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

Expert's answer

**Solution**

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

Finding the homogeneous solution y_{0}: k^{2}-12k+20=0 => k_{1,2} = 6±4; k_{1} = 10, k_{2} = 2 => y_{0}=Ae^{10x}+Be^{2x}

Partial nonhomogeneous solution y_{1}: y_{1}=Ce^{3x}; substitution into equation gives C(9-12*3+20)=4 => C= -4/7

General solution y(x) = y_{0}(x)+ y_{1}(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 y_{0}: k^{2}+5k+6=0 => k_{1,2} = -2.5±0.5; k_{1} = -2, k_{2} = -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)=3e^{3} => (3-2e)A=3e^{3}+6 =>

A=3(e^{3}+2)/( 3-2e) = -27.193; B=2-3(e^{3}+2)/( 3-2e) = 29.193

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

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