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lecture_4 - MA 36600 LECTURE NOTES WEDNESDAY JANUARY 21...

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MA 36600 LECTURE NOTES: WEDNESDAY, JANUARY 21 First Order Linear Equations Definitions. So far we have considered initial value problems in the form dy dt = - b + a y, y (0) = y 0 where a and b are constants. We showed that the solution is y ( t ) = b a + C e at where C = y 0 - b a . In general, an ordinary differential equation is said to be a first order equation if it is a relation in the form F ( t, y, y 0 ) = 0 . Upon expressing the derivative y 0 in terms of the other variables, we find an expression in the form dy dt = G ( t, y ) for some function G ( t, y ). We say that a function y = y ( t ) is a solution if it satisfies this equation. If y ( t 0 ) = y 0 is an initial value, we think of the graph of y = y ( t ) as passing through the point ( t 0 , y 0 ). We say that this first order equation is linear if G ( t, y ) = g ( t ) - p ( t ) y for some functions g ( t ) and p ( t ). That is, if the differential equation is in the form dy dt + p ( t ) y = g ( t ) . Integrating Factors. Consider the first order linear differential equation dy dt + p ( t ) y = g ( t ) . Choose a function μ = μ ( t ). Multiply the first order differential equation above by this function: μ ( t ) dy dt + μ ( t ) p ( t ) y = μ ( t ) g ( t ) . Say that we can choose a function μ = μ ( t ) such that i. μ ( t ) is not identically zero, and ii. μ 0 ( t ) = p ( t ) μ ( t ). Then we would have μ ( t ) dy dt + μ ( t ) p ( t ) y = μ ( t ) · dy dt + dt · y ( t ) = d dt [ μ ( t ) y ( t )] This gives the differential equation d dt [ μ ( t ) y ( t )] = μ ( t ) g ( t ) = μ ( t ) y ( t ) = Z t μ ( τ ) g ( τ
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