Problem_1.5soln

Problem_1.5soln - First Order ODE Problems Problem_1.5 Find...

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First Order ODE Problems Problem_1.5 Find the general solution to the following ODE (1) y ÅÅÅÅÅÅÅ t + 4 y = t 2 Solution Before we begin it is worthwhile to inspect the form of this equation. It is a linear first order ODE with constant coefficients and an inhomogeneous term. To solve this equation we will make use of an integrating factor m (t) such that (2) m H t L J y ÅÅÅÅÅÅÅ t + 4 y N = m H t L t 2 Note all we have done is multiply both sides of Eqn(1) by m (t). The trick now is to choose m H t L so that (3) m H t L J y ÅÅÅÅÅÅÅ t + 4 y N = ÅÅÅÅÅÅÅ t H m H t L y L If we expand the RHS we find (4) ÅÅÅÅÅÅÅ t H m H t L y L = „m ÅÅÅÅÅÅÅ t y + m y ÅÅÅÅÅÅÅ t Thus Eqn(3) becomes (5) m H t L J y ÅÅÅÅÅÅÅ t + 4 y N = „m ÅÅÅÅÅÅÅ t y + m y ÅÅÅÅÅÅÅ t ï m H t L 4 y = „m ÅÅÅÅÅÅÅ t y Cancelling the factor of y, it then follows that m (t) is defined by the following ODE (6) „m ÅÅÅÅÅÅÅ t = 4 m fl m H t L = Exp H 4 t L Now that we have found an integrating factor we can write the original ODE as Subscribe to view the full document. 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

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