User:Egm6321.f10.team2.oztekin/HW2

=Problem 1: Verify Mdx+Ndy=0 is general N1_ODE=

Given
In general form a N1_ODE can be written as follows:


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$$M\left ( x,y \right )dx+N\left ( x,y \right )dy=0 \displaystyle$$ (Eq 1.1)
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Find
Show Eq1 has a general form.

Solution
We can write Eq as


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$$M\left ( x,y \right )+N\left ( x,y \right )\frac{dy}{dx}=M\left ( x,y \right )+N\left ( x,y \right )y^{'}\displaystyle$$ (Eq 1.2)
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Now let consider about general form as follows :


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$$P\left ( x \right )y{}'+Q\left ( x \right )y=R\left ( x \right )\displaystyle$$ (Eq 1.3)
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You can see from the lecturer notes from the 10-2

If we divide both sides with P(x);


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$$y{}'+\frac{Q(x)}{P(x)}y=\frac{R(x)}{P(x)}\displaystyle$$ (Eq 1.4)
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Then we can say that ;
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$$\bar{N}(x,y)=1\displaystyle$$ (Eq 1.5)
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$$\bar{M}(x,y)=\frac{Q(x)}{P(x)}y-\frac{R(x)}{P(x)}\displaystyle$$ (Eq 1.6) So we concluded that
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$$\bar{M}(x,y)+\bar{N}(x,y)y{}'=0\displaystyle$$ (Eq 1.7)
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Another example for this porblem we can give in book that it says in the page 511 for the Variables Separable kind of differential equations.

It is shown as :


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$$g(y)\frac{dy}{dt}=f(t)\displaystyle$$ (Eq 1.8)
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To solve this ;


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$$\int g(y)\frac{dy}{dt}=\int f(t)\displaystyle$$ (Eq 1.9)
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$$\int g(y)dy=\int f(t)dt\displaystyle$$ (Eq 1.10)
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We can rearrange this


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$$g(y)\frac{dy}{dt}-f\left ( t \right )=0\displaystyle$$ (Eq 1.11)
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And again


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$$\bar{N}(t,y)=g(y),\bar{M}(t,y)=-f(t)\displaystyle$$ (Eq 1.12)
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$$\bar{N}(t,y)\frac{dy}{dt}+\bar{M}(t,y)=0\displaystyle$$ (Eq 1.13)
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An ordinary differential equation of the form


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$$y'+ P(x)y = Q(x)y^n\displaystyle$$ (Eq 1.14)
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is called a Bernoulli equation when n≠1, 0. Bernoulli equations are special because they are nonlinear differential equations with known exact solutions. Dividing by $$y^n$$ yields
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$$\frac{y'}{y^{n}} + \frac{P(x)}{y^{n-1}} = Q(x)\displaystyle$$ (Eq 1.15)
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If we use changing of variables method...
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$$w=\frac{1}{y^{n-1}}\displaystyle$$ (Eq 1.16)
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$$w'=\frac{(1-n)}{y^{n}}y'\displaystyle$$ (Eq 1.17)
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$$\frac{w'}{1-n} + P(x)w = Q(x)\displaystyle$$ (Eq 1.18)
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The substituted equation can be solved using the integrating factor
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$$h(x)= e^{(1-n)\int P(x)dx}\displaystyle$$ (Eq 1.19)
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By showing all of these now we can say that for every type first order nonlinear differansial equations can be written in Eq 1.1 type by some arrangements.

=Problem 2 : Show why Eq 7 in page (7-1) is first order and non linear=

Given
Eq 7 in page 7-1
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$$\left ( 4*x^7+sin(y) \right )+\left ( x^2*y^3 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=0 \displaystyle$$ (Eq 2.1)
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Find
Show why the equation above is first order and non linear

Solution
Since we define our first order differential $$F\left ( x,y,y{}' \right )=0$$ there must be only first degree differential of dependent variable in the function.If we look at the example we can notice that there is only first degree differential of dependent variable multiplied by N(x,y) so we conclude that this is first order differential equation.


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$$F\left ( x,y \right )=\left ( 4*x^7+sin(y) \right )+\left ( x^2*y^3 \right )\frac{\mathrm{d} y}{\mathrm{d} x}=0\displaystyle $$ (Eq 2.2)
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$$F\left ( x,(\alpha u+\beta v) \right )=\left ( 4*x^7+sin(\alpha u+\beta v) \right )+\left ( x^2*(\alpha u+\beta v)^3 \right )(\alpha \frac{\mathrm{d} u}{\mathrm{d} x}+\beta \frac{\mathrm{d} v}{\mathrm{d} x})=0\displaystyle $$ (Eq 2.3)
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$$F\left ( x,\alpha u \right )+F\left ( x,\beta v \right )=\left ( 4*x^7+sin(\alpha u) \right )+\left ( x^2*(\alpha u)^3 \right )\alpha \frac{\mathrm{d} u}{\mathrm{d} x}+\left ( 4*x^7+sin(\beta v) \right )+\left ( x^2*(\beta v)^3 \right )\beta \frac{\mathrm{d} v}{\mathrm{d} x} \displaystyle $$ (Eq 2.4)
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$$F\left ( x,\alpha u \right )+F\left ( x,\beta v \right )\neq F\left ( x,(\alpha u+\beta v) \right )\displaystyle$$ (Eq 2.5)
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=Problem3:Plot Homogeneous Solutions And Show independence=

