Answer to Question #191752 in Civil and Environmental Engineering for michael

Question

Solve the initial value problem y'' -12y' + 20y = 4e^3x ; y(0) = 3; y'(0) = 0 by first finding

the homogeneous solution, base solution, general solution and then finally the particular

solution.

Solve the boundary value problem  y''+5y'+6y=0; y(0)=2; y'(1)=3

Answer 1

Solution

1. Initial value problem y'' -12y' + 20y = 4e^3x ; y(0) = 3; y'(0) = 0

Finding the homogeneous solution y₀: k²-12k+20=0  => k_1,2 = 6±4; k₁ = 10, k₂ = 2 => y₀=Ae^10x+Be^2x

Partial nonhomogeneous solution y₁:  y₁=Ce^3x; substitution into equation gives C(9-12*3+20)=4 => C= -4/7   

General solution y(x) = y₀(x)+ y₁(x) = Ae^10x+Be^2x -4 e^3x /7

Particular solution : 

y(0) = 3 => A+B-4/7=3  => B=17/7-A

y'(0) = 0 => 10A+2B-12/7=0 => (10-17/7)A=12/7 => A = 12/53 = 0.226; B = 17/7-12/53 = 2.202

Therefore y(x) = 0.226e^10x+2.202e^2x -4 e^3x /7

2. Boundary value problem  y''+5y'+6y=0; y(0)=2; y'(1)=3

Finding the general solution y₀: k²+5k+6=0  => k_1,2 = -2.5±0.5; k₁ = -2, k₂ = -3 => y(x)=Ae^-2x+Be^-3x

y(0)=2 => A+B=2 => B=2-A

 y'(1)=3 => -2Ae^-2-3Be^-3=3 => -2Ae-3(2-A)=3e³ => (3-2e)A=3e³+6 =>

A=3(e³+2)/( 3-2e) = -27.193; B=2-3(e³+2)/( 3-2e) = 29.193

Finally y(x)= -27.193e^-2x+29.193e^-3x