Question #126298
Solve the initial value problem

y′′+4y′+4y=0, y(−1)=4, y′(−1)=4
1
Expert's answer
2020-07-16T19:36:04-0400

y+4y+4y=0y''+4y'+4y=0

initial values:

y(1)=4y(-1)=4

y(1)=4y'(-1)=4


Step by step solution: d2y(x)dx2+4dy(x)dx+4y(x)=0\tfrac{d^2y(x)}{dx^2} + 4 \tfrac{dy(x)}{dx} + 4y(x) = 0

1-step: Assume a solution will be proportioanl to eλxe^{\lambda x} for some constant λ\lambda . Substitute y(x)=eλxy(x) = e^{\lambda x} into the differential equation:


d2(eλx)dx2+4d(eλx)dx+4eλx=0\tfrac{d^2 (e^{\lambda x})}{dx^2} + 4 \tfrac{d (e^{\lambda x})}{dx} + 4e^{\lambda x} = 0


2-step: Substiute d2(eλx)dx2=λ2eλx\tfrac{d^2 (e^{\lambda x})}{dx^2} = \lambda ^2 e^{\lambda x} and d(eλx)dx=λeλx\tfrac{d (e^{\lambda x})}{dx} = \lambda e^{\lambda x}


λ2eλx+4λeλx+4eλx=0\lambda^2 e^{\lambda x} + 4\lambda e^{\lambda x} + 4e^{\lambda x} = 0


3-step: Factor out eλxe^{\lambda x}


(λ2+4λ+4)eλx=0(\lambda^2+4\lambda+4)e^{\lambda x} = 0


4-step: Since eλx=/0e^{\lambda x} {=}\mathllap{/\,} 0 for any finite λ\lambda , the zeros must come from the polynomial λ2+4λ+4=0\lambda^2+4\lambda+4=0

5-step: Factor


(λ+2)2=0(\lambda + 2)^2 = 0


6-step: Solve for λ\lambda


λ=2\lambda = -2 or λ=2\lambda = -2


7-step: The multiplicity of the root λ=2\lambda = -2 is 2 which gives y1(x) = C1 e2xe^{-2x} , y2(x) = C2 e2xxe^{-2x} x as solutions, where C1 and C2 are arbitrary constants.

The general solution is the sum of the above solutions:


y(x) = y1(x) + y2(x) = C1e2xe^{-2x} + C2e2xxe^{-2x}x


8-step: Solve for the unknown constants using the given initial conditions

Compute dy(x)dx:\tfrac{dy(x)}{dx}:


dy(x)dx=ddx\tfrac{dy(x)}{dx} = \tfrac{d }{dx} (C1e2x+e^{-2x} + C2e2xxe^{-2x}x ) = - 2C1e2x+e^{-2x} + C2e2xe^{-2x} - 2C2e2xxe^{-2x} x



9-step: Substitute y(-1) = 4 into y(x) = C1e2x+e2xxe^{-2x} + e^{-2x}xC2


e2e^{2} C1 - e2e^{2} C2 = 4


10-step: Substitute y'(-1) = 4 into dy(x)dx=2e2x\tfrac{dy(x)}{dx} = -2e^{-2x} C1 + e2xe^{-2x} C2 - 2e2xx2e^{-2x}x C2


-2C1e2+3e2e^{2} + 3e^{2} C2 = 4


11-step: Solve the system:


C1 = 16e2\tfrac{16}{e^{2}}


C2 = 12e2\tfrac{12}{e^{2}}


Substitute C1 = 16e2\tfrac{16}{e^{2}} and C2 = 12e2\tfrac{12}{e^{2}} into y(x) = e2xe^{-2x} C1 + e2xxe^{-2x}x C2


Answer:y(x)=4e2(x+1)(3x+4)y(x) = 4e^{-2(x+1)}(3x+4)

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