Given

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$$y_{H}^{1}(x)=x\displaystyle $$ (Eq 3.1)
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$$y_{H}^{2}(x)=\frac{x}{2}*log\left ( \frac{(1+x)}{(1-x)} \right )-1 \displaystyle $$ (Eq 3.2)
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Linear Dependence and Independence
Let us consider defined vector space called V. And x1,x2,x3,....xk are the vectors in this space. E={x1,x2,.....xk} is the cluster of these vectors in this vector space. Now look at the condition of c1*x1+c2*x2+......ck*xk=0.

If the coefficients of x1,x2,...xk which are c1,c2,....ck can not be all together zero then we can say that x1,x2,....xk vectors are linearly dependent to each other. But if the statement c1*x1+c2*x2+.....ck*xk=0 can be valid for only for only where all coefficients are zero which is c1=c2=......ck=0 then we can say that vectors in V space are 'linearly independent'.

Find
Plot $$ y_{H}^{1}(x)$$and$$y_{H}^{2}(x)$$ for visual introduction.

Show that we can not derive $$ y_{H}^{1}(x)$$ from multiplying $$y_{H}^{2}(x)$$ with $$\alpha$$

Solution
IMPORTANT NOTE:For plotting we define the interval of x between -1 to 1 to avoid imaginary parts of $$y_{H}^{2}(x)$$

 Matlab Code: 



Since for any X


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$$ y_{H}^{1}(x)-\alpha*y_{H}^{2}(x)=x-\alpha *\frac{x}{2}*log\left ( \frac{(1+x)}{(1-x)} \right )-1\neq 0 \displaystyle $$ (Eq 3.3)
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we can sat that $$ y_{H}^{1}(x) $$ and $$ y_{H}^{2}(x)$$ are independent homogeneous solutions.

=Problem 4=

Given

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$$ \phi (x,y)=x^{2}y^{3/2}+log(x^{3}y^{2}) \displaystyle $$ (Eq 4.1)
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Find
$$ F(x,y,{y}')=\frac{\partial \phi (x,y)}{\partial x} \displaystyle $$, verify that $$ F(x,y,{y}') \displaystyle $$ is an exact non-linear ODE and invent three more non-linear ODE's

Solution
$$ F(x,y,{y}')=\frac{\partial \phi (x,y)}{\partial x}=M(x,y)+N(x,y){y}' \displaystyle $$

where:
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$$ M=\frac{\partial \phi }{\partial y} \displaystyle $$ (Eq 4.2) and
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$$ N=\frac{\partial \phi }{\partial x} \displaystyle $$ (Eq 4.3) Solving for M and N we get:
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$$ M=2xy^{3/2}+\frac{3}{xlog10} \displaystyle $$ (Eq 4.4)
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$$ N=\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10} \displaystyle $$ (Eq 4.5) Plugging into the equation we get:
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$$ F(x,y,{y}')=(2xy^{3/2}+\frac{3}{xlog10})+(\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}){y}' \displaystyle $$ (Eq 4.6) In order for an equation to be an exact N1-ODE the following 2 conditions must be satisified:
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Condition 1
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$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (Eq 4.7) Condition 2
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$$ M_{y}(x,y)=0=N_{x}(x,y) \displaystyle $$ (Eq 4.8)
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For condition 1:
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$$ M(x,y)+N(x,y){y}'=0 \displaystyle $$ (Eq 4.9) Using $$ M(x,y) \displaystyle $$ and $$ N(x,y) \displaystyle $$ from above we can plug in and we find:
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$$ (2xy^{3/2}+\frac{3}{xlog10})+(\frac{3}{2}x^{2}y^{1/2}+\frac{2}{ylog10}){y}'=0 \displaystyle $$ (Eq 4.10) For Condition 2:
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x} \displaystyle $$ (Eq 4.11)
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$$ \frac{\partial M(x,y)}{\partial y}=3xy^{1/2} \displaystyle $$ (Eq 4.12)
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$$ \frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (Eq 4.13) Therefore:
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}=3xy^{1/2} \displaystyle $$ (Eq 4.14) and $$ F(x,y,{y}') \displaystyle $$ is an exact N1-ODE.
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Three exact N1-ODE's

