Unformatted text preview: 3 V 2 / 3 . The diﬀerential equation becomes, dV dt = r S = r 2 V 2 / 3 where the constant (36 π ) 1 / 3 has been absorbed into r and written as a new constant r 2 . 1.3 8 Verify both functions are solutions to the diﬀerential equation, y 00 + 2 y3 y = 0 (1) y 1 ( t ) = e3 t ⇒ y 1 =3 e3 t , y 1 00 = 9 e3 t (2) Substituting (2) into (1), 9 e3 t + 2(3 e3 t )3( e3 t ) = 0 (3) 0 = 0 (4) y 2 ( t ) = e t ⇒ y 2 = e t , y 2 00 = e t (5) Substituting (5) into (1), e t + 2 e t3 e t = 0 (6) 0 = 0 (7) 1...
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 Spring '07
 TERRELL,R
 Derivative, Slope, Drop

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