y′′−2y′+y=4x2−1+x−1exy″−2y′+y=4x2−1+x−1ex
Question
y′′−2y′+y=4x2−1+x−1exy″−2y′+y=4x2−1+x−1ex
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Similar Questions
x2 y′′ − 2x y′ − 4y = x2 + 2 log x
The general solution of the given differential equation y′′−2y′+y=4x2−1+x−1exy″−2y′+y=4x2−1+x−1ex isy(x)=C1ex+C2□ex+4x2+16x+ex□lnx+□.
x2y′′−3xy′+4y=x3
y'^2 - 4y' + 4 - 4x' + 4 + 4y' - 8 + 8 = 0Simplifying:
Solve y′′ + 2y′ + 4y = 13cos(4x − 2) using method of undetermined coefficients.
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