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The Solution of the Equation for the Differential Equation x2y′′−2xy′+2y=0x ^ { 2 } y ^ { \prime \prime } - 2 x y ^ { \prime } + 2 y = 0

Question 15

Multiple Choice

The solution of the equation for the differential equation x2y′′−2xy′+2y=0x ^ { 2 } y ^ { \prime \prime } - 2 x y ^ { \prime } + 2 y = 0 is


A) y=c1xcos⁡(ln⁡x) +c2xsin⁡(ln⁡x) y = c _ { 1 } x \cos ( \ln x ) + c _ { 2 } x \sin ( \ln x )
B) y=c1x1/2cos⁡(3ln⁡x/2) +c2x1/2sin⁡(3ln⁡x/2) y = c _ { 1 } x ^ { 1 / 2 } \cos ( \sqrt { 3 } \ln x / 2 ) + c _ { 2 } x ^ { 1 / 2 } \sin ( \sqrt { 3 } \ln x / 2 )
C) y=c1x(1+3) /2+c2x(1−3) /2y = c _ { 1 } x ^ { ( 1 + \sqrt { 3 } ) / 2 } + c _ { 2 } x ^ { ( 1 - \sqrt { 3 } ) / 2 }
D) y=c1x+c2xln⁡xy = c _ { 1 } x + c _ { 2 } x \ln x
E) y=c1x+c2x2y = c _ { 1 } x + c _ { 2 } x ^ { 2 }

Correct Answer:

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