User:Egm6321.f10.team2.oztekin/hw3

Given
$$\displaystyle a(x)=\sin x^{3}$$ $$\displaystyle b(x)=\cos x$$ $$\displaystyle c(y)=\exp(2y)$$

Find
1. Find an N1-ODE of the form (1) p.13-2 that is either exact or can be made exact by IFM. (See R3.1)
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(3.3.1)
 * $$ \displaystyle \underbrace{\bar b(x,y) c(y)}_{\displaystyle\color{blue}{N(x,y)}} \, y' + \underbrace{a(x) \bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0 $$
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2. Find the first integral

$$\phi(x,y)=k $$

Solution
1. From definitions Eq. (2) p.13-2 and (1) p.13-3,


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 * $$ \displaystyle \bar{b} \left( x\right) = \int\limits_ – ^{x}{{b} \left( s\right)}ds = \int\limits_ – ^{x}{cos(s)}ds = sin x $$
 * $$ \displaystyle \bar{b} \left( x\right) = \int\limits_ – ^{x}{{b} \left( s\right)}ds = \int\limits_ – ^{x}{cos(s)}ds = sin x $$

(3.3.2)
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(3.3.3)
 * $$ \displaystyle \bar{c} \left( y\right) = \int\limits_ – ^{y}{{c} \left( s\right)}ds = \int\limits_ – ^{y}{e^{2s}} ds = \frac{1}{2}e^{2y} $$
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Let substituting $$\displaystyle a(x),\bar{b}(x), \bar{c}(y), c(y) $$ into Eq. 3.3.1, we have


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(3.3.4)
 * $$ sin (x) e^{2y} y' + sin(x^{3}) \frac{1}{2} e^{2y} = 0 $$
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We can cancel Eq. 3.3.4 by $$ \displaystyle e^{2y} $$, then


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(3.3.5)
 * $$ sin (x)y' + \frac{1}{2} sin(x^{3})  = 0 $$
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 * 
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i. First exactness condition,

$$\displaystyle N(x,y)y' + M(x,y) = 0$$

with


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$$\displaystyle M(x) = \frac{1}{2} sin(x^{3}) $$

(3.3.6)
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and


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$$\displaystyle N(x) = sin (x) $$

(3.3.7)
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so it satisfies the first condition.

ii. Second exactness condition


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$$\displaystyle M_{y} (x,y)= \frac{\partial M }{\partial y} = 0 $$ (3.3.8)
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(3.3.9) so
 * $$\displaystyle N_{x} (x,y)= \frac{\partial N }{\partial x} = cos(x) $$
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 * 
 * }

$$\displaystyle M_{y}(x,y) \ne N_{x}(x,y) $$ Therefore, Eq (3.3.5) is not exact.

iii. Finding an integrating factor h(x,y) such that the following exact N1-ODE:


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(3.3.10)
 * $$h(x)\left[\underbrace{sin(x)}_{N(x,y)}{y}'+\underbrace{\frac{1}{2}sin({{x}^{3}})}_{M(x,y)}\right]=0$$
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 * 
 * }

satisfies the condition (2) p,11-2 :


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(3.3.11)
 * $$\displaystyle \frac{h_{x}}{h}= -\frac{1}{N}(N_{x}-M_{y})=: n(x)$$
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then

Substituting $$\displaystyle N$$, $$\displaystyle N_{x}$$, and  $$\displaystyle M_{y}$$  into  Eq (3.2.11)


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(3.3.12)
 * $$\displaystyle n(x)= -\frac {1}{sin x}[cos x - 0]$$
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(3.3.13) Changing the variable of function,
 * $$\displaystyle n(x)= -\frac {cos x}{sin x} $$
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(3.3.14)
 * $$\displaystyle n(s)= -\frac {cos (s)}{sin (s)} $$
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Definition from $$\displaystyle h(x)$$


