y” –3 y’ + 2y = 4 , y(0) = 1, y’(0) = 0
Question
y” –3 y’ + 2y = 4 , y(0) = 1, y’(0) = 0
Solution
The given equation is a second order linear homogeneous differential equation with constant coefficients. The general solution of such an equation is given by:
y = Ae^(m1x) + Be^(m2x)
where m1 and m2 are the roots of the characteristic equation associated with the differential equation. The characteristic equation is obtained by replacing each derivative in the differential equation by a power of m:
m^2 - 3m + 2 = 0
Solving this quadratic equation gives m1 = 1 and m2 = 2. So the general solution of the differential equation is:
y = Ae^(x) + Be^(2x)
We can find the constants A and B by using the initial conditions y(0) = 1 and y’(0) = 0. Substituting x = 0 into the general solution gives:
1 = A + B
Differentiating the general solution gives:
y’ = Ae^(x) + 2Be^(2x)
Substituting x = 0 into this equation gives:
0 = A + 2B
Solving these two equations simultaneously gives A = 2 and B = -1. So the solution of the differential equation that satisfies the initial conditions is:
y = 2e^(x) - e^(2x)
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