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January 31, 2015

January 31, 2015

Posted by **Alice** on Sunday, November 9, 2008 at 4:54pm.

a) (dy/dx)^2 - x^4 = 0 given the initial value y(1) = 0

b) dy/dx = ysinx given the initial value y(pi) = 1

- Math -
**bobpursley**, Sunday, November 9, 2008 at 5:38pma) move the x^4 over, take the square root of each side.

dy/dx=x^2

y= 1/3 x^3 +C or y=-1/3 x^3-C

Specific solution? It looks to me that both will work.

dy/dx=x^2 dy/dx=-x^2

y'^2=x^4 y'^2=x^4

and if you pursue it, there are two imaginary solutions also.

b)dy/y= dx*sinx

lny=- cosx

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