User:EGM6321.F11.TEAM1.Prob R*4.3 Bens

=Problem R*4.3: Find $$\displaystyle m$$ and $$\displaystyle n$$ of the integrating factor $$\displaystyle H(x,y) = x^{m}y^{n} $$ such that the given N2-ODE is exact then solve for $$\displaystyle y(x)$$.=

Given
A function is given by


 * $$\displaystyle x^{m}y^{n} \lbrack \sqrt{x}y''+2xy'+3y \rbrack = 0$$

$$\displaystyle (1) p.21-3 $$

Per Lecture notes: [[media:pea1.f11.mtg21.djvu|Mtg 21 (c)]] where


 * $$\displaystyle h(x,y) = x^{m}y^{n}$$

$$\displaystyle (4.3.1) $$

Find
Part A: Find $$\displaystyle (m,n) $$ such that $$\displaystyle (1) p.21-3 $$ is exact. 

Part B: Show that the first integral is a L1-ODE-VC and solve


 * $$\displaystyle \phi(x,y,p) = xp+(2x^{\frac{3}{2}}-1)y = k$$ $$\displaystyle (2)p.21-3

$$

for $$\displaystyle y(x)$$.

Solution

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Part A:

The first exactness criteria for second order non-liner ODE is
 * $$\displaystyle F(x,y,p) = f(x,y,p)y'' + g(x,y,p) $$

$$\displaystyle (2) p.16-4 $$

where


 * $$\displaystyle p = y'$$

Comparing (1) p.21-3 with the form of (2) p.16-4, the functions $$\displaystyle f $$ and $$\displaystyle g $$ can be defined as


 * $$\displaystyle f(x,y,p)=x^{0.5+m}y^{n}$$$$\displaystyle (4.3.2) $$


 * $$\displaystyle g(x,y,p)=2x^{m+1}y^{n}p+3x^{m}y^{n+1}$$$$\displaystyle (4.3.3) $$

So the form of the 1st exactness condition is met.

For a second order non-liner ODE, the first part of the second exactness condition is
 * $$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_{y} $$$$\displaystyle (1) p.16-5 $$

For a second order non-liner ODE, the second part of the second exactness condition is
 * $$\displaystyle f_{xp} + pf_{yp} + 2f_{y} = g_{pp}$$$$\displaystyle (2) p.16-5 $$

Using the second part of the second exactness condition given above in $$\displaystyle (2) p.16-5 $$ and recognizing the given$$\displaystyle f(x,y,p) $$ from equation $$\displaystyle (4.3.2) $$  is not a function of $$\displaystyle p$$, so the first two terms are zero. Additionally $$\displaystyle g_p$$ is constant wrt. $$\displaystyle p$$ and therefore $$\displaystyle g_{pp}$$ is zero. $$\displaystyle (2) p.16-5 $$ can be simplified as shown below


 * $$\displaystyle 2f_y=g_{pp}=0$$$$\displaystyle (4.3.4)$$

Taking the derivative .wrt $$\displaystyle y $$ of $$\displaystyle f(x,y,p) = f_y $$ in equation $$\displaystyle (4.3.2) $$ and substituting into $$\displaystyle (4.3.4)$$ yields


 * $$\displaystyle 2\left[x^{0.5+m}ny^{n-1}\right]=0$$$$\displaystyle (4.3.5) $$

In order for equation $$\displaystyle (4.3.5) $$ to be true $$\displaystyle m=0$$


 * $$\displaystyle n=0$$$$\displaystyle (4.3.6) $$

For clarity we can now re-write equations $$\displaystyle (4.3.2) $$$$\displaystyle (4.3.3) $$ with $$\displaystyle n=0$$ as shown below:


 * $$\displaystyle f=x^{0.5+m}$$ $$\displaystyle (4.3.7) $$


 * $$\displaystyle g=2x^{m+1}p+3x^{m}y $$ $$\displaystyle (4.3.8) $$

Next, use the first part of the the second exactness condition given in equation $$\displaystyle (1) p.16-5  $$ with the following derivatives of $$\displaystyle f $$ and $$\displaystyle g $$ plugged in.


