User:Egm6321.f09.Team6.dianafoster/HW2

Problem 4:
Statement: 

Show the following is an exact N1-ODE:

$$(\frac{1}{3})x^3y^4y'+(5x^3+2)(\frac{1}{5})y^5=0 $$

Solution: 

First, we need to see if the ODE has the form

$$M+Ny'=0$$

It does, so the first condition for exactness is satisfied.

Next, we need to see if

$$M_y=N_x$$

We know that

$$M_y=a(x)c(y)$$ $$N_x=b(x)c(y)$$

From the problem statement, we can determine the following: $$a(x)=5x^3+2$$ $$bi(x)=(\frac{1}{3})x^3$$ $$b(x)=x^2$$ $$c(y)=y^4$$ $$ci(y)=(\frac{1}{5})y^5$$

Therefore:

$$M_y=5x^3y^4+2y^4$$ $$N_x=x^2y^4$$

These are not equal! Thus, we must use the Euler integrating factor method. This class of exact N1-ODE is related to case 1. That means $$h_y$$ is equal to 0. Thus: $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)c(y)})(b(x)c(y)-a(x)c(y))$$ $$(\frac{1}{N})(N_x-M_y)=(\frac{1}{bi(x)})(b(x)-a(x))$$ $$(\frac{1}{N})(N_x-M_y)=-f(x)$$

We now know that h(x,y) is only a function of x. Therefore: $$h(x)=exp\int{f(x)dx}$$

where: $$f(x)=(\frac{1}{bi(x)})(b(x)-a(x))$$

So: $$h(x)=exp\int{\frac{x^2-5x^3-2}{\frac{1}{3}x^3}}dx$$

$$h(x)=exp\frac{3(x^2ln(x)-5x^3+1)}{x^2}$$

$$h(x)=abs(x^3)exp(\frac{3}{x^2}-15x)$$

The initial equation is now exactly integratable by the Euler integrating factor method: $$hM+hNy'=0$$

Problem 5:
Statement: 

Show that the second condition is satisfied for the following N2-ODE:

$$(xy)y''+x(y')^2+yy'=0$$

Solution: 

We know that in order for the second condition of exactness to be satisfied, the following two equations must hold true:

$$f_{xx}+2pf_{xy}+p^2f_{yy}=g_{xp}+pg_{yp}-g_y$$ $$f_{xp}+pf_{yp}+2f_y=g_{pp}$$

Solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xx}=0$$ $$f_x=y$$ → $$f_{xy}=1$$ $$f_y=x$$ → $$f_{yy}=0$$ $$g_x=p^2$$ → $$g_{xp}=2p$$ $$g_y=p$$ → $$g_{yp}=1$$ $$g_y=p$$

When entered into the first equation, we get: $$0+2p(1)+p^2(0)=2p+p(1)-p$$ $$2p=2p$$

The first equation holds true! Now let's check the second equation. First, solving for the partial differential equations gives: $$f_x=y$$ → $$f_{xp}=0$$ $$f_y=x$$ → $$f_{yp}=0$$ $$f_y=x$$ $$g_p=2xp+y$$ → $$g_{pp}=2x$$

When entered into the second equation, we get: $$0+p(0)+2x=2x$$ $$2x=2x$$

The second equation also holds true!

Thus, $$(xy)y''+x(y')^2+yy'=0$$ satisfies the second condition of exactness for N2-ODEs.