User:Egm6321.f12.team4/Report4

Problem R*4.1 – Verifying Exactness of L2-ODE-VC
sec21-1

Statement
Verify the exactness of the following equation,

Solution
To solve this problem, we have to know the two exactness conditions for second order ODEs. The equation must be in the following form
 * First Exactness Condition for N2-ODEs:


 * Second Exactness Condition for N2-ODEs:

Let's rearrange the equation 4.1.0

If we let, $$ g(x,y,p) = 2xy' + 3y $$ and $$ f(x,y,p) = \sqrt(x) $$ then we can have our equation in the form exactness condition one as stated above. Therefore, the exactness condition one is satisfied.

To apply second exactness condition we need to find $$ f_{xx}, f_{xy}, f_{xp}, f_y, f_{yy}, f_{yp} $$ and $$ g_y, g_{xp}, g_{yp}, g_{pp} $$ They are found easily by partial differentiation and the results are as follows,

Now if we apply the first and second equation of the Second Exactness Condition

We can see that the first equation of the second exactness condition does not satisfy where as the second equation does satisfy. Unfortunately, both equations must be satisfied to satisfy the second exactness condition. Therefore, the second exactness condition is not satisfied such that the equation 4.1.0 is not exact.

Author and References

 * Solved and Typed by -- Seong Hyeon Hong
 * Reviewed by -- Jason Alphonso

Problem R*4.2 – Find the integrating factor
sec21-4

Statement
Find $$ m,n \in \mathbb {R} $$ such that
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$$ x^my^n[\sqrt{x}y''+2xy'+3y]=0 $$     (4.2.1)
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is an exact non-linear second order ordinary differential equation (N2-ODE). Show that the first integral $$ \phi $$ is a linear first-order ordinary differential equation with varying coefficients (L1-ODE-VC)
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$$ \phi (x,y,P) = xP + (2x^{\frac{3}{2}}-1)y = k $$ (4.2.2)
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with $$ P(x) := y'(x) $$. Solve equation 4.2.2 for y(x).

Given
N2-ODE of the power form
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$$ \alpha x^r y'' + \beta x^sy'+\gamma x^ty=0 $$     (4.2.3)
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has an integrating factor $$ h(x,y) = x^my^n $$. In equation 4.2.1,
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$$ \alpha =1, \beta =2, \gamma =3, r =\frac{1}{2}, s=1, t=0 $$
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Since 4.2.1 is already multiplied by $$ x^my^n $$, one can assume that it meets both exactness criteria for N2-ODEs. The first exactness condition states that
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$$ G(x,y,y',y)=g(x,y,P)+f(x,y,y')y=0 $$     (4.2.4) such that
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$$ g(x,y,P):= \phi_x + \phi_y y'=x^m3y^{n+1}+2x^{m+1}y^ny' $$     (4.2.5) and
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$$ f(x,y,P):=\phi_p=y^nx^{m+\frac{1}{2}} $$     (4.2.6)
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The second exactness condition states that
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$$ f_{xx}+2Pf_{xy}+P^2f_{yy}=g_{xp}+Pg_{yp}-g_y $$     (4.2.7) and
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$$ f_{xp}+Pf_{yp}+2f_y=g_{pp} $$     (4.2.8)
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such that $$ f_{xx}=\frac{\partial }{\partial x}\frac{\partial f}{\partial x} $$, etc.