 Example 1 

Let's consider following equation:


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$$\phi \left ( x,y \right )=\frac{x^{2}}{y^{3}}-\frac{1}{y}+c \displaystyle $$ (Eq 4.15)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=\frac{2x}{y^{3}}\displaystyle $$ (Eq 4.16)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=\frac{y^{2}-3x^{2}}{y^{4}}\displaystyle $$ (Eq 4.17)
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Providing second condition of exactness :


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$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=\frac{-6x}{y^{4}}\displaystyle $$ (Eq 4.18)
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Now we can write :


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$$M(x,y)dx+N(x,y)dy=\frac{2x}{y^{3}}dx+\frac{y^{2}-3x^{2}}{y^{4}}dy=0\displaystyle $$ (Eq 4.19)  Example 2 
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Let's consider following equation:


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$$\phi \left ( x,y \right )=e^xsiny \displaystyle $$ (Eq 4.20)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=e^xsiny\displaystyle $$ (Eq 4.21)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=-e^xcosy\displaystyle $$ (Eq 4.22)
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Providing second condition of exactness :


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$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=-e^xcosy\displaystyle $$ (Eq 4.23)
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Now we can write :


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$$M(x,y)dx+N(x,y)dy=e^xsinydx+-e^xcosydy=0\displaystyle $$ (Eq 4.24)
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 Example 3 

Let's consider following equation:


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$$\phi \left ( x,y \right )=x^{1/2}lny \displaystyle $$ (Eq 4.25)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial x}=M(x,y)=\frac{x^{-1/2}lny}{2}\displaystyle $$ (Eq 4.26)
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$$\frac{\partial \phi \left ( x,y \right )}{\partial y}=N(x,y)=\frac{x^{1/2}}{y}\displaystyle $$ (Eq 4.27)
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Providing second condition of exactness :


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$$\frac{\partial N\left ( x,y \right )}{\partial x}=\frac{\partial M\left ( x,y \right )}{\partial y}=\frac{x^{-1/2}}{2y}\displaystyle $$ (Eq 4.28)
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Now we can write :


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$$M(x,y)dx+N(x,y)dy=\frac{x^{-1/2lny}}{2}dx+\frac{x^{1/2}}{y}dy=0\displaystyle $$ (Eq 4.29)
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= Problem 5: Giving example of ODEs whcih cannot satisfy the first condition of exactness =

Given

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$$ \displaystyle M(x,y)+N(x,y)f(y')=0$$ (Eq 5.1)
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Find
Find $$\displaystyle f(y')$$ so that there is no analytical solution to $$\displaystyle f(y')=-\frac{M}{N}$$. (i.e., such ODE cannot be exact.)

Solution
If we were to create the following parts of such an equation that is described in Eq 5.1, we can get the following functions out of an infinite possibilities pool.


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$$ \displaystyle
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M(x,y)=5xy$$

(Eq 5.2)
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$$ \displaystyle N(x,y)={{x}^{2}}{{y}^{3}}$$ (Eq 5.3)
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$$ \displaystyle f(y')=10y'{{e}^{y'}}$$ (Eq 5.4)
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Then plugging in all of the equations into Eq 5.1 will get us the following:


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$$ \displaystyle 5xy+{{x}^{2}}{{y}^{3}}(10y'{{e}^{y'}})=0$$ (Eq 5.5)
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Reorganizing the equation to solve for y’ will yield


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$$ \displaystyle 5xy+{{x}^{2}}{{y}^{3}}(10y'{{e}^{y'}})=0$$ (Eq 5.6)
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$$ \displaystyle {{x}^{2}}{{y}^{3}}(10y'{{e}^{y'}})=-5xy$$ (Eq 5.7)
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$$ \displaystyle y'{{e}^{y'}}=-\frac{5xy}{10{{x}^{2}}{{y}^{3}}}=-\frac{1}{2xy}$$ (Eq 5.8)
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Finally taking the natural logarithms of both sides will yield


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$$ \displaystyle \ln (y')+y'=\ln (-\frac{1}{2xy})$$ (Eq 5.9)
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Therefore, it can be concluded that $$\displaystyle f(y')=10y'{{e}^{y'}}$$ will not yield an exact equation for the given $$\displaystyle M(x,y)$$ & $$\displaystyle N(x,y)$$.

= Problem 6: Verifying the solution of a N1-ODE =

Given

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$$ \displaystyle y(x)={{\sin }^{-1}}(k-15{{x}^{5}})$$, (Eq 6.1)
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$$ \displaystyle M(x,y)+N(x,y)y'=0$$, (Eq 6.2)
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where $$\displaystyle \left\{ \begin{align} & M(x,y)=75{{x}^{4}} \\ & N(x,y)=\cos y \\ \end{align} \right.$$.