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(3.3.15) Substituting $$\displaystyle n(s) $$ into Eq (2.18.10)
 * $$\displaystyle h(x) = exp \left[\int^x n(s)ds + k \right]$$
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(3.3.16)
 * $$\displaystyle h(x) = exp \left[\int^x (-\frac {cos (s)}{sin (s)})ds + k \right]$$
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(3.3.17)
 * $$\displaystyle h(x) = exp \left[-ln(sin (x))+ k \right]$$
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(3.3.18)
 * $$\displaystyle h(x) = \frac{1}{sin x} $$
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Now, we can find exact N1-ODE because we have the integrating factor h(x),


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(3.3.19)
 * $$ \displaystyle \underbrace {(hN)}_{\displaystyle \color{blue}{\bar N}} \, y' + \underbrace{(hM)}_{\displaystyle \color{blue}{\bar M}}=0 $$
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So then


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(3.3.20)
 * $$\left[\underbrace{\frac {1}{sin x}sin(x)}_{\bar N(x,y)} \, y' +\underbrace{\frac {1}{sin x}\frac{1}{2}sin({{x}^{3}})}_{\bar M(x,y)}\right]=0$$
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If we rearrange and rewrite it, we will have


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(3.3.21)
 * $$\left[\underbrace{1}_{\bar N(x,y)} \, y' +\underbrace{\frac{sin (x^3)}{2 sin (x)}}_{\bar M(x,y)}\right]=0$$
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iv. If we test second exactness condition for Eq. 3.3.


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$$\displaystyle \bar M_{y} (x,y)= \frac{\partial \bar M }{\partial y} = 0 $$ (3.3.22)
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(3.3.23) so
 * $$\displaystyle \bar N_{x} (x,y)= \frac{\partial \bar N }{\partial x} = 0 $$
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$$\displaystyle \bar M_{y}(x,y) = \bar N_{x}(x,y) $$ Therefore, Eq (3.3.) is exact.

2. To find $$\displaystyle\phi(x,y)=k $$, we need to use the exact N1-ODE we have found in Eq. 3.3

From Eq. (2) p. 8-5 and (1) p. 8-6 respectively,
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(3.3.24)
 * $$\bar M(x,y)=\frac{\partial \phi{(x,y)}}{\partial {x}}=:\phi_{x}{(x,y)}$$
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(3.3.25)
 * $$\bar N(x,y)=\frac{\partial \phi{(x,y)}}{\partial {y}}=:\phi_{y}{(x,y)}$$
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If we integrate Eq. 3.3.24 and Eq. 3.3.25 with respect to x, y, respectively, we will get


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

(3.3.26)
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 * }


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 * $$\,\int \phi_{y}{(x,y)}dy = \phi {(x,y)} = \int \bar N(x,y) dy + k(x)$$
 * $$\,\int \phi_{y}{(x,y)}dy = \phi {(x,y)} = \int \bar N(x,y) dy + k(x)$$

(3.3.27)
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If we differentiate Eq. 3.3. with respect to y, we will have


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 * $$\displaystyle \frac{\partial \phi{(x,y)}}{\partial {y}} = \frac{d ( \int \bar M(x,y) dx) }{dy} + k'(y)$$
 * $$\displaystyle \frac{\partial \phi{(x,y)}}{\partial {y}} = \frac{d ( \int \bar M(x,y) dx) }{dy} + k'(y)$$

(3.3.28)
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 * }


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 * $$ \displaystyle \bar N(x,y) = 0 + k'(y)$$
 * $$ \displaystyle \bar N(x,y) = 0 + k'(y)$$

(3.3.29)
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 * $$ \displaystyle 1 = k'(y)$$
 * $$ \displaystyle 1 = k'(y)$$

(3.3.30)
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Integrate Eq. 3.3.30 with respect to y,


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 * $$ \displaystyle y + c = k(y)$$
 * $$ \displaystyle y + c = k(y)$$

(3.3.31)
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Substituting Eq. 3.3.31 into Eq. 3.3.26 to get $$\displaystyle\phi(x,y)$$


