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Find the general solution of the differential equation: d^2x/dt^2 + 5dx/dt + 6x = 2cos(t) - sin(t)

First solve complementary function, i.e. d 2 x/dt 2 + 5dx/dt + 6x = 0. To do so, let x = e mt , where m = arbitrary constant. Differentiating gives dx/dt = m e mt and d 2 x/dt 2 = m 2 e mt . Substituting int...
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Answered by Mick G. • Maths tutor
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Using the product rule, differentiate: y = (x^2 - 1)(x^3 + 3).

y=(x 2 -1)(x 3 +3) let u=x 2 -1 u'=2x let v=x 3 +3 v'=3x 2 uv'+u'v=(x 2 -1)(3x 2 )+(2x)(x 3 +3) =(3x 4 -3x 2 )+(2x 4 +6X) = 5x 4 -3x 2 +6x
Answered by • Maths tutor
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integral of (tan(x))dx using the substitution u = cos(x)

given u = cos(x), therefore du/dx=-sin(x), as tan(x)=sin(x)/cos(x), can rewrite tan(x)=(-du/dx)/u, therefore integral can become [(-1/u)du], after inegrating you are left with -ln(u)+c, therefore ln(1/u)+c, ...
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Answered by Frederick R. • Maths tutor
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f(x) = x^x, find f'(3).

Therefore, y = x x can then natural log both sides leaving ln(y) = xln(x) then differentiating both sides wrst to x d/dx(ln(y)=xln(x))we are then left with this expression (dy/dx)(1/y)=ln(x)+1 multiplying up...
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Answered by Frederick R. • Maths tutor
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A 10 kilogram block slides down a 30 degree inclined slope, the slope has a coefficient of friction of 0.2. Calculcate the blocks acceleration down the slope.

Resolving all forces in the vertical gives the normal force as 10gcos(30). Resolving all forces normal to the slope gives the frictional force as 2gcos(30) (given Friction = mu*R). Using second law, F=ma. 10...
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Answered by Frederick R. • Maths tutor
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