Find
Verify that Eq 6.1 satisfies Eq 6.2.

Step 1
When we plug in the given $$\displaystyle M(x,y)$$ & $$\displaystyle N(x,y)$$ values into Eq 6.2, we get:


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$$ \displaystyle 75{{x}^{4}}+(\cos y)y'=0$$. (Eq 6.3)
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Step 2
So now let’s look at Eq 6.1 and rewrite it in a different from such that


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$$ \displaystyle y(x)={{\sin }^{-1}}(k-15{{x}^{5}})\Rightarrow \sin y=(k-15{{x}^{5}})$$. (Eq 6.4)
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Step 3
Taking the derivative of both sides of the new form we get the following:


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$$ \displaystyle (\sin y)'=(k-15{{x}^{5}})'$$, (Eq 6.5)
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$$ \displaystyle (\cos y)y'=-75{{x}^{4}}$$. (Eq 6.6)
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Step 4
Rearranging the result from Step 3 will give us $$\displaystyle 75{{x}^{4}}+(\cos y)y'=0$$; thereby, proof is complete.

= Problem 7: Solving General Non-Homogeneous Linear First Order Ordinary Differential Equations =

Given

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

$$ \displaystyle {y}'+xy=2x+3$$, (Eq 7.1)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle {y}'+\frac{{{a}_{0}}(x)}{{{a}_{1}}(x)}y=\frac{b(x)}{{{a}_{1}}(x)}$$, (Eq 7.2)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where we assume that for any $$\displaystyle x $$, $$\displaystyle {{a}_{1}}(x)\ne 0 $$.


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

$$ \displaystyle ({{x}^{2}}+1){y}'+xy=2x$$. (Eq 7.3)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Find
Solutions regarding Eq 7.1 ~ Eq 7.3.

Solution
Before we get started, we should find the method of deriving solutions of non-homogeneous 1st order ODEs.

Method by King et al.
Consider a general non-homogeneous 1st order ODE like:


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

$$ \displaystyle {{a}_{1}}(x){y}'+{{a}_{0}}(x)y=b(x)$$, (Eq 7.4)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

If for any $$\displaystyle x $$, $$\displaystyle {{a}_{1}}(x)\ne 0 $$, then we get another form of $$Eq 7.4 $$,


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

$$ \displaystyle {y}'+P(x)y=Q(x)$$, (Eq 7.5)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$P(x)=\frac{{{a}_{0}}(x)}{{{a}_{1}}(x)}$$ and $$Q(x)=\frac{b(x)}{{{a}_{1}}(x)}$$.

If we let factor $$\displaystyle \exp \left\{ \int_ – ^{x}{P(s)ds} \right\}$$ times each sides of Eq 7.5, we derive,


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

$$ \displaystyle \exp \left\{ \int_ – ^{x}{P(s)ds} \right\}\frac{dy}{dx}+\exp \left\{ \int_ – ^{x}{P(s)ds} \right\}P(x)y=Q(x)\exp \left\{ \int_ – ^{x}{P(s)ds} \right\}$$. (Eq 7.6)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We can see that the left side of the equation can be simplified, thus,


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

$$ \displaystyle {{h}^{1}}\frac{dy}{dx}+y\frac{d{{h}^{1}}}{dx}=Q{{h}^{1}}$$, (Eq 7.7)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

By integrating Eq 7.7 we have the solution,


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

$$ \displaystyle y(x)=\frac{1}{{{h}^{1}}(x)}\int_ – ^{x}{{{h}^{1}}(s)Q(s)}ds$$, (Eq 7.8)
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * <p style="text-align:right">
 * }

where $$\displaystyle {{h}^{^{1}}}(x)=\exp \left\{ \int_ – ^{x}{P(s)ds} \right\} $$.

Method by finding exact total differential
We can also get the solution fully under the idea of using exact total differential to eliminate the $$\displaystyle {{y}'} $$ and utility of line integral. By doing this, Eq 7.5 can be also be written as,


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

$$ \displaystyle \left[ P(x)y-Q(x) \right]dx+dy=0$$. (Eq 7.9)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

If we find a function $$\displaystyle \mu (x,y)$$ whose total differential is same as the left side of Eq 7.9, then we will have,


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

$$ \displaystyle \mu (x,y)=C$$, (Eq 7.10)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle C$$ is some constant, then we get the implicit or explicit solution from Eq 7.10.

