4.直线x+2y-1=0的斜率为-1/2,
曲线y=y(x)在原点处的切线垂直于上述直线,
所以y'(0)=2.
A.y'=[(1/4)cos2x+(x/4)cos2x-(x/2)sin2x]e^x,
y'(0)=1/4,弃A.
B.y'=[(5/4+x/4)sin2x+(2+x/2)cos2x]e^x,
y'(0)=2.
y''=[-(7/4+3x/4)sin2x+(5/2+x)cos2x]e^x,
y''-2y'+5y=[-(7/4+3x/4)sin2x+(5/2+x)cos2x
..................-(5/2+x/2)sin2x-(4+x)cos2x+(5+5x/4)sin2x]e^x
=[(3/4)sin2x-(3/2)cos2x]e^x,弃B.
C.y'=e^x*(3sin2x+cos2x+1),
y'(0)=2.
y''=(sin2x+7cos2x+1)e^x,
y''-2y'+5y=[sin2x+7cos2x+1
.................-6sin2x-2cos2x-2
.................+5sin2x-5cos2x+5)e^x
=4e^x,弃C.
选D.
y'(0)=2.
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