User:Egm6321.f11.team4.shin.js/HW6

= Problem R*6.3 - Invalid solution = From [[media:Pea1.f11.mtg35.djvu|Mtg 35-4]]

Given
The first homogeneous solution: Trial solution: Characteristic equation:

Find
Explain why $$\displaystyle r_2(x)$$ is not a valid root, i.e., $$\displaystyle u_2(x) = e^{xr_{2}(x)}$$ is not a valid solution.

Solution
- Solved on our own Validity of $$\displaystyle u_2(x) = e^{xr_2(x)}$$ can be shown by substituting $$\displaystyle y= u_2(x) $$ into the Eq. (4.1). Before the substitution, compute the first and the second derivatives of $$\displaystyle y $$. Substitute $$\displaystyle y, y^{'}, y^{''}$$ in the Eq. (3.6) into the Eq. (3.1), - As shown in the Eq. (3.7), $$\displaystyle u_2(x) $$ is not a valid homogeneous solution.

Author
Contributed by Shin

= Problem R*6.4 - 2nd homogeneous solution by variation of parameters = From [[media:Pea1.f11.mtg35.djvu|Mtg 35-4]]

Given
Application: King 2003 p.28

Find
For the L2-ODE-VC Eq. (4.1), select a valid homogeneous solution, and call it $$\displaystyle u_1 $$. Find the 2nd homegeneous solution $$\displaystyle u_2(x) $$ by variation of parameters, and compare to $$\displaystyle e^{xr_2(x)} $$.

Solution
- Solved on our own Let trial solution be: Then, characteristic equation is: After factoring, the Eq. (4.3) becomes: Therefore, the first homogeneous solution is: Since it was proven that $$\displaystyle r_2(x) $$ is not a valid root in the problem R*6.3, $$\displaystyle r_2(x) $$ is ignored. The second homogeneous solution can be obtained using the equation (4) in the class note [[media:Pea1.f11.mtg34.djvu|34-5]]. The Eq. (3.6) depicts the equation (4) in the class note [[media:Pea1.f11.mtg34.djvu|34-5]]. The term $$\displaystyle a_1(x)$$ is determined by changing the form of the Eq. (4.1) as follows. First, the term $$\displaystyle \frac{1}{h(x)} $$ is first computed in order to solve the Eq. (4.7). Substitute $$\displaystyle a_1(x) $$ in the Eq. (4.9) into the Eq. (4.8), then compute $$\displaystyle \frac{1}{h(x)} $$, Then, substitute $$\displaystyle \frac{1}{h(x)} $$ from the Eq. (4.10) into the Eq. (4.7), - Since the second homogeneous solution has the form of $$ \displaystyle u_2(x) = -x $$, $$ \displaystyle u_2(x) = e^{x r_2(x)} $$ is not valid solution.

Author
Contributed by Shin