it can be made exact by using Integrating factor method.Refer to Eq(7), page 10-1 of the lecture notes, and let's multyply Eq.(9-6) by h(x) in order to have an exact differential equation.


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

h(x) = exp[ \int^{x} {(-\frac {1}{N}(N_{x} -M_{y}))dx]} $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }
 * We can rewrite Eq.9.6 by cancelling $$\displaystyle e^{2y}$$ from both term as:
 * {| style="width:100%" border="0"

$$\underbrace{sin(x)}_{N(x,y)}{y}'+\underbrace{\frac{1}{2}sin({{x}^{3}})}_{M(x,y)}=0$$ $$ Now:
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }
 * {| style="width:100%" border="0"

h(x) = exp[ \int^{x} {(-\frac {1}{sin(x)}(cos(x) - 0))dx]} $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }


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

= exp[ \int^{x} {(-\frac {cos(x)}{sin(x)})dx]} $$ $$
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }


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

= exp[ -ln{(sin(x))}] $$ $$
 * $$\displaystyle
 * $$\displaystyle
 * <p style="text-align:right;">$$\displaystyle
 * style= |
 * }


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

= \frac {1}{sin(x)}$$
 * <p style="text-align:right">$$\,(Eq.9.11)$$
 * }

Now Exact differential equation is given by
 * {|style="width:100%" border="0"

$$ That is
 * $$h(x)[sin(x) e^{2y} y' + \frac{1}{2}sin(x^{3}) e^{(2y)}] = 0
 * $$h(x)[sin(x) e^{2y} y' + \frac{1}{2}sin(x^{3}) e^{(2y)}] = 0
 * style= |
 * }


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

$$sin(x)\frac{1}{sin(x)}y' + \frac{1}{2}sin(x^3)\frac{1}{sin(x)} = 0$$
 * style= |
 * }


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

$$ y' + \frac{1}{2} \frac{sin(x^3)}{sin(x)}= 0$$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right">$$\,(Eq.9.12)$$
 * }

Part 2


 * Find $$\displaystyle\phi(x,y)=k

$$
 * We know that
 * {| style="width:100%" border="0"

$$\,M(x,y)=\phi_{x}{(x,y)}$$ $$
 * <p style="text-align:right;">$$\,
 * <p style="text-align:right">$$\,(Eq.9.13)$$
 * }


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

$$\,N(x,y)=\phi_{y}{(x,y)}$$
 * <p style="text-align:right">$$\,(Eq.9.14)$$
 * }


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

$$\,\phi_{x}{(x,y)} = \frac{\partial \phi{(x,y)}}{\partial {x}}$$
 * <p style="text-align:right">$$\,(Eq.9.15)$$
 * }
 * {| style="width:100%" border="0"

$$\,\phi_{y}{(x,y)} = \frac{\partial \phi{(x,y)}}{\partial {y}dx}$$
 * <p style="text-align:right">$$\,(Eq.9.16)$$
 * }

By integrating Eq.9.15 and 9.16 with respect to x, y, respectively, we can have equations as given below.


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

$$\,\int \phi_{x}{(x,y)}dx = \phi {(x,y)} = \int M(x,y) dx + k(y)$$


 * <p style="text-align:right">$$\,(Eq.9.17)$$
 * }


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

$$\,\int \phi_{y}{(x,y)}dy = \phi {(x,y)} = \int N(x,y) dy + k(x)$$
 * <p style="text-align:right">$$\,(Eq.9.18)$$
 * }

Now we can calculate $$\displaystyle k(y)$$ by differentiating Eq.9.17 with respect to y and equating to Eq.9.14=>


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

$${{k}^{'}}\left( y \right)=N\left( x,y \right)=1$$ $$
 * <p style="text-align:right;">$$\,
 * style= |
 * }
 * Now integrate the above equation =>


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

$$\displaystyle k(y)=y+k1$$ $$ Then $$\displaystyle\phi(x,y)$$ is obtained by substituting $$\displaystyle k(y)$$ in Eq.9.17:
 * <p style="text-align:right;">$$\,
 * style= |
 * }
 * {| style="width:100%" border="0"

$$\phi \left( x,y \right)=\int{M\left( x,y \right)dx+y+k1}$$ $$
 * <p style="text-align:right;">$$\,
 * style= |
 * }

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

$$\phi \left( x,y \right)=\frac{1}{2}\int{\frac{\sin \left( {{x}^{3}} \right)}{\sin x}dx+y=k}$$ $$
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * style="width:2%; padding:1px; border:2px solid #FF0000" |
 * <p style="text-align:right;">$$\,
 * style= |
 * }

Contributing Members

 * Solved Part 2 and proof read by-Egm6321.f10.team3.Sudheesh 04:24, 22 September 2010 (UTC)
 * Solved and posted by-Egm6321.f10.team3.Hong SJ 22:30, 21 September 2010 (UTC)

=Contributing Team Members=
 * Adam Franklin 18:56, 12 September 2010 (UTC) solved 1 & 8 and proofed 1, 2 , 7 & 8
 * Sudheesh 18:11, 14 September 2010 (UTC), solved 1 & 4, posted 4 and proofed 1 & 5
 * Jimmy Roark 14:26, 9 September 2010 (UTC)
 * Michele Sigilló 22:07, 12 September 2010 (UTC), solved 1 & 5, posted 1,5 and 6, proof read 2,3,4,6
 * Kurt Schulze 01:10, 13 September 2010 (UTC), solved 6 & 7, proofed 2 & 4
 * Chris Cook 11:06, 14 September 2010 (UTC)
 * Hong SJ 11:12, 14 September 2010 (UTC)

=Notes and references=