verify that y = e ^ -2x is a solution of the differential equation
y"' + 4 y " + 3 y ' - 2y = 0 .
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y = e^-2x
y'= -2e^-2x
y"' = -8 e ^-2x
substituting these derivatives in given diff. eq. we get
lhs = y'" + 4y" + 3y'-2y. = -8e^-2x + 4(4e ^-2x ) + 3(-2e^-2x ) - 2e^-2x
= 0
y'= -2e^-2x
y"' = -8 e ^-2x
substituting these derivatives in given diff. eq. we get
lhs = y'" + 4y" + 3y'-2y. = -8e^-2x + 4(4e ^-2x ) + 3(-2e^-2x ) - 2e^-2x
= 0
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