User:Egm6321.f12.team4.harris/Homework4 R*4.2

Problem R*4.2 –
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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To solve equation 4.2.2 for $$ y $$, it must be of the form
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$$ G(x,y,y',y'')=0=xP + (2x^{\frac{3}{2}}-1)y-k $$     (4.2.25)
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From here, one can define $$ M(x,y) = (2x^{\frac{3}{2}}-1)y-k $$ and $$ N(x,y)=x $$. Thus it can be shown that $$ M_y=\frac {\partial M}{\partial y} = 2x^{\frac{3}{2}}-1 $$ and $$ N_x =\frac {\partial N}{\partial x}=1 $$. Then it follows that
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$$ \frac{h_x}{h} = -\frac{1}{N}[N_x - M_y] = -\frac{1}{x}[1-2x^{\frac{3}{2}}+1]=2x^{\frac{1}{2}}-\frac{2}{x} $$     (4.2.26)
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Integrating equation 4.2.26 yields
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$$ \int \frac{h_x}{h}dx = \int 2x^{\frac{1}{2}}dx - \int \frac{2}{x}dx $$
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$$ ln h(x) = \frac{4}{3}x^{\frac{3}{2}} - 2lnx + k_1 $$
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As $$ k_1 $$ is not necessary,
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$$ h(x) = \frac{1}{x^2}exp[\frac{4}{3}x^{\frac{3}{2}}] $$     (4.2.27)
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One can solve for $$ y(x) $$ such that
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$$ y(x) = \frac{1}{h(x)}[\int^x h(s)n(s)+k] $$     (4.2.28)
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Author and References

 * Solved and Typed by -- Kaitlin Harris