2. Y = e^-2x . Sin 3x show that
d^2y/dx^2 = -13y - 4 dy/dx
2.
y=e−2xsin3xy=e^{-2x}sin3xy=e−2xsin3x
y′=3e−2xcos3x−2e−2xsin3xy'=3e^{-2x}cos3x-2e^{-2x}sin3xy′=3e−2xcos3x−2e−2xsin3x
y′′=−9e−2xsin3x−6e−2xcos3x−6e−2xcos3x+4e−2xsin3x=y''=-9e^{-2x}sin3x-6e^{-2x}cos3x-6e^{-2x}cos3x+4e^{-2x}sin3x=y′′=−9e−2xsin3x−6e−2xcos3x−6e−2xcos3x+4e−2xsin3x=
=−5e−2xsin3x−12e−2xcos3x=-5e^{-2x}sin3x-12e^{-2x}cos3x=−5e−2xsin3x−12e−2xcos3x
Then:
−13y−4y′=−13e−2xsin3x−4(3e−2xcos3x−2e−2xsin3x)=-13y - 4y'=-13e^{-2x}sin3x-4(3e^{-2x}cos3x-2e^{-2x}sin3x)=−13y−4y′=−13e−2xsin3x−4(3e−2xcos3x−2e−2xsin3x)=
=−13e−2xsin3x−12e−2xcos3x+8e−2xsin3x==-13e^{-2x}sin3x-12e^{-2x}cos3x+8e^{-2x}sin3x==−13e−2xsin3x−12e−2xcos3x+8e−2xsin3x=
=−5e−2xsin3x−12e−2xcos3x=y′′=-5e^{-2x}sin3x-12e^{-2x}cos3x=y''=−5e−2xsin3x−12e−2xcos3x=y′′
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