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

Question #191752

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

1
2021-05-31T06:39:39-0400

Solution

1. Initial value problem y'' -12y' + 20y = 4e3x ; y(0) = 3; y'(0) = 0

Finding the homogeneous solution y0: k2-12k+20=0  => k1,2 = 6±4; k1 = 10, k2 = 2 => y0=Ae10x+Be2x

Partial nonhomogeneous solution y1:  y1=Ce3x; substitution into equation gives C(9-12*3+20)=4 => C= -4/7

General solution y(x) = y0(x)+ y1(x) = Ae10x+Be2x -4 e3x /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) = 0.226e10x+2.202e2x -4 e3x /7

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

Finding the general solution y0: k2+5k+6=0  => k1,2 = -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

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