y′′+4y′+4y=0
initial values:
y(−1)=4
y′(−1)=4
1-step: Assume a solution will be proportioanl to eλx for some constant λ . Substitute y(x)=eλx into the differential equation:
dx2d2(eλx)+4dxd(eλx)+4eλx=0
2-step: Substiute dx2d2(eλx)=λ2eλx and dxd(eλx)=λeλx
λ2eλx+4λeλx+4eλx=0
3-step: Factor out eλx
(λ2+4λ+4)eλx=0
4-step: Since eλx=/0 for any finite λ , the zeros must come from the polynomial λ2+4λ+4=0
5-step: Factor
(λ+2)2=0
6-step: Solve for λ
λ=−2 or λ=−2
7-step: The multiplicity of the root λ=−2 is 2 which gives y1(x) = C1 e−2x , y2(x) = C2 e−2xx 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) = C1e−2x + C2e−2xx
8-step: Solve for the unknown constants using the given initial conditions
Compute dxdy(x):
dxdy(x)=dxd (C1e−2x+ C2e−2xx ) = - 2C1e−2x+ C2e−2x− 2C2e−2xx
9-step: Substitute y(-1) = 4 into y(x) = C1e−2x+e−2xxC2
e2 C1 - e2 C2 = 4
10-step: Substitute y'(-1) = 4 into dxdy(x)=−2e−2x C1 + e−2x C2 - 2e−2xx C2
-2C1e2+3e2 C2 = 4
11-step: Solve the system:
C1 = e216
C2 = e212
Substitute C1 = e216 and C2 = e212 into y(x) = e−2x C1 + e−2xx C2
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