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solve equation for xsin^2 x=du/dx-cos^2x*dv/dx u=2x^3+3x^2 v=x^2
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To solve the equation, we'll have to differentiate u and v, with respect to x.
du/dx = d(2x^3+3x^2)/dx
du/dx = 6x^2 + 6x
dv/dx = d(x^2)/dx
dv/dx = 2x
We'll substitute du/dx and dv/dx by their expression.
(sin x)^2 = 6x^2 + 6x - 2x*(cos x)^2
We'll move 2x*(cos x)^2 to the left side.
(sin x)^2 + 2x*(cos x)^2 = 6x(x+1)
But, (cos x)^2 = 1 - (sin x)^2
(sin x)^2 + 2x - 2x(sin x)^2 = 6x(x+1)
(sin x)^2(1 - 2x) = 2x(3x+3-1)
(sin x)^2(1 - 2x) = 2x(3x+2)
(sin x)^2 = 2x(3x+2)/(1 - 2x)
sin x = 0 for x = 0
2x(3x+2)/(1 - 2x)= 0 for x = 0
The common solution of the given equation is x = 0.
Posted by giorgiana1976 on February 25, 2011 at 8:50 PM (Answer #1)
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