prob9sol - Math 151 Spring 2008 Problem Set # 9 Separable...

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Math 151 Spring 2008 Problem Set # 9 Separable Di f erential Equations: Solutions 1. a) dy dx = y 2 1 y 2 dy dx = 1 Z 1 y 2 dy dx dx = Z dx Z 1 y 2 dy = x + C ⇒− 1 y = x + C y ( x )= 1 x C (the general solution can be expressed as y ( x 1 x + C as well). b) We have y (0) = 1 1 0 C =1 C = 1 . Therefore, the solution of the initial value problem is y ( x 1 x ( 1) = 1 x +1 . Note that we will obtain the same result if we express the general solution as y ( x 1 x + C . In this case, y (0) = 1 1 C C , so that y ( x 1 x , as before. y ( x 1 x 1
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2. a) dy dx = y 2 +1 1 y 2 dy dx =1 Z 1 y 2 dy dx dx = Z dx Z 1 y 2 dy = x + C arctan ( y )= x + C y =tan( x + C ) . Alternatively, dy dx = y 2 1 y 2 dy = dx Z 1 y 2 dy = Z dx arctan ( y x + C y x + C ) b) We have y ³ π 4 ´ =0 tan ³ π 4 + C ´ . We can set π 4 + C =arctan(0)=0 C = π 4 . Therefore, the solution of the given initial value problem is y ( x )=tan ³ x π 4 ´ . Since tangent is periodic with period π , the solution can be expressed as tan ³ x π 4 ± ´ , where n is an arbitrary positive integer. 3. a) dy dx = p 1+ y 2 1 p y 2 dy dx Z 1 p y 2 dy dx dx = Z dx Z 1 p y 2 dy = x + C arcsinh ( y x + C y =sinh( x + C ) . 2
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b) We have y (0) = 2 sinh (0 + C )=2 C =arcsinh(2) . Therefore, the solution of the given initial value problem is y ( x )=s inh( x +arcsinh(2)) 4. a) dy dx + y 2 sin ( x )=0 1 y 2 dy dx = sin ( x ) Z 1 y 2 dy dx dx = Z sin ( x ) dx Z 1 y 2 dy =cos( x )+ C ⇒− 1 y x C y ( x )= 1 cos ( x C b) We have y ³ π 3 ´ = 1 2 ⇔− 1 cos ³ π 3 ´ + C = 1 2 1 1 2 + C = 1 2 C = 5 2 .
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prob9sol - Math 151 Spring 2008 Problem Set # 9 Separable...

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