lecture_10 (dragged) 3

lecture_10 (dragged) 3 - dτ ² ∂ ∂t µ μ N = − p...

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4 MA 36600 LECTURE NOTES: WEDNESDAY, FEBRUARY 4 We seek a function μ = μ ( t, y ) such that when we multiply this equation by this function we do have an exact equation. Consider the function μ ( t, y )=exp °± t p ( τ ) ² . Then we have the products μ ( t, y ) M ( t, y )= ³ g ( t ) p ( t ) y ´ exp °± t p ( τ ) ² and μ ( t, y ) N ( t, y )= exp °± t p ( τ ) ² . This gives the partial derivatives ∂y µ μM = p ( t )exp °± t p ( τ )
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Unformatted text preview: dτ ² , ∂ ∂t µ μ N ¶ = − p ( t ) exp °± t p ( τ ) dτ ² = ⇒ ∂ ∂y µ μ M ¶ = ∂ ∂t µ μ N ¶ . In particular, the diFerential equation ³ μ ( t, y ) M ( t, y ) ´ dt + ³ μ ( t, y ) N ( t, y ) ´ dy = 0 is an exact equation. Hence the de±nitions of integrating factor are consistent....
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue.

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