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