User:Egm6321.f11.team4.allen/HW4

= Problem R*4.3 Solve L2-ODE-VC via the Power Form = From [[media:Pea1.f11.mtg21.djvu|Mtg 21-3]]

Given
$$\displaystyle (x^my^n)[\sqrt{x}y''+2xy'+3y]=0$$

Find
$$\displaystyle \mathbf a)$$ Find $$\displaystyle m, n $$ to render the above equation, equation (1) p.21-3, exact. $$\displaystyle \mathbf b)$$ Show that the first integral is a L1-ODE-VC as shown below, equation (2) p.21-3. $$\displaystyle \phi(x,y,p)=xp+(2x^\frac{3}{2}-1)y=k $$ with $$\displaystyle p(x):=y'(x) $$. $$\displaystyle \mathbf c)$$ Solve equation (2) p.21-3 for $$\displaystyle y(x) $$.

Solution Part A
$$\displaystyle \mathbf a)$$ In order to determine $$\displaystyle m $$ and $$\displaystyle n $$, we must analyze the two conditions of exactness for N2-ODEs. The first condition requires the N2-ODE to fit the following particular form of equation (2) p.16-4. $$\displaystyle G(x,y,y',y)=g(x,y,p)+f(x,y,p)y=0 $$ The second condition requires the N2-ODE to satisfy the following equations, (1)-(2) p.16-5. $$\displaystyle f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y $$ $$\displaystyle f_{xp}+pf_{yp}+2f_y=g_{pp} $$ First, we must determine the new $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ that include the Integrating Factor, $$\displaystyle h(x,y)=x^m y^n $$. Factor through by the integrating factor to yield the following: $$\displaystyle g(x,y,p)=2x^{m+1}y^np+3x^m y^{n+1} $$ $$\displaystyle f(x,y,p)=x^{0.5+m}y^n $$ This satisfies the first exactness condition. Since $$\displaystyle f(x,y,p) $$ is not a function of $$\displaystyle p $$, the second equation of the second exactness condition significantly simplifies to $$\displaystyle \cancel {f_{xp}}+\cancel {pf_{yp}}+2f_y=\cancel {g_{pp}} $$ $$\displaystyle 2f_y=0 $$. Differentiate $$\displaystyle f(x,y,p)=x^{0.5+m}y^n $$ with respect to $$\displaystyle y $$ to obtain $$\displaystyle f_y(x,y,p)=nx^{0.5+m}y^{n-1} $$. Plugging in $$\displaystyle f_y(x,y,p) $$ into $$\displaystyle 2f_y=0 $$ yields $$\displaystyle 2f_y=2nx^{0.5+m}y^{n-1}=0 $$. The only way to solve the problem is to set $$\displaystyle n=0 $$. Since we solved for $$\displaystyle n $$, we can simplify $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ to $$\displaystyle f(x,y,p)=x^{0.5+m} $$. $$\displaystyle g(x,y,p)=2x^{m+1} p+3x^m y $$. Now we turn our attention to the first equation of the second exactness condition after we simply. $$\displaystyle f_{xx}+\cancel {2pf_{xy}}+\cancel {p^2f_{yy}}=g_{xp}+\cancel {pg_{yp}}-g_y $$ Next, solve for the partial differentials of $$\displaystyle f(x,y,p) $$ and $$\displaystyle g(x,y,p) $$ with the results shown below. $$\displaystyle f_{xx}(x,y,p)=(m+0.5)(m-0.5)x^{m-0.5} $$. $$\displaystyle g_{xp}(x,y,p)=2(m+1)x^m $$. $$\displaystyle g_y(x,y,p)=3x^m $$. Plug these three equations back into the first equation of the second exactness condition to obtain $$\displaystyle (m+0.5)(m-0.5)x^{m-0.5}=2(m+1)x^m-3x^m $$. After a visual inspection, one can deduce the LHS to be zero due to no other like terms. $$\displaystyle 2(m+1)\cancel {x^m}-3 \cancel {x^m}=0 $$ The $$\displaystyle x^m $$ terms cancel out leaving a simple algegra problem. $$\displaystyle 2(m+1)-3=0 $$ So $$\displaystyle m=0.5 $$. The integrating factor for this L2-ODE-VC is $$\displaystyle h(x,y)=x^{0.5} $$.

