Solving simultaneous equation for an ODE BVP.

[leehufford]

Homework Statement

Hello,
We are getting a quick taste of Boundary Value Problems in my ODE class and the application is deflection of beams. Basically I get to the point where I need to solve for the constants, but each constant appears in each equation but the powers of L are throwing me off.

Homework Equations

c₁L² + c₂L³ + w_oL⁴/24EI = 0

2c₁L + 3c₂L² + w_oL³/6EI = 0

The Attempt at a Solution

I've solved many systems with 2 equations and 2 unknowns but the L's and the powers are throwing me off. This is the only step holding me back from getting the correct answers for this section. Any help much appreciated! Thanks,

Lee

 

[lippyka]

To solve this system of equations, you can first rewrite them in terms of the constants, c1 and c2. You can then use the elimination method to solve for the constants. The first equation can be rewritten as:c1 = -(woL4/24EI)/L2 - (c2L3/L2)The second equation can be rewritten as:c2 = -(woL3/6EI)/L2 - (2c1L/L2)Substituting c1 into c2, we get:c2 = -(woL3/6EI)/L2 - (2(-woL4/24EI)/L2 - (c2L3/L2))L/L2Simplifying, we get:c2 = -(woL3/6EI)/L2 + (woL4/12EI)/L2 + (c2L3/L2)Now, we can solve for c2 by equating the coefficients of c2 on both sides of the equation:1 = 1 + (L3/L2)Therefore, L3/L2 = 0, and c2 = -(woL3/6EI)/L2. Substituting this value of c2 into the first equation, we get:c1 = -(woL4/24EI)/L2 - (-(woL3/6EI)/L2)(L3/L2)Simplifying the equation, we get:c1 = -(woL4/24EI)/L2 + (woL3/12EI)/L2Hence, the constants c1 and c2 can be expressed as:c1 = -(woL4/24EI)/L2 + (woL3/12EI)/L2c2 = -(woL3/6EI)/L2

 

[lippyka]

To solve this system of equations, you can use the elimination method. First, multiply both equations by the appropriate factors to make the coefficients of one variable the same in both equations. Then, subtract the two equations from each other to eliminate that variable, leaving you with a single equation in the two remaining variables. Solve this equation for either of the two variables and then substitute it back into either of the original equations to solve for the other variable.