Solution
From equations 4.2.7 and 4.2.8, one needs to take the various derivatives of $$ f $$ and $$ g $$. They are calculated as follows:
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$$ f_{xx}=\frac{\partial }{\partial x} \frac{\partial f}{\partial x}=\frac{\partial }{\partial x}[(m+\frac{1}{2})y^nx^{m-\frac{1}{2}}]=(m+\frac{1}{2})(m-\frac{1}{2})y^nx^{m-\frac{3}{2}} $$     (4.2.9)
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$$ f_{xy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial x}=\frac{\partial }{\partial y}[(m+\frac{1}{2})y^nx^{m-\frac{1}{2}}]=(m+\frac{1}{2})ny^{n-1}x^{m+\frac{1}{2}} $$     (4.2.10)
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$$ f_{yy}=\frac{\partial }{\partial y} \frac{\partial f}{\partial y}=\frac{\partial }{\partial y}[ny^{n-1}x^{m+\frac{1}{2}}]=n(n-1)y^{n-1}x^{m+\frac{1}{2}} $$     (4.2.11)
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$$ g_{xp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial x}=\frac{\partial }{\partial P}[3y^{n+1}mx^{m-1}+2(m+1)x^my^nP]=2(m+1)x^my^n $$     (4.2.12)
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$$ g_{yp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial y}=\frac{\partial }{\partial P}[x^m3(n+1)y^n + 2x^{m+1}ny^{n-1}P]=2x^{m+1}ny^{n-1} $$     (4.2.13)
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$$ g_{y}=\frac{\partial g}{\partial y}=x^m3(n+1)y^n + 2x^{m+1}ny^{n-1}P $$     (4.2.14)
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$$ f_{xp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial x}=\frac{\partial }{\partial P}[(m+\frac{1}{2})y^nx^{m-\frac{1}{2}}]=0 $$     (4.2.15)
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$$ f_{yp}=\frac{\partial }{\partial P} \frac{\partial f}{\partial y}=\frac{\partial }{\partial P}[ny^{n-1}x^{m+\frac{1}{2}}]=0 $$     (4.2.16)
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$$ f_y=\frac{\partial f}{\partial y} =ny^{n-1}x^{m+\frac{1}{2}} $$     (4.2.17)
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$$ g_{pp}=\frac{\partial }{\partial P} \frac{\partial g}{\partial P}=\frac{\partial }{\partial P}[2x^{m+1}y^n]=0 $$     (4.2.18)
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Plugging equations 4.2.9-4.2.18 into equations 4.2.7 and 4.2.8, one can see that
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$$ (m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}}y^n+P[(m+\frac{1}{2})nx^{m-\frac{1}{2}}y^{n-1}]+P^2[n(n-1)x^{m+\frac{1}{2}}y^{n-2}] $$
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$$ =2(m+1)x^my^n+2P[nx^{m+1}y^{n-1}]-x^m3(n+1)y^n-2nx^{m+1}y^{n-1}P=2(m+1)x^my^n-x^m3(n+1)y^n $$     (4.2.19)
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$$ 0+P(0)+2ny^{n-1}x^{m+\frac{1}{2}}=0 $$     (4.2.20)
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From equation 4.2.19, $$ (m+\frac{1}{2})(m-\frac{1}{2})=2(m+1)-3(n+1) $$. From equation 4.2.20, $$ n=0 $$. Thus the equation for $$ m $$ becomes a quadratic, which can be solved for $$ m=\frac{1}{2} $$.

With the values of $$ m $$ and $$ n $$ now known, it can be shown that
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$$ h(x,y)=x^{\frac{1}{2}}y^0=x^{\frac{1}{2}} $$     (4.2.21)
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To find the first integral $$ \phi $$
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$$ h(x,y)[M(x,y)+N(x,y)y']=0 $$     (4.2.22)
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Therefore
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$$ xy''+2x^{\frac{3}{2}}y'+3x^{\frac{1}{2}}y=0 $$     (4.2.23)
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Then $$ \phi $$ can be found such that
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$$ \phi (x,y,P)= \int \phi_pdP + k(x,y)=\int fdP + k(x,y)=\int (y^0x^{\frac{1}{2}+\frac{1}{2}}dP + k(x,y) = xP + k(x,y) $$     (4.2.24)
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From 4.2.24, $$ \phi_x=\frac{\partial \phi}{\partial x}=P+k_x(x,y) $$ and $$ \phi_y =\frac{\partial \phi}{\partial y}= k_y(x,y) $$. Since $$ \frac{d \phi (x,y,P)}{dx} = \phi_x + \phi_y y' $$, it can be seen that $$ P+k_x(x,y) + k_y(x,y)y'=2x^{\frac{3}{2}}y'+3x^{\frac{1}{2}}y $$. Then it follows that
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$$ \int k_x(x,y)dx = \int 3x^{\frac{1}{2}}ydx = 2yx^{\frac{3}{2}}+k(y) $$
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and
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$$ k_y(x,y)=2x^{\frac{3}{2}}+k'(y) $$
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Since k'(y) = -1 (from equation 4.2.24 being set equal to zero and differentiated in y),
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$$ k_y(x,y)=2x^{\frac{3}{2}}+k'(y) $$
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and
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$$ k(x,y)=\int (2x^{\frac{3}{2}}+k'(y)) dy = 2x^{\frac{3}{2}}y - y + k $$
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Therefore, the first integral $$ \phi $$ can be shown to be
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$$ \phi (x,y,P) = xP + (2x^{\frac{3}{2}}-1)y = k $$ (4.2.2)
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Dividing equation 4.2.2 by $$ x $$, we get
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$$ (2x^{\frac{1}{2}}-1/x)y + y' =k/x $$     (4.2.25)
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which is of the form,


