User:Egm6321.f11.team2.Xia/RP4.3

=R*4.3 - Integrating Factor Methods =

Given
Giving the following N2-ODE $$\displaystyle \sqrt{x}y''+2xy'+3y=0$$ (4.3.1)

Find
find $$ \displaystyle m,n\in\mathbb{R} $$ such that 4.3.2 is exact: $$ \displaystyle (x^{m}y^{n})\left[\sqrt{x}y''+2xy'+3y\right]=0 $$ (4.3.2) show that the first integral is a L1-ODE-VC $$ \displaystyle \phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y=k $$ (4.3.3) with $$ \displaystyle p(x):=y'(x) $$ Solve 4.3.3 for $$ \displaystyle y $$.

Solution
The $$ 2^{nd} $$ exactness condition is , $$\displaystyle f_{xx} + 2pf_{xy}+p^2f_{yy} = g_{xp} + pg_{yp} - g_{y}$$  (4.3.4) $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp} $$  (4.3.5) From 4.3.2, we have $$ \displaystyle \underbrace{(x^{m+\frac{1}{2}}y^{n})}_{f(x,y,p)}y''+\underbrace{2(x^{m+1}y^{n})p+3x^{m}y^{n+1}}_{g(x,y,p)}=0 $$  (4.3.5) Then $$\displaystyle f_{x}=(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n} $$ $$\displaystyle f_{xx}=(m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}}y^{n} $$ $$\displaystyle f_{xy}=n(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n-1} $$ $$\displaystyle g_{pp}=0 $$ $$\displaystyle f_{yy}=n(n-1)x^{m-\frac{1}{2}}y^{n-2} $$ $$\displaystyle f_{xp}=0,\qquad f_{yp}=0 $$ $$\displaystyle g_{x}=2(m+1)x^{m}y^{n}p+3mx^{m-1}y^{n+1} $$ $$\displaystyle g_{y}=2nx^{m+1}y^{n-1}p+3(n+1)x^{m}y^{n} $$ $$\displaystyle g_{xp}=2(m+1)x^{m}y^{n}+3mx^{m-1}y^{n+1} $$ $$\displaystyle g_{yp}=2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^{n} $$ $$\displaystyle g_{pp}=0 $$

Substitute the above equation to 4.3.4 and 4.3.5, we have $$\displaystyle (m+\frac{1}{2})(m-\frac{1}{2})x^{m-\frac{3}{2}}y^{n}+2pn(m+\frac{1}{2})x^{m-\frac{1}{2}}y^{n-1}+p^{2}n(n-1)x^{m-\frac{1}{2}}y^{n-2}=2(m+1)x^{m}y^{n} $$ $$\displaystyle +3mx^{m-1}y^{n+1}+p\left(2nx^{m+1}y^{n-1}+3(n+1)x^{m}y^{n}\right)-2nx^{m+1}y^{n-1}p-3(n+1)x^{m}y^{n} $$  (4.3.6) $$\displaystyle 2nx^{m+\frac{1}{2}}y^{n-1}=0 $$  (4.3.7) From 4.3.7 we know $$\displaystyle n=0 $$ , then exam 4.3.6 we have $$\displaystyle m= \frac{1}{2} $$. Thus $$\displaystyle f(x,y,p)=\underbrace{x}_{\phi_{p}},\quad g(x,y,p)=\underbrace{3x^{\frac{1}{2}}y}_{\phi_{x}}+\underbrace{2x^{\frac{3}{2}}}_{\phi_{y}}p $$  (4.3.8) Integrating 4.3.8, we have $$\displaystyle \begin{array}{l} \phi=xp+a(x,y)\\ \phi=2x^{\frac{3}{2}}y+b(y,p)\\ \phi=2x^{\frac{3}{2}}y+c(x,p)\end{array} $$ Thus $$ \displaystyle \phi(x,y,p)=xp+(2x^{\frac{3}{2}}-1)y=k $$  (4.3.9) From 4.3.9, we have $$\displaystyle \underbrace{1}_{a_{1}(x)}\cdot y'+\underbrace{(2x^{\frac{1}{2}}-\frac{1}{x})}_{a_{0}(x)}y=\underbrace{\frac{k}{x}}_{b(x)} $$ So