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# lecture_4 (dragged) 2 - MA 36600 LECTURE NOTES WEDNESDAY...

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MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21 3 Separable Equations. We introduce a trick to solve separable di ff erential equations. Divide both sides by the function Y = Y ( y ) so that all of the y terms are to one side: dy dt = Y ( y ) T ( t ) = 1 Y ( y ) dy dt = T ( t ) . Now compute the antiderivatives f 1 ( t ) = t T ( τ ) d τ f 2 ( y ) = y 1 Y ( σ ) d σ = d dt f 1 ( t ) = T ( t ) d dt f 2 ( y ) = 1 Y ( y ) dy dt We see that the function f ( t, y ) = f 1 ( t ) f 2 ( y ) satisfies d dt f ( t, y ) = d dt f 1 ( t ) d dt f 2 ( y ) = T ( t ) 1 Y ( y ) dy dt = 0 , so that f ( t, y ) must be a constant. We conclude that the general solution to the di ff erential equation y = G ( t, y ) in terms of G ( t, y ) = Y ( y ) T ( t ) must be f ( t, y ) = C where f ( t, y ) = t T ( τ ) d τ y 1 Y ( σ ) d σ . This is an implicit solution to the di ff erential equation y = G ( t, y ) because we have not explicitly written y = y ( t ) as a function of time. We also call these integral curves
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