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The solution of the given equation y′′−4y′+3y=0\large y''-4 y'+3 y=0y′′−4y′+3y=0 satisfying the indicated initial conditions y∣x=0=6,y′∣x=0=10y|_{x=0}=6,y'|_{x=0}=10y∣x=0=6,y′∣x=0=10 is y=aebx+cedxy= ae^{bx}+c e^{d x}y=aebx+cedx.Then the value of a×b×c×da\times b\times c\times da×b×c×d is
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