Math for CSTutorial 91 Second Order Linear Differential Equations, part II

Math for CSTutorial 92 Homogeneous Linear Equations with Constant Coefficients Consider a nonhomogeneous equation with constant coefficients: where a, b and c are constants and g(x) is an exponent, polynom or harmonic function (e wx ; a 0 x n +…a n ; sin(wx) or cos(wx) ) or their product. This type of equation is very useful in many applied problems (physics, electrical engineering, etc..). Then one should make an intelligent guess about the form of the solution, up to the constant multipliers, and then substitute this guess into the equation to calculate the values of the multipliers.

Math for CSTutorial 93 Example 1 Consider a nonhomogeneous equation with constant coefficients: Suppose, the solution is y=A·sin(x)+B·cos(x), then: y`=A·cos(x)-B·sin(x) y``=-A·sin(x)-B·cos(x) Substituting, we obtain: (-A-3B-4A)cos(x)+(-B+3A-4B)sin(x)=2sin(x) From where we obtain -5A-3B=0 3A-5B=2 A=3/17;B=-5/17; Y=1/17(3cos(x)-5sin(y))

Math for CSTutorial 94 Example 2 Consider a nonhomogeneous equation with constant coefficients: Suppose, the solution is y=Ax 2, then: y`=2Ax y``=2A Substituting, we obtain: 2A-6Ax-4Ax 2 =4x 2 We see that there are no solutions in Ax 2. Now, try Ax 2 +Bx+C. And obtain y=-x2+3/2x-13/8.

Math for CSTutorial 95 Example 3 Consider a nonhomogeneous equation with constant coefficients: Two linearly independent solutions of the homogeneous equation are y 1 =cos(x) and y 2 =sin(x) For a particular solution y p =u 1 cos(x)+u 2 sin(x) Then y p =[-u 1 sin(x)+u 2 cos(x)]+[…=0] Differentiating again, and substituting: u 1 ’(x)cos(x)+u 2 ’(x)sin(x)=0; -u 1 ’(x)sin(x)+u 2 ’(x)cos(x)=sec(x). Solving, we obtain:

Math for CSTutorial 96 Example 3 (Solution) u 1 ’(x)=-tan(x); u 2 ’(x)=1; u 1 (x)=ln(cos(x)); u 2 (x)=x; Therefore, the particular solution is y p (x)=xsinx+cos(x)ln(cos(x)) And the general solution is y=c 1 cos(x)+c 2 sin(x)+xsin(x)+cos(x)ln(cos(x))

Math for CSTutorial 97 Example 4 Solve the system: Solution: Assuming that x=ae rt, we obtain the system of algebraic equations, whose determinant is:

Math for CSTutorial 98 Example 4. Solution. Therefore r 1 =1, r 2 =2, r 3 =-1. The corresponding eigenvectors are: The general solution is