User:Egm6321.f12.team7.parikh.a/R5.13

=R*5.13=

Problem: Verify exactness and Solutions of L2-ODE-VC
Problem R*5.13 from lecture notes section 27.

Given: 2 L2-ODE-VC
Given the following two linear, 2nd order, ordinary differential equations with varying coefficients:

Find: Exactness, Integrating factors and Solution
Verify the exactness of the two equations. If ($$) is not exact, check whether it is in power form and can be made exact using IFM with $$h(x,y) = x^m y^n$$. Also, verify the first few Hermite polynomials as shown below are homogenous solutions to ($$).

Solution

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On our honor, we did this assignment on our own, without looking at the solutions in previous semesters or other online solutions.
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Setting:

then it is apparent that ($$) satisfies the first exactness condition, i.e., that it is of the form:

Similarly, setting

makes it apparent that ($$) satisfies the first exactness condition.

The second exactness condition is checked using relations (1) and (2) in lecture notes section 16-5 and also by relation (1) in lecture notes section 22-3.

The partial derivatives of ($$) and ($$) are given below.

Plugging into (1) and (2) of 16-5

The two sides of ($$) are only equal if $$n = 0$$.

For relation (1) of 22-3,

Again, the relation is only true for $$n = 0$$. Therefore ($$) is only exact for $$n = 0$$.

The partial derivatives of ($$) and ($$) are given below.

Plugging into (1) and (2) of 16-5

Since ($$) is never true, ($$) is not exact.

For relation (1) of 22-3,

Again, the relation not true and therefore ($$) is not exact.

($$) can be considered to be in the power form given in (3) of 21-2 if

Therefore, an integrating factor of the form $$h(x,y) = x^m y^n$$ can be used. This results in, where $$n$$ in the original equation is replaced with $$s$$ to avoid confusion:

The power coefficients, $$m$$ and $$n$$ must now be determined. The partial derivatives are now given by:

Relation (2) of 16-5 becomes

Therefore, $$n=0$$. Relation (1) of 16-5 now becomes

This implies

Which is satisfied when $$m = 1, s = -2$$. This means the integrating factor that makes the original equation exact is:

Another solution is the first Hermite polynomial:

Which implies $$n=0$$. Therefore

And the differential equation is satisfied.

For $$n=1$$, the Hermite polynomial solution is:

Therefore,

The equation is satisfied.

For $$n=2$$, the Hermite polynomial solution is:

Therefore,

Which indeed is a homogenous solution.