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Unformatted text preview: from both sides:
C = − ln x − x2 + y 2 or
ln x − x2 + y 2 = −C. Exponentiate. Let K be the new constant e−C . We have
x−
so x − K = x2 + y 2 = K ; x2 + y 2 and (x − K )2 = x2 + y 2 and ± (x − K )2 − x2 = y. Check: We check y = +
y′ = (x − K )2 − x2 , with K ≤ x . We see that 2(x − K ) − 2x
=
2 ( x − K ) 2 − x2 −K
.
( x − K ) 2 − x2 So, yy ′ + x = −K + x . On the other hand,
x2 + y 2 =
Thus, yy ′ + x = x2 + ( x − K ) 2 − x2 = y 2 + x2 as required. (x − K )2 = x − K....
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This document was uploaded on 03/23/2014 for the course MATH 242 at South Carolina.
 Fall '11
 Staff
 Differential Equations, Equations

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