However the prerequisite of existence of such $$\displaystyle \mu (x,y)$$ is not assured, so we need a factor multipled to the Eq 7.9 to make it satisfy the second condition of exactness. Let's call this factor $$\displaystyle {{h}^{2}}$$ and assume it a function only of $$\displaystyle x$$. We have,


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

$$ \displaystyle {{h}^{2}}(x)\left[ P(x)y-Q(x) \right]dx+{{h}^{2}}(x)dy=0$$. (Eq 7.11)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Eq 7.11 must satisfy the second condition of exactness, hencce we obtain,


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

$$ \displaystyle \frac{\partial \left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right] \right\}}{\partial y}=\frac{\partial {{h}^{2}}(x)}{\partial x}$$ (Eq 7.12)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \Rightarrow {{h}^{2}}(x)=\exp \int_ – ^{x}{P(s)}ds$$ (Eq 7.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

We can see that the assumption "$$\displaystyle {{h}^{2}}(x)$$ is a function only of $$\displaystyle x$$" is automatically satisfied and $$\displaystyle {{h}^{2}}(x)$$ is actually equal to $$\displaystyle {{h}^{1}}(x)$$.

Now we can concentrate on getting the critical important function: $$\displaystyle \mu (x,y)$$.

Theorem 7-1
Let $$\displaystyle I(x,y)$$ and $$\displaystyle J(x,y)$$ be both smooth functions in its domain and such relationship is satisfied:


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

$$ \displaystyle \frac{\partial I}{\partial y}=\frac{\partial J}{\partial x}$$. (Eq 7.14)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Then the total differential of function $$\displaystyle u(x,y)=\int_{({{x}_{0}},{{y}_{0}})}^{(x,y)}{I(x,y)dx+J(x,y)dy}$$ equals $$\displaystyle I(x,y)dx+J(x,y)dy$$ in which the $$\displaystyle u(x,y)$$ is a line integral and $$\displaystyle ({{x}_{0}},{{y}_{0}})$$ is an arbitrary point.

Verification:

To illustrate this theorem, we just need to prove relationships below:


 * $$\displaystyle \frac{\partial u}{\partial x}=I(x,y)$$, $$\displaystyle \frac{\partial u}{\partial y}=J(x,y)$$

Then we have,


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

$$ \displaystyle \frac{\partial u}{\partial x}=\underset{\Delta x\to 0}{\mathop{\lim }}\,\frac{u(x+\Delta x,y)-u(x,y)}{\Delta x}$$ (Eq 7.15)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The Eq 7.14 satisfied the condition of path independence of line integrals(please see Green's theorem), therefore, line integral $$\displaystyle u(x+\Delta x,y)=\int_{({{x}_{0}},{{y}_{0}})}^{(x+\Delta x,y)}{I(x,y)dx+J(x,y)dy}$$ can be designated onto 2 line segments:


 * $$\displaystyle ({{x}_{0}},{{y}_{0}})\to (x,y)\to (x+\Delta x,y)$$. We obtain,


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

$$ \displaystyle u(x+\Delta x,y)=u(x,y)+\int_{(x,y)}^{(x+\Delta x,y)}{I(x,y)dx+J(x,y)dy}$$, (Eq 7.16)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \Rightarrow u(x+\Delta x,y)-u(x,y)=\int_{x}^{x+\Delta x}{I(x,y)dx}$$, (Eq 7.17)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }


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

$$ \displaystyle \Rightarrow u(x+\Delta x,y)-u(x,y)=I(x+\omega *\Delta x,y)\Delta x$$, (Eq 7.18)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle 0\le \omega \le 1$$(please see Intermediate value theorem).

Then we substituting Eq 7.18 back to Eq 7.15, we derive,


 * $$\frac{\partial u}{\partial x}=I(x,y)$$.

Similarly we have,


 * $$\frac{\partial u}{\partial y}=J(x,y)$$.

Thus the whole theorem is proved.

Now we go back to the Eq 6.11. Because Eq 7.12 meets the prerequisite of Eq 7.14, we can apply Theorem 7-1 on Eq 7.11, we have,


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

$$ \displaystyle \begin{align} u(x,y) & =\int_{({{x}_{0}},{{y}_{0}})}^{(x,y)}{\left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right]dx+{{h}^{2}}(x)dy \right\}} \\ & =\int_{({{x}_{0}},{{y}_{0}})}^{(x,{{y}_{0}})}{\left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right] \right\}dx}+\int_{(x,{{y}_{0}})}^{(x,y)}{{{h}^{2}}(x)dy} \end{align}$$, (Eq 7.19)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle ({{x}_{0}},{{y}_{0}})$$ is an arbitrary point. And if combined with Eq 7.10, we are able to get the solution from the equation below:


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

$$ \displaystyle \int_{({{x}_{0}},{{y}_{0}})}^{(x,{{y}_{0}})}{\left\{ {{h}^{2}}(x)\left[ P(x)y-Q(x) \right] \right\}dx}+\int_{(x,{{y}_{0}})}^{(x,y)}{{{h}^{2}}(x)dy}=C$$, (Eq 7.20)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle C$$ is some constant.