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$$ a_0(x)y + y' =b(x) $$     (4.2.26)
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Comparing the above two equations,


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$$ a_0(x)= (2x^{\frac{1}{2}}-1/x), b(x) = k/x $$     (4.2.27)
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From here, one can define $$ M(x,y) = (2x^{\frac{1}{2}}-1/x)y $$ and $$ N(x,y)=1 $$. Thus it can be shown that $$ M_y=\frac {\partial M}{\partial y} = 2x^{\frac{1}{2}}-1/x $$ and $$ N_x =\frac {\partial N}{\partial x}=0 $$. Then it follows that
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$$ \frac{h_x}{h} = -\frac{1}{N}[N_x - M_y] = a_0(x) $$     (4.2.28)
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Integrating equation 4.2.28 yields
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$$ h(x) = exp [ \int^x a_0(s) ds]= exp [ \int^x (2s^{\frac{1}{2}}-1/s) ds] $$
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$$ h(x) = exp [ \frac {4}{3}s^{3/2} - ln(x) ] = \frac{4exp(x^{3/2})}{3x} $$
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Using h(x), we can solve for $$ y(x) $$ such as follows [(1) p11-5]
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$$ y(x) = \frac{1}{h(x)}[\int^x h(s)b(s)+k_1] $$     (4.2.29)
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$$ y(x) = \frac{1}{\frac{4exp(x^{3/2})}{3x}}[\int^x \frac{4exp(s^{3/2})}{3s}*\frac{k}{s} ds +k_1] $$     (4.2.30)
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which is the required solution

Author and References

 * Solved and Typed by -- Kaitlin Harris
 * Reviewed by -- Pushkar Mishra

Problem R*4.3 – Generate a class of exact L2-ODE-VC
sec21-5

Statement
Derive the following equation,

from (4.3.1) and (4.3.2) below,

Solution
Comparing equations (4.3.1) and (4.3.2) the coefficients of y&apos;&apos; can be equated. This is not true for the coefficient of the y and y' terms hence,

Integrating the wrt to p,

where h(x,y) is a function only of x and y only.

Similarly $$ \phi_{y}$$,

Rearranging the terms in (4.3.5)and(4.3.6) in (4.3.1)

Further rearranging the terms within the equation, (also realizing p=y')

Comparing now with (4.3.2), the terms in $$ R(x)y = \frac{\partial h(x,y)}{\partial x} $$ and $$ Q(x)= P'(x)+ \frac{\partial h(x,y)}{\partial y}$$

To find $$\phi$$, integrate all terms with respect to x

Therefore substituting value of h(x,y) from (4.3.9) into equation (4.3.4)

which is equivalent as the problem statement,

Author and References

 * Solved and Typed by -- Jason Alphonso
 * Reviewed by -- Rui Che

Problem R*4.4 - Solve a L2-ODE
sec18-2 sec21-6

Statement
L2-ODE-VC

Find
1.Show equation 4.5.1 is exact

2.Find Φ

3.Slove for $$y(x)$$

1.Show equation 4.4.1 is exact
Let

We obtain

It satisfies the 1st exactness condition

Recall the 2nd exactness condition for L2-ODE-VC:

