UNIT9

# UNIT9 - Unit 9: First Order Dierential Equations with...

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Unit 9: First Order Diferential Equations with Applications First Order Diﬀerential Equations By a diferential equation, we mean an equation which involves the deriva- tive oF some unknown Functon, say y , and the task at hand is to ±nd an explicit expression For y . We will study two types oF ±rst order diferential equation, namely separable ,and linear diferential equations ( shortened to D.E.’s For the obvious reason)). Each will have its own prescribed method oF solution. Defnition 9.1. By a separable diferential equation we will mean an equa- tion which may be put into one oF the Following Forms: dy dx = f ( x ) g ( y ) or dy dx = g ( y ) f ( x ) or dy dx = f ( x ) g ( y ) To solve a diferential equation is to ±nd all possible solutions ( remember C in integration?). Strictly speaking we have already dealt with some D.E.’s, as the Following example will show. We ±rst outline some basic steps For solving such a problem. 1

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Steps For Solving a separable Diferential Equation: Assuming that our D.E. is given in terms o± dy dx with the generalization being clear: (1) Recognize the problem as a separable D.E. (2) Seperate the equation into the ±orm (terms not involving x)( dy ) = (terms not involving y)( dx ) (3) Integrate both sides, adding one arbitrary constant, say C, to the x-side. This is done since adding an arbitrary constant to both sides a±ter integrating is equivalent to adding just a single arbiratary constant to one side. (4) Solve ±or y i± possible. I± we can solve ±or y we get what is called the general solution , and i± we can not solve ±or y we get what is called an implicit solution . Example 1. Solve: dy dx = x 2 Solution: F i r s twerecogn izetha twehaveasepa rab leD .E .o ±e i the ro ± the ²rst two ±orms listed above (with g(y)=1), and so proceed: dy = x 2 dx This is step (2) R dy = R x 2 dx Begin step(3) y = x 3 3 + C step(3) and step (4) So we have in this case ±ound the general solution, y = x 3 3 + C 2
Example 2. Solve: dy dx = 4 x 3 y 3 Solution: We frst recognize this as a separable D.E., and then move on: 3 y 3 dy =4 xdx step (2) R 3 y 3 dy = R 4 xdx Begin step(3) 3 y 4 4 x 2 2 + C step(3) and step (4) 3 y 4 4 =2 x 2 + C y 4 = 4 3 2 x 2 2 + 4 3 C y 4 = 8 3 x 2 + 4 3 C y 4 = 8 3 x 2 + C y = 4 q 8 3 x 2 + C So once again we have a general solution, namely y = 4 q 8 3 x 2 + C Example 3. Solve: e y dy dx x 3 +2 x Solution: We frst recognize this as a separable D.E., and then move on:

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## This note was uploaded on 04/15/2010 for the course MATH 20C taught by Professor Lit during the Spring '10 term at UCLA.

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UNIT9 - Unit 9: First Order Dierential Equations with...

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