 * $$\displaystyle {f_{xx} = (m+0.5)(m-0.5)x^{-1.5+n}}$$


 * $$\displaystyle{f_{xy} = 0}$$


 * $$\displaystyle{f_{yy} = 0}$$


 * $$\displaystyle{g_{xp} = 2(m+1)x^{m}}$$


 * $$\displaystyle{g_{yp} = 0}$$


 * $$\displaystyle{g_y=3x^{m}}$$

Which yeilds:
 * $$\displaystyle(m+0.5)(m-0.5)x^{-1.5+m} = 2(m+1)x^{m}-3x^{m}$$ $$\displaystyle (4.3.9) $$

Solving this equation for $$ \displaystyle m $$ yields


 * $$ \displaystyle m = \frac{1}{2}$$ $$\displaystyle (4.3.10)$$

So the integrating factor to make $$\displaystyle (1) p.21-3 $$ exact is


 * $$\displaystyle h(x,y) = x^{0.5} $$ <p style="text-align:right;">$$\displaystyle (4.3.11) $$

Part B:

So using the integrating factor found above and multiplying through to equation $$\displaystyle (1) p.21-3 $$ to get and exact N2-ODE one can use equation $$\displaystyle (3) p.16-5 $$ to obtain the first integral given in $$\displaystyle (2)p.21-3 $$

Rearranging equation $$\displaystyle (2)p.21-3 $$ and solving for $$\displaystyle y(x)$$ by dividing through by x to put the ODE into the general form:


 * $$ \displaystyle y' + a_0(x)y = b(x) $$<p style="text-align:right;">$$\displaystyle (3)p.11-3 $$

Yields:


 * $$ p + (2x^{\frac{1}{2}}-{\frac{1}{x}} )y=k $$

Where:
 * $$ a_o(x)=\frac{(2x^{\frac{3}{2}}-1)}{x}=2x^{\frac{1}{2}}-x^{-1}$$ <p style="text-align:right;">$$\displaystyle (4.3.12) $$
 * $$ b(x)=\frac{k}{x}$$ <p style="text-align:right;">$$\displaystyle (4.3.13) $$

The ODE can be solved by finding an integrating factor $$h(x,y)$$ per


 * $$h(x)=exp(\int a_o(x)dx)$$ <p style="text-align:right;">$$\displaystyle (3)p.11-4 $$


 * $$y(x)= \frac{1}{h(x)}\int h(x)b(x)dx+C_1 $$ <p style="text-align:right;">$$\displaystyle (1)p.11-5 $$

Substituting in $$h(x)=exp(\int a_o(x)dx)$$ gives:


 * $$y(x)=\frac{\int exp(\int a_0(x)dx)b(x)+C_1}{exp(\int a_0(x)dx)}$$ <p style="text-align:right;">$$\displaystyle (4.3.14) $$

Integrating $$a_0(x)$$ which is given in $$\displaystyle (4.3.12) $$
 * $$\int a_0(x)dx=\int 2x^{\frac{1}{2}}-x^{-1}=\frac{4}{3}x^{\frac{3}{2}}-ln(x)+C_2$$ <p style="text-align:right;">$$\displaystyle (4.3.15) $$

Plugging in $$\displaystyle (4.3.15) $$ and $$\displaystyle (4.3.13) $$ into $$\displaystyle (4.3.14) $$ yields:
 * $$y(x)=\frac{\int C_3x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})\frac{k}{x}+C_1}{ C_3x^{-1}exp(\frac{4}{3}x^{\frac{3}{2}})}$$ <p style="text-align:right;">$$\displaystyle (4.3.16) $$

where $$C_3 = exp(C_2) $$

Equation $$\displaystyle (4.3.16) $$ is the solution for $$\displaystyle y(x) $$