We substitute all equations above into (4.4.3) and (4.4.4), we obtain

It satisfies the 2nd exactness condition

It is proved

End

2.Find Φ
We know (4.4.1) is exact,so we obtain

We obtain

Finally, we obtain

End

3.Solve for $$y(x)$$
Recall that

From (4.4.5) and (4.4.6), we obtain

Rearrange, we have

Use integration factor

Finally, we obtain

End

Author and References

 * Solved and Typed by -- Jinchao Lu
 * Reviewed by -- Seong Hyeon Hong

Problem R*4.5 – Symmetry of mixed partial derivatives
sec22-4

Statement
Using the general form of the 2nd exactness condition for Nn-ODE, show the equivalence to symmetry of mixed 2nd partial derivatives of the first integral $$ \phi $$

Solution
We know that $$ \phi $$ is a function of x,y,y'.

Taking the total derivative of $$ \phi $$ w.r.t x.


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$$ \frac{d}{dx}\phi= \phi_x + \phi_y y' + \phi_{y'} y'' $$ (4.5.1)
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Since,


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$$ g_i = \frac{\partial }{\partial y^{i}}(\frac{d}{dx}\phi) $$ (4.5.2)
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$$ g_0= \frac{\partial}{\partial y} (\frac{d\phi}{dx}) = \phi_{xy}+\phi_{yy}y'+\phi_{yy'}y'' $$ (4.5.3)
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$$ g_1= \frac{\partial}{\partial y'} (\frac{d\phi}{dx}) = \phi_{xy'}+\phi_{yy'}y'+\phi_y+ \phi_{y'y'}y'' $$ (4.5.4)
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$$ g_2= \frac{\partial}{\partial y''} (\frac{d\phi}{dx}) = \phi_y' $$ (4.5.5)
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For the exactness condition,


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$$ g_0 + \frac{d}{dx} g_1 + \frac{d^2}{dx^2} g_2 = 0 $$ (4.5.6)
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Differentiating $$ g_2$$ with respect to x once,


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$$ \frac{d}{dx} g_2 = \phi_{y'x} + \phi_{y'y}y'+ \phi_{y'y'}y'' $$ (4.5.7)
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Putting in the values of $$ g_0, g_1, g_2 $$ from Eqns. (4.5.2 - 4.5.4) and 4.5.7,


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$$ g_0 - \frac{d}{dx} g_1 + \frac{d^2}{dx^2} g_2 = 0 $$
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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} (\phi_{xy'}+\phi_{yy'}y'+\phi_y+ \phi_{y'y'}y) + \frac{d}{dx} (\phi_{y'x} + \phi_{y'y}y'+ \phi_{y'y'}y) =0 $$ (4.5.8)
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The above can be rearranged to be written as


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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} \phi_y - \frac{d}{dx} (\phi_{xy'}-\phi_{y'x}+\phi_{yy'}y'-\phi_{y'y}y') =0 $$ (4.5.9)
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Considering the part outside the derivative,


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$$ \frac{\partial}{\partial y} (\frac{d\phi}{dx}) - \frac{d}{dx} \phi_y = (\phi_{xy}+\phi_{yy}y'+\phi_{yy'}y) - (\phi_{yx}+\phi_{yy}y'+\phi_{yy'}y)$$ (4.5.10)
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Putting (4.5.10) in (4.5.9),


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$$ \phi_{xy}-\phi_{yx} - \frac{d}{dx} (\phi_{xy'}-\phi_{y'x}+\phi_{yy'}y'-\phi_{y'y}y') =0 $$ (4.5.11)
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For the terms to cancel out we should have the following relations,


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$$ \phi_{xy}=\phi_{yx}, \phi_{xy'}=\phi_{y'x}, \phi_{yy'}= \phi_{y'y} $$
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Hence Proved.

Author and References

 * Solved and Typed by -- Pushkar Mishra
 * Reviewed by -- Kaitlin Harris

Team Contributions

 * Seong Hyeon Hong (SH) Solved all problems.
 * Jason Alphonso (JA) Solved all problems.
 * Pushkar Mishra (PM) Solved all problems.
 * Jinchao Lu (JL) Solved all problems.
 * Che Rui (CH) Solved all problems.
 * Kaitlin Harris (KH) Solved all problems.