微分方程 $y''-4y'=0$ 的通解为（ ）.
A: $y=Ce^{4x}$
B: $y=C_1x+C_2e^{4x}$
C: $y=(C_1+C_2x)e^{4x}$
D: $y=C_1+C_2e^{4x}$
A: $y=Ce^{4x}$
B: $y=C_1x+C_2e^{4x}$
C: $y=(C_1+C_2x)e^{4x}$
D: $y=C_1+C_2e^{4x}$
举一反三
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