Solution for Eq 7.1
We use the first method.

From Eq 7.1 we know,

$$\displaystyle P(x)=x$$, $$\displaystyle Q(x)=2x+3$$.

The integrating factor is,


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

$$ \displaystyle \begin{align} h= & \exp \int_ – ^{x}{sds} \\ & ={{e}^{\tfrac{2}}} \end{align}$$. (Eq 7.21)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

So,


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

$$ \displaystyle \begin{align} y & =\frac{1}{h(x)}\int_ – ^{s}{h(s)Q(s)ds} \\ & ={{e}^{-\frac{2}}}\int_ – ^{x}(2s+3)ds \\ & ={{e}^{-\frac{2}}}\left( 2\int_ – ^{x}{{{e}^{\frac{2}}}d\frac{2}+3\int_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}ds}} \right) \\ & =2+{{e}^{-\frac{2}}}*3\int_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}ds} \\ & =2+\frac{3\sqrt{\pi }}{2}{{e}^{-\frac{2}}}*\left( erfi(x)+C \right) \end{align}$$, (Eq 7.22)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle C$$ is some constant.

The term $$\displaystyle \int_ – ^{x}{{{e}^{\frac{{{s}^{2}}}{2}}}ds}$$ is unable to be described by the combination of elementary functions and their integrals (please see Lists of integrals). But it can be described by imaginary error function(the term $$\displaystyle erfi(x)$$ in Eq 7.22), see also Error Function.

Solution for Eq 7.2
From Eq 7.2 we know,

$$P(x)=\frac{{{a}_{0}}(x)}{{{a}_{1}}(x)}$$ and $$Q(x)=\frac{b(x)}{{{a}_{1}}(x)}$$.

Substituting these two into Eq 7.8, we dereive,


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

$$ \displaystyle y(x)=\frac{1}{\exp \left\{ \int_ – ^{x}{\frac{{{a}_{0}}(s)}{{{a}_{1}}(s)}ds} \right\}}\int_ – ^{x}{\exp \left\{ \int_ – ^{x}{\frac{{{a}_{0}}(s)}{{{a}_{1}}(s)}ds} \right\}\frac{b(s)}{{{a}_{1}}(s)}ds}$$ (Eq 7.23)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Solution for Eq 7.3
We use second method.

As $$\displaystyle {{a}_{1}}=({{x}^{2}}+1)\ne 0$$, Eq 7.3 can de written as:


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

$$ \displaystyle {y}'+\frac{x}{{{x}^{2}}+1}y=\frac{2x}{{{x}^{2}}+1}$$ (Eq 7.24)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

The integation factor is,


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

$$ \displaystyle h=\exp \int_ – ^{x}{\frac{s}{{{s}^{2}}+1}ds}={{\left( {{x}^{2}}+1 \right)}^{\frac{1}{2}}}$$ (Eq 7.25)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Hence Eq 7.24 can be written as,


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

$$ \displaystyle \frac{\left( y-2 \right)x}{\sqrt{{{x}^{2}}+1}}dx+\sqrt{{{x}^{2}}+1}dy=0$$ (Eq 7.26)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

Integrate Eq 7.26, we have,


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

$$ \displaystyle \begin{align} & \int_{(0,0)}^{(x,y)}{\left\{ \frac{\left( y-2 \right)x}{\sqrt{{{x}^{2}}+1}}dx+\sqrt{{{x}^{2}}+1}dy \right\}=0} \\ & \Rightarrow \int_{(0,0)}^{(x,0)}{\frac{\left( y-2 \right)x}{\sqrt{{{x}^{2}}+1}}dx}+\int_{(x,0)}^{(x,y)}{\sqrt{{{x}^{2}}+1}dy}=0 \\ & \Rightarrow -2\sqrt{{{x}^{2}}+1}+\sqrt{{{x}^{2}}+1}*y=C \\ & \Rightarrow y=\frac{C}{\sqrt[{}]{{{x}^{2}}+1}}+2 \\ & \\ \end{align}$$, (Eq 7.27)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

where $$\displaystyle C$$ is some constant.