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(3.3.32)
 * $$\phi ( x,y )=\int{\bar M\left( x,y \right)dx + y + c}$$
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As a result,


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(3.3.33)
 * $$\phi ( x,y )=\int{\frac{sin (x^3)}{2 sin (x)}dx + y = k}$$
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Given

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$$\displaystyle \left( \frac {1}{3}x^3+d_1\right)\left(y^4\right)y' + (5x^3+2)\left(\frac {1}{5}y^5+ sinx + d_2\right) = 0$$ (3.2.1)
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Find
Show that the Equation (3.2.1) is exact or not. If it is not, can it be made exact by using the IFM? Find the IFM h(x,y).

Solution
1. First exactness condition,

$$\displaystyle N(x,y)y' + M(x,y) = 0$$

with


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$$\displaystyle M(x,y) = (5x^3+2)\left(\frac {1}{5}y^5+ sinx + d_2\right)$$

(3.2.2)
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and


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$$\displaystyle N(x,y) = \left( \frac {1}{3}x^3+d_1\right)\left(y^4\right)$$

(3.2.3)
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so it satisfies the first condition.

2. Second exactness condition


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$$\displaystyle M_{y} (x,y)= a(x).c(y)= (5x^3+2) (y^4) $$ (3.2.4)
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(3.2.5) so
 * $$\displaystyle N_{x} (x,y)= b(x).c(y)= (x^2) (y^4) $$
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$$\displaystyle M_{y}(x,y) \ne N_{x}(x,y) $$ Therefore, Eq (3.2.1) is not exact.

3. Finding an integrating factor h(x,y) such that the following exact N1-ODE:


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

(3.2.5)
 * $$ \displaystyle h(x,y) \left[\underbrace{\bar b(x,y) c(y)}_{\displaystyle\color{blue}{N(x,y)}} \, y' + \underbrace{a(x) \bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}\right]=0 $$
 * <p style="text-align:right">
 * <p style="text-align:right">
 * }

satisfies the condition (2) p,11-2 :


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(3.2.6)
 * $$\displaystyle \frac{h_{x}}{h}= -\frac{1}{N}(N_{x}-M_{y})=: n(x)$$
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that an integrating factor h(x) can be found to render it exact, only if $$ \displaystyle k_1(y)=d_1 $$ (constant)

then

Substituting $$\displaystyle N$$ Eq (3.2.3), $$\displaystyle N_{x}$$  Eq (3.2.5),  and  $$\displaystyle M_{y}$$  Eq (3.2.4)  into  Eq (3.2.6)


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(3.2.7)
 * $$\displaystyle n(x)= -\frac {1}{\bar b(x)\cancel{c(y)}}\left[ b(x)\cancel{c(y)}-a(x)\cancel{c(y)}\right] = -\frac {1}{\frac{1}{3} x^3}[x^2-(5x^3+2)]$$
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(3.2.8) Changing the variable of function,
 * $$\displaystyle n(x)= 3(\frac{2}{x^3}-\frac{1}{x}+5) $$
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(3.2.9)
 * $$\displaystyle n(s)= 3(\frac{2}{s^3}-\frac{1}{s}+5)$$
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Definition from $$\displaystyle h(x)$$


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(3.2.10) Substituting $$\displaystyle n(s) $$ into Eq (2.18.10)
 * $$\displaystyle h(x) = exp \left[\int^x n(s)ds + k \right]$$
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(3.2.11)
 * $$\displaystyle h(x) = exp \left[3 \int^x (\frac{2}{s^3}-\frac{1}{s}+5)ds + k \right]$$
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 * }
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(3.2.12)
 * $$\displaystyle h(x) = exp \left[-\frac{3}{x^2}-3ln(x)+15x+ k \right]$$
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 * }
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(3.2.13)
 * $$\displaystyle h(x) = \frac{1}{x^3}exp \left[-\frac{3}{x^2}+15x + k \right]$$
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 * }