If y=e2xsinxy=e^{2x}\sin xy=e2xsinx then y′=e2x(2sinx+cosx)y'=e^{2x}(2\sin x+\cos x)y′=e2x(2sinx+cosx) and y′′=2e2x(2sinx+cosx)+e2x(2cosx−sinx)y''=2e^{2x}(2\sin x+\cos x)+e^{2x}(2\cos x-sin x)y′′=2e2x(2sinx+cosx)+e2x(2cosx−sinx) hence y′′−4y′+5y=0y''-4y'+5y=0y′′−4y′+5y=0
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