= Problem 8: Ordinary Differential Equations with Varying Coefficients (ODE_VC) =

Given
Show that the constant, k1, is not necessary when the following expression is evaluated:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=\frac{1}{h\left( x \right)}\int_ – ^{x}{h\left( s \right)b\left( s \right)ds}

$$ (eq 8.1) [1]
 * <p style="text-align:right">
 * }

Solution
The function h(x), given in Equation 8.1, is further expressed as:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

h\left( x \right)={{e}^{\int_ – ^{x}{{{a}_{0}}\left( s \right)ds}}}

$$ (eq 8.2) [2]
 * <p style="text-align:right">
 * }

By converting the integral using the following principle:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int_ – ^{x}{f\left( s \right)ds}=\int{f\left( x \right)dx}+k $$ (eq 8.3) [2]
 * <p style="text-align:right">
 * }

The function h can be expressed as follows:
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

h\left( x \right)={{e}^{\left( \int{{{a}_{0}}\left( x \right)dx} \right)+k}}={{e}^{k}}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}} $$ (eq 8.4)
 * <p style="text-align:right">
 * }

Setting the constant ek to k yields the following:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

h\left( x \right)={k}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}} $$ (eq 8.5)
 * <p style="text-align:right">
 * }

Plugging Equation 8.4 into both instances of h(x) in Equation 8.1 yields the following:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=\frac{1}\int_ – ^{x}{{{k}_{2}}{{e}^{\int{{{a}_{0}}\left( x \right)dx}}}b\left( s \right)ds}=\frac{{e}^{-\int{{{a}_{0}}\left( x \right)dx}}}\int_ – ^{x}{{{e}^{\int{{{a}_{0}}\left( x \right)dx}}}b\left( s \right)ds}

$$ (eq 8.6)
 * <p style="text-align:right">
 * }

Let the constant k2=k3k1 which therefore makes k1 unnecessary.

Given
Show that Equation 8.1 agrees with the following:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=A{{y}_{h}}\left( x \right)+{{y}_{p}}\left( x \right)

$$ (eq 8.7) [3]
 * <p style="text-align:right">
 * }

Solution
Applying the principle in Equation 7.3 directly into Equation 8.1 yields the following:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=h{{\left( x \right)}^{-1}}\left[ \int{h\left( x \right)b\left( x \right)dx}+k \right]

$$ (eq 8.8)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y\left( x \right)=kh{{\left( x \right)}^{-1}}+h{{\left( x \right)}^{-1}}\int{h\left( x \right)b\left( x \right)dx}

$$ (eq 8.9)
 * <p style="text-align:right">
 * }

It is shown that the two equations agree by letting A=k and making the following substitutions:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{h}}\left( x \right)=h{{\left( x \right)}^{-1}}

$$ (eq 8.10)
 * <p style="text-align:right">
 * }


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{p}}\left( x \right)=h{{\left( x \right)}^{-1}}\int{h\left( x \right)b\left( x \right)dx}

$$ (eq 8.11)
 * <p style="text-align:right">
 * }

Given
Find yH independently by solving the following equation:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

y'+{{a}_{0}}y=0

$$ (eq 8.12)
 * <p style="text-align:right">
 * }

Solution
Equation 8.12 can be converted into the following form:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\frac{dx}+{{a}_{0}}{{y}_{H}}=0

$$ (eq 8.13)
 * <p style="text-align:right">
 * }

In solving for yH, it follows that


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{}^{d{{y}_{H}}}\!\!\diagup\!\!{}_\;=-{{a}_{0}}dx

$$
 * <p style="text-align:right">
 * }


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\int{{}^{d{{y}_{H}}}\!\!\diagup\!\!{}_\;}=-\int{{{a}_{0}}dx} $$
 * <p style="text-align:right">
 * }


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\ln \left( {{y}_{H}} \right)=-\int{{{a}_{0}}dx}+k $$
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{y}_{H}}=K{{e}^{-\int{{{a}_{0}}dx}}}$$
 * <p style="text-align:right">
 * }

= Problem 9: Exact Differential Equations =

Given
Find a N1_ODE that is either exact or can be made exact using the following equation:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\bar{b}\left( x \right)c\left( y \right)y'+a\left( x \right)\bar{c}\left( y \right)=0

$$ (eq 9.1)
 * <p style="text-align:right">
 * }

With the following values
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

a\left( x \right)=\sin \left( {{x}^{3}} \right) $$ (eq 9.2)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

b\left( x \right)=\cos \left( x \right) $$ (eq 9.3)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

c\left( y \right)={e}^{2y} $$ (eq 9.4)
 * <p style="text-align:right">
 * }

Solution
The functions with superscript bars are integrals of the original functions and thus the following equations are derived:


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

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

b\left( x \right)=\cos \left( x \right)\Rightarrow \bar{b}\left( x \right)=\sin \left( x \right) $$ (eq 9.5)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

c\left( x \right)={{e}^{2y}}\Rightarrow \bar{c}\left( x \right)={}^{1}\!\!\diagup\!\!{}_{2}\;{{e}^{2y}} $$ (eq 9.6) Inserting all of the values into Equation 9.1 yields the following:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

\underbrace{\sin \left( x \right)\left( {{e}^{2y}} \right)}_{N\left( x,y \right)}y'+\underbrace{\sin \left( {{x}^{3}} \right)\left( {}^{1}\!\!\diagup\!\!{}_{2}\;{{e}^{2y}} \right)}_{M(x,y)}=0 $$ (eq 9.7) Equation 9.7 meets the first condition of exactness which states that:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

N\left( x,y \right)y'+M\left( x,y \right)=0 $$ (eq 9.8)
 * <p style="text-align:right">
 * }

Divide Equation 9.7 by e2y.

