Boundary Value Problems and Differential Equations

The video tutorial explains how to solve a boundary value problem involving a second-order linear homogeneous differential equation with constant coefficients. It begins by stating the problem and then derives the general solution using the characteristic equation. The tutorial applies initial conditions to find the specific solution, ultimately concluding that there is only one solution to the problem. The solution is expressed in terms of sine functions, and the process is explained step-by-step for clarity.

What is the initial condition given for X at zero?

• X of zero equals one

• X of zero equals zero

• X prime of zero equals one

• X prime of zero equals zero

What type of differential equation is being solved in this problem?

• Second order nonlinear

• First order nonlinear

• Second order linear

• First order linear

What is the characteristic equation derived from the given differential equation?

• r squared minus one equals zero

• r squared plus one equals zero

• r squared plus two equals zero

• r squared minus two equals zero

What are the roots of the characteristic equation?

• Plus or minus one

• Plus or minus two

• Plus or minus i

• Plus or minus zero

What form does the general solution take given the roots of the characteristic equation?

• X of T equals a e to the t plus b e to the minus t

• X of T equals a t plus b

• X of T equals a cosine t plus b sine t

• X of T equals a sine t plus b cosine t

What does the first boundary condition X of zero equals zero imply about the constant 'a'?

• a equals zero

• a equals two

• a equals one

• a equals negative one

What is the derivative of the general solution X of T equals B sine t?

• Negative B sine t

• B cosine t

• Negative B cosine t

• B sine t

What does the second boundary condition X prime of pi equals one imply about the constant 'b'?

• b equals two

• b equals negative one

• b equals zero

• b equals one

What is the final solution to the boundary value problem?

• X of T equals negative sine t

• X of T equals cosine t

• X of T equals negative cosine t

• X of T equals sine t

What is the value of cosine pi used in the second boundary condition?

• Two

• Negative one

• One

• Zero