Solution Part B
$$\displaystyle \mathbf b) $$ We can solve for the first integral using equation (3) p.16-5, $$\displaystyle \phi(x,y,p)=h(x,y)+\int f(x,y,p) dp $$. Since $$\displaystyle f(x,y,p)=x $$ we can simply equation (3) p.16-5 to $$\displaystyle \phi(x,y,p)=h(x,y)+xp $$. Solve this for the partials as shown below. $$\displaystyle \phi_x=h_x+p $$. $$\displaystyle \phi_y=h_y $$. Next, we use the first equation of the second exactness condition, $$\displaystyle g(x,y,p):=\phi_x+\phi_yp $$, to obtain the following: $$\displaystyle 3x^{0.5}y+2x^{1.5}p=h_x+p+h_yp $$. One can equate the following: $$\displaystyle h_x=3x^{0.5} $$. $$\displaystyle h_y+1=2x^{1.5} $$. Integrating these yields the following: $$\displaystyle h=2x^{1.5}y+k_1(y) $$. $$\displaystyle h=(2x^{1.5}-1)y+k_2(x) $$. After differentiating the equation you obtain $$\displaystyle h_y=\cancel {2x^{1.5}}+k_1'(y)=\cancel {2x^{1.5}}-1 $$. $$\displaystyle h_x=\cancel {3x^{1.5}y}+k_x'(x)=\cancel {3x^{0.5}y} $$. Integrating yields the next result: $$\displaystyle h_1(y)=-y+k_1 $$. $$\displaystyle h_2(x)=k_2 $$. Plug these back into the integrating function equation to yield $$\displaystyle h=2x^{1.5}y-y+k_1 $$. $$\displaystyle h=(2x^{1.5}-1)y+k_2 $$. These equation are the same. Now that we have solved for the integrating function equation, we can input this into the original equation for the final solution. $$\displaystyle \phi(x,y,p)=xp+(2x^{1.5}-1)y=k $$ with $$\displaystyle k=k_3-k_1 $$.

Solution Part C
$$\displaystyle \mathbf c) $$ Next we need to solve for $$\displaystyle y(x) $$ starting from the first integral equation above. The N1-ODE-VC can be rearranged to the following form: $$\displaystyle P(x)y'+Q(x)y=R(x) $$. $$\displaystyle xy'+(2x^{1.5}-1)y=k $$. Divide through by $$\displaystyle x $$ to obtain $$\displaystyle y'+\frac{Q(x)}{P(x)}y=\frac{R(x)}{P(x)} $$. $$\displaystyle y'+(2x^{0.5}-\frac{1}{x})y=\frac{k}{x} $$. Using equation (3) p.11-4, we can solve for the integrating factor. Knowing $$\displaystyle a_0(x)=\frac{Q(x)}{P(x)} $$ and solving yields $$\displaystyle h(x)=\exp[\int^x a_0(s)ds+k_1] $$ $$\displaystyle h(x)=\exp[\int^x \frac{Q(s)}{P(s)} ds+k_1] $$ $$\displaystyle h(x)=\exp[\int^x (2s^{0.5}-\frac{1}{s})ds+k_1] $$ $$\displaystyle h(x)=\exp[\frac{4}{3}x^{1.5}-\ln{x}+k_1] $$ $$\displaystyle h(x)=\frac{1}{x}\exp[{\frac{4}{3}x^{1.5}}+k_1] $$. Next we use equation (1) p. 11-5 to solve for $$\displaystyle y(x) $$. $$\displaystyle y(x)=\frac{1}{h(x)}\left[\int^x h(s)b(s)ds+k_2\right] $$ $$\displaystyle b(x)=\frac{R(x)}{P(x)}=\frac{k}{x} $$ $$\displaystyle y(x)=\frac{1}{\frac{1}{x}\exp[{\frac{4}{3}x^{1.5}}+k_1]}*\left[\int^x (\frac{1}{s}\exp[{\frac{4}{3}s^{1.5}}+k_1])(\frac{k}{s})ds+k_2 \right]$$

Author
Egm6341.f11.team4.allen 03:45, 18 October 2011 (UTC)