Next we can check to see if the equation passes the second condition of exactness in which My must equal Nx
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{M}_{y}}\left( x,y \right)=0 $$ (eq 9.9)
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{N}_{x}}\left( x,y \right)=\cos \left( x \right) $$ (eq 9.10) From the previous equations it is evident that
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:95%" |
 * style="width:95%" |

{{M}_{y}}\left( x,y \right)\ne {{N}_{x}}\left( x,y \right) $$ (eq 9.11) Therefore, in order to make the equation exact, Euler's integrating factor, h(x) must be derived as follows:
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle h\left( x \right)={{e}^{-\int{\frac{1}{N}\left( {{N}_{x}}-{{M}_{y}} \right)dx}}} $$ (eq 9.12)4
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

$$ \displaystyle ={{e}^{-\int{\frac{1}{\sin \left( x \right)}\left( \cos \left( x \right)-0 \right)dx}}} $$
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }
 * {| style="width:100%" border="0"

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


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

$$ \displaystyle h\left( x \right)=\frac{1}{\sin \left( x \right)} $$ (eq 9.13)
 * style="width:95%" |
 * style="width:95%" |
 * <p style="text-align:right">
 * }

In order to verify that the derived integrating factor is correct it is necessary to multiply Equation 8.8 by h(x) and denote the new values for M and N as follows:
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$$ \displaystyle \underbrace{\left( \frac{1}{\sin \left( x \right)} \right)\left( \sin \left( x \right) \right)}_{\bar{N}=hN\left( x,y \right)}y'+\underbrace{\left( \frac{1}{\sin \left( x \right)} \right)\left( \frac{\sin \left( {{x}^{3}} \right)}{2} \right)}_{\bar{M}=hM\left( x,y \right)}=0 $$ (eq 9.14)
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Again we check to see if the second condition of exactness has been met as follows:


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$$ \displaystyle {{\bar{M}}_{y}}\left( x,y \right)=0 $$ (eq 9.15)
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$$ \displaystyle {{\bar{N}}_{x}}\left( x,y \right)=0 $$ (eq 9.16)
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The N1_ODE has been made exact since


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$$ \displaystyle {{\bar{M}}_{y}}\left( x,y \right)={{\bar{N}}_{x}}\left( x,y \right) $$ (eq 9.17)
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Given
Find the first integral of the N1_ODE used in the previous problem so that


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$$ \displaystyle
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\phi \left( x,y \right)=k

$$ (eq 9.18)
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Solution
Considering the following equations:
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$$ \displaystyle
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M\left( x,y \right)={{\phi }_{x}}\left( x,y \right) $$ (9.19)5
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$$ \displaystyle
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N\left( x,y \right)={{\phi }_{y}}\left( x,y \right) $$ (9.20)5
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The following equations can be derived:
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$$ \displaystyle
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\int{M\left( x,y \right)dx}+w\left( y \right)=\phi \left( x,y \right) $$ (9.21)6
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$$ \displaystyle
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\int{N\left( x,y \right)dy}+v\left( x \right)=\phi \left( x,y \right) $$ (9.22)6
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By setting equations 8.21 and 8.22 equal to each other, the functions w(y) and v(x) can be found. From there, equation 8.13 can be solved.
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$$ \displaystyle \int{M}\left( x,y \right)dx+w\left( y \right)=\int{N}\left( x,y \right)dy+v\left( x \right) $$ (eq 9.23)
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$$ \displaystyle \int{\frac{\sin \left( {{x}^{3}} \right)}{2\sin \left( x \right)}dx}+w\left( y \right)=y+v\left( x \right) $$
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$$ \displaystyle \phi \left( x,y \right)=\int{\frac{\sin \left( {{x}^{3}} \right)}{2\sin \left( x \right)}dx}+y=k $$
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=Contributions Of Team Members=

Pages
Patrick's Page Zou's Page Patterson's Page

=References=

1Vu-Quoc,L. EGM 6321.F10, 10-3, Equation (6)

2Vu-Quoc,L. EGM 6321.F10, 11-1:11-2

3King, A.C. et al. Differential Equations: Linear, Nonlinear, Ordinary, Partial. Cambridge: Cambridge University Press

4Vu-Quoc,L. EGM 6321.F10, 10-1, Equation (6)&(7)

5Vu-Quoc,L. EGM 6321.F10, 8-2, Equations (1)&(2)

6Zill, D., and Cullen, M. Differential Equations with Boundary-Value Problems. Pacific Grove, CA: Brooks/ Cole,