C y c x 2 x y 2 y y 2 x y x c x 2 2 c x 2 x 2 cx 2 c

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= C y = C x 2 ; x y ' = 2 y ; y ' = 2 x y x ( C x 2 ) ' = 2 ( C x 2 ) x ( 2 Cx )= 2 ( C x 2 ) 2 C x 2 = 2 C x 2 General solution is y = C x 2 3. For the below ordinary differential equation with initial conditions, state the order and determine if the equation is linear or nonlinear. Then find the solution of the ordinary differential equation, and apply the initial conditions. Verify your solution.
x 2 y 2 3 dy dx = 1 2 y y ( 1 ) = 2 The equation is nonlinear because: dy dx + P ( x ) y ≠Q ( x ) dy dx + Q ( x ) y ≠P ( x ) Solve: x 2 y 2 3 dy dx = y 2 y y ( ¿¿ 2 3 )= dx x 2 + ln C dy ¿ 1 3 1 y dy + 1 3 y y 2 3 = dx x 2 + ln C 1 3 ln y + 1 6 2 y y 2 3 = 1 x + ln C ¿ y 2 3 ¿ = 1 x + ln C 1 3 ln y + 1 6 ln ¿ ¿ y 1 3 ( y 2 3 ) 1 6 C ¿ = 1 x ln ¿ y 1 3 ( y 2 3 ) 1 6 = C e 1 x is the general solution Obtain C: y ( 1 )= 2 ( 2 ) 1 3 (( 2 ) 2 3 ) 1 6 = C e 1 x = Ce 1 = C = e 3 2 Desired Solution:
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