User:EGM6321.f12.team4.Rui.C/Homework2 R2.10

Statement
The non-homogeneous L1-ODE-VC is



y'+\frac{1}{x}y=x^2 $$

Show that



h(x)=x $$ and $$ y(x)=\frac{x^3}{4}+\frac{k}{x} $$

Solution
We first check the exactness condition 1:

the original equation can be written as:

which is in the form:

Here $$M(x,y)=\frac{1}{x}y-x^2$$ 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:

If we let the constant $$k=\frac{C_2}{C_1}$$, then we can get the required form of $$y(x)$$:

For the required expression of $$h(x)=C_1x$$, we can let the integral constant $$C_1=1$$, then:

Thus, we have shown the expression of both $$h(x)$$ and $$y(x)$$ as required.

Author and References

 * Solved and Typed by -- Rui Che