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**Unformatted text preview: **(7) „ ÅÅÅÅÅÅÅ „ t H m H t L y L = m H t L t 2 This equation can be integrated directly to give (8) m y = ‡ m H t L t 2 „ t + C Substituting into the RHS the expression for m (t) gives (9) m y = ‡ Exp H 4 t L t 2 „ t + C Hence the general solution is (10) y H t L = Exp @-4 t D 9 ‡ Exp @ 4 t D t 2 „ t + C = Using Integration tables we can express (11) ‡ Exp @ 4 t D t 2 „ t = Exp @ 4 t D i k j j 1 ÅÅÅÅÅÅÅ 32-t ÅÅÅÅ 8 + t 2 ÅÅÅÅÅÅ 4 y { z z Thus the general solution is (12) y H t L = i k j j 1 ÅÅÅÅÅÅÅ 32-t ÅÅÅÅ 8 + t 2 ÅÅÅÅÅÅ 4 y { z z + C Exp @-4 t D where C is the constant of integration ‡ Mathematica Solution We will use DSolve DSolve @ y' @ t D + 4 y @ t D ã t 2 , y @ t D , t D 99 y @ t D Ø 1 ÅÅÅÅÅÅÅ 32 H 1-4 t + 8 t 2 L + ‰-4 t C @ 1 D== 2 Problem_1.5.nb...

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- Summer '13
- LeThuy
- Derivative, Trigraph, Hm HtL yL, J ÅÅÅÅÅÅÅÅ