User:Egm6321.f12.team4.harris/Homework3 R*3.7

Homework 3 Problem R*3.7 – Finding the first integral of an equation made exact by use of the Integrating Factor Method
sec13-4

Statement
Find a non-linear ordinary differential equation (N1-ODE) of the form
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$$ \bar{b}(x,y)c(y)y'+a(x)\bar{c}(x,y)=0 $$      (3.7.1) such that
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$$ \bar{b}(x,y):=\int^xb(s)ds+k_1(y) $$      (3.7.2)
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$$ \bar{c}(x,y):=\int^xc(s)ds+k_2(x) $$      (3.7.3)
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that is either exact or can be made exact through use of the Integrating Factor Method. Find the first integral $$ \phi (x,y)=k $$.

Given

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$$ a(x)=sin(x^3) $$      (3.7.4)
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$$ b(x)=cosx $$      (3.7.5)
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$$ c(y)=exp(2y) $$      (3.7.6)
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Solution
Solving equations 3.7.2 and 3.7.3 from the given equations 3.7.5 and 3.7.6
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$$ \bar{b}(x,y)=\int^xcos(x)dx+k_1(y)= -sinx+k_1(y) $$      (3.7.7)
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$$ \bar{c}(x,y)=\int^xexp(2y)+k_2(x)=\frac{1}{2}exp(2y)+k_2(x) $$      (3.7.8)
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Recalling homework problem R*3.5, $$ k_1(y)=d_1=constant $$. Then substituting equations 3.7.7 and 3.7.8 into equation 3.7.1 yields
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$$ (-sinx+d_1)(exp(2y))y'+sin(x^3)(\frac{1}{2}exp(2y)+k_2(x))=0 $$      (3.7.9)
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To satisfy the first exactness condition of N1-ODEs, equation 3.7.9 must be of the form
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$$ M(x,y)+N(x,y)y'=0 $$
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By defining
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$$ M(x,y)=sin(x^3)(\frac{1}{2}exp(2y)+k_2)) $$      (3.7.10)
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$$ N(x,y)=(-sinx+d_1)(exp(2y)) $$      (3.7.11)
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and rearranging, one can see that equation 3.7.9 can be written as
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$$ M(x,y)+N(x,y)y'=0 $$
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As such, it is shown that the above equations 3.7.4-3.7.6, when manipulated according to the formula of equation 3.7.1, can be rearranged to form an equation that satisfies the first exactness condition of N1-ODEs.
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The second exactness condition states that
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x} $$ Using the definitions of $$ M(x,y) $$ and $$ N(x,y) $$ from equations 3.7.10 and 3.7.11
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$$ \frac{\partial M(x,y)}{\partial y}=\frac{\partial}{\partial y} [(sinx^3)(\frac{1}{2}exp(2y)+k_2)] $$
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$$ \frac{\partial}{\partial y}[\frac{sinx^3}{2}exp(2y)+sinx^3(k_2)] $$

$$ sinx^3exp(2y) $$     (3.7.12)
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$$ \frac{\partial N(x,y)}{\partial x} = \frac{\partial}{\partial x} [(-sinx+d_1)(exp(2y))] $$ $$ \frac{\partial}{\partial x}[(-sinx+d_1)(exp(2y)d_1)] $$
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$$ -exp(2y)cosx $$     (3.7.13)
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It can be seen that
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$$ sinx^3exp(2y)\ne -exp(2y)cosx $$
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$$ \therefore \frac{\partial M(x,y)}{\partial y} \ne \frac{\partial N(x,y)}{\partial y} $$


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Thus equation 3.7.9 does not satisfy the second exactness condition and is therefore not exact. The Integrating Factor Method must be used to find the exact solution and $$ \phi (x,y) $$.

The first integral $$ \phi (x,y) $$ is found according to the equation
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$$ \frac{d\phi (x,y)}{dx} = \frac{\partial \phi (x,y)}{\partial x}+\frac{\partial \phi (x,y)}{\partial y} \frac{dy}{dx}=0 $$
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such that
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$$ \bar{M}(x,y)= \frac{\partial \phi (x,y)}{\partial x} $$
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$$ \bar{N}(x,y)=\frac{\partial \phi (x,y)}{\partial y} $$


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Define $$ \bar{M}(x,y) $$ and $$ \bar{N}(x,y) $$ as $$ h(x) $$ multiplied by equations 3.7.12 and 3.7.13, where $$ h(x) $$ is the integrating factor that would make equation 3.7.9 exact.

Integrating $$ d\phi $$ yields
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$$ \int d\phi = \int \frac{\partial \phi}{\partial x} dx+\int \frac{\partial \phi}{\partial y} \underbrace{\frac{dy}{dx}dx}_{dy} $$
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$$ \phi =\int \bar{M}(x,y)dx +\int \bar{N}(x,y)dy $$

$$ \phi = \int h(x)[\frac{sinx^3}{2}exp(2y)+k_2]dx+ \int h(x)[(-sinx+d_1)exp(2y)]dy $$


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Since $$ exp(2y)\ne f(x) $$ and $$ -sinx+d_1 \ne f(y) $$, they can be taken out of the first and second integrals, respectively, on the right hand side of the equation. Thus one is left with
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$$ \phi = \frac{exp(2y)}{2}\int h(x)sinx^3dx+ \int h(x)k_2dx+ (-sinx+d_1)\underbrace{\int h(x)exp(2y)dy}_{h(x)\frac{1}{2}exp(2y)} $$     (3.7.14)
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In equation 3.7.14, there is no direct integral for the function of $$ sinx^3 $$. However, one can redefine $$ k=\int h(x)k_2dx $$ and rearrange equation 3.7.14 to be
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$$ \phi = \frac{exp(2y)}{2}[\int h(x)sinx^3dx+h(x)(-sinx+d_1)+k] $$     (3.7.15)
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Equation 3.7.15 is the first integral $$ \phi (x,y) $$ for the set of given equations 3.7.4-3.7.6 when used in equation 3.7.1.

Authors and References

 * Solved and Typed by -- Kaitlin Harris