User:EGM6321.f10.team6.yoshioka/HW5

= Problem 3 Solution of the Euler L2_ODE_VC and L2_ODE_CC =

Given
The Euler L2_ODE_VC:

The Euler L2_ODE_CC:

And the characteristic equation

Find
1.1) $${{a}_{2}}$$, $${{a}_{1}}$$ and $${{a}_{0}}$$ such that equation 3.3 is the characteristic equation of 3.1.

1.2) The 1st homogeneous solution

1.3) The second homogeneous solution

1.4) The general homogeneous solution

2.1) $${{b}_{2}}$$, $${{b}_{1}}$$ and $${{b}_{0}}$$ such that equation 3.3 is the characteristic equation of 3.2.

2.2) The 1st homogeneous solution

2.3) The 2nd homogeneous solution

2.4) The general homogeneous solution

1.1) The coefficients of 3.1 can be found by using the trial solution method. The trial solution for the Euler equation with variable coefficient is:
The derivatives of the trial solution is defined as:

Substituting the equations 3.1.1, 3.1.2 and 3.1.3 into 3.1 we get:

Therefore we need to solve the following equation to find the coefficients:

Comparing the equation 3.1.7 to the equation 3.3 we have:

Therefore, comparing the coefficients of both side of the equation 3.1.8 we get:

1.3) Since $$\lambda $$ has two identical roots the second homogeneous solution can be calculate using the form:
Calculating the derivatives of 3.1.17:

Substituting 3.1.17, 3.1.18 and 3.1.19 into 3.1 we have:

Rearranging the terms of 3.1.20:

The equation 3.1.21 is simplified as:

Substituting the coefficients $${{a}_{2}}$$ (equation 3.1.10), $${{a}_{1}}$$ (equation 3.1.13), $${{a}_{0}}$$ (equation 3.1.14) and the 1st homogeneous solution with the respective derivatives presented in $${{y}_{1}}$$ (equation 3.1.1), $${{y}_{1}}'$$ (equation 3.1.2) and $${{y}_{1}}''$$ (equation 3.1.13), where r is replaced by $$\lambda $$ into 3.1.22:

Therefore we will solve for:

Using the reduction of order:

Substituting 3.1.28 into 3.1.29:

Integrating both sides of 3.1.32:

Using the exponential function in 3.1.34

Where $${{k}_{2}}$$ can be written as:

Therefore, z(x) can be written as:

Since the order of U(x) was reduced, we can get U(x) by integrating 3.1.38

Finally, substituting 3.1.42 and 3.1.16 into 3.1.17 the 2nd homogeneous solution is defined as:

1.4) The generalized homogeneous solution can be calculated by using the following equation:
Substituting equations 3.1.16 and 3.1.43 into 3.1.44 we have:

Rearranging the constants from equation 3.1.47, the generalized homogeneous solution can be expressed as:

2.1) The coefficients of 3.2 can be found by using the trial solution method. The trial solution for the Euler equation with constant coefficient is:
The derivatives of the trial solution is defined as:

Substituting the equations 3.2.1, 3.2.2 and 3.2.3 into 3.2 we get:

Therefore we need to solve the following equation to find the coefficients:

Comparing the equation 3.2.6 to the equation 3.3 we have:

Therefore, comparing the coefficients of both side of the equation 3.2.7 we get:

2.3) Since $$\lambda $$ has two identical roots the 2nd homogeneous solution can be calculate using the form:
Calculating the derivatives of 3.2.15:

Substituting 3.1.15, 3.1.16 and 3.1.17 into 3.2 we have:

Rearranging the terms of 3.2.18:

The equation 3.2.19 is simplified as:

Substituting the coefficients $${{b}_{2}}$$ (equation 3.2.9), $${{b}_{1}}$$ (equation 3.2.11), $${{b}_{0}}$$ (equation 3.2.12) and the 1st homogeneous solution with the respective derivatives presented in $${{y}_{1}}$$ (equation 3.2.1), $${{y}_{1}}'$$ (equation 3.2.2) and $${{y}_{1}}''$$ (equation 3.2.3), where r is replaced by $$\lambda $$, into 3.2.20:

Therefore we need to solve for:

Integrating both sides of 3.2.24 we have:

Integrating 3.2.25 we have U(x) as:

Finally, substituting 3.2.26 and 3.2.14 into 3.2.15 the 2nd homogeneous solution is defined as:

2.4) The generalized homogeneous solution can be calculated by using the following equation:
Substituting equations 3.2.14 and 3.2.27 into 3.2.28 we have:

Rearranging the constants from equation 3.2.31, the generalized homogeneous solution can be expressed as:

= Problem 7 =

= Find =

The 2 homogeneous solution $${{u}_{1}}$$ and $${{u}_{2}}$$ for (7.1) using the trial solution:

= Solution =

First we need to calculate the derivatives of the trial solution(7.1):

Substituting the equations 7.2, 7.3 and 7.4 into 7.1 we get:

Rearranging equation 7.7 we get:

Solving the equation 7.8 for r we get:

From equation 7.13 the root $${{r}_{1}}$$ can be calculated by:

From equation 7.13 the root $${{r}_{2}}$$ can be calculated by:

With the root $${{r}_{1}}$$ from equation 7.18 we can calculate the first homogeneous solution $${{u}_{1}}$$ by:

With the root $${{r}_{2}}$$ from the equation 7.23 we can calculate the second homogeneous solution $${{u}_{2}}$$ by: