User:EGM6321.f12.team4.Rui.C/Homework2 R2.11

Statement
The general L1-ODE-VC is



a_1(x)y'+a_0(x)=b(x) $$


 * (1) $$a_1(x)=1, a_0(x)=x+1, b(x)=x^2+4$$


 * (2) Solve $$y(x)$$ in terms of $$a_1(x),a_0(x)$$ and $$b(x)$$


 * (3) $$a_1(x)=x^2+1, a_0(x)=x^3, b(x)=x^4$$

Solve these two specific L1-ODEs-VC.

Solution(1)
For (1):

We first check the exactness condition 1:

the original equation can be written as:

which is in the form:

Here $$M(x,y)=(x+1)y-(x^2+4)$$ and $$N(x,y)=1$$. So the first exactness condition holds.

For exactness condition 2:

Then $$M_y \neq N_x$$. So the second exactness condition does not hold.

Assume a factor $$h(x,y)$$. Let

and

Take the derivative of $$\overline{M}(x,y)$$ and $$\overline{N}(x,y)$$ with respect to $$y$$ and $$x$$ respectively:

Let the above two terms equal to each other:

We further assume that $$h=h(x)$$, namely $$h$$ is only the function of $$x$$. Then after plugging the expression of $$M_y$$ and $$N_x$$ into the equation, we have:

Recall that

Then the equation becomes:

We integrate both side of the equation:

where $$C$$ is the integral constant.

Then

Note that

Plug it into the first term of the equation to replace $$h(x)$$ with $$xh_x(x)$$:

We reorganize it to be:

Then we can integrate both side of the equation with respect to x:

This integration is difficult to be solved. Finally we use a symbolic integration software to integrate it out:

Solution(2)
The general L1-ODE-VC is

which is in the form:

Dividing Eq (2.11.27) throughout by $$ a_1(x) $$, we get,

Substituting $$\frac{a_0(x)}{a_1(x)}$$ as $$ a_0'(x)$$ and $$\frac{b(x)}{a_0(x)}$$ as $$b'(x)$$ we get,

which is of the same form as Eq. (1)p11.4 [p11.4].

For the second exactness condition to hold true as in Eq 4 p11.2 [p11.2].

Since $$h$$ is a function of x only,

It can be easily seen that, $$\frac{-1}{N}(N_x-M_y)= a_0'(x)$$. Integrating both sides of Eq. (2.11.30)

Since $$ h=ah_x $$ and using the above in Eq. (2.11.28), we get

The above can be integrated to get, the solution $$ y(x) $$

Using the values of $$h(x)$$ and replacing $$a_0'(x)$$ with $$\frac{a_0(x)}{a_1(x)}$$ and $$b'(x)$$ with $$\frac{b(x)}{a_1(x)} $$

Solution(3)
For (3):

We first check the exactness condition 1:

the original equation can be written as:

which is in the form:

Here $$M(x,y)=\frac{x^3y-x^4}{x^2+1}$$ and $$N(x,y)=1$$. So the first exactness condition holds.

For exactness condition 2:

Then $$M_y \neq N_x$$. So the second exactness condition does not hold.

Assume a factor $$h(x,y)$$. Let

and

Take the derivative of $$\overline{M}(x,y)$$ and $$\overline{N}(x,y)$$ with respect to $$y$$ and $$x$$ respectively:

Let the above two terms equal to each other:

We further assume that $$h=h(x)$$, namely $$h$$ is only the function of $$x$$. Then after plugging the expression of $$M_y$$ and $$N_x$$ into the equation, we have:

Recall that

Then the equation becomes:

We integrate both side of the equation:

where $$C$$ is the integral constant.

Then

Note that

Plug it into the first term of the equation to replace $$h(x)$$ with $$xh_x(x)$$:

We reorganize it to be:

Then we can integrate both side of the equation with respect to x:

At this time, even the software cannot solve this complex integration analytically. We can assume the result to be:

So the final result can be written as:

where $$k=\frac{C_2}{C_1}$$.

Author and References

 * Solved and Typed by -- Rui Che