Solving for x: Differential Equation with Sin & Cos

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iii. \frac{dx}{dt} + x - -3sin2t + 4cos2t

Can anyone help with this. All i know is to set

x = a\cos2t + b\sin2t

I don't even know if that is right:confused:
 
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Is this a differential equation? I'm missing the equation part, in that case. Do you mean the following?

\frac{dx}{dt} + x = -3\sin(2t) + 4\cos(2t)
 
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oh sorry, yeah that's what i meant to write :redface:
 
In that case, your proposal was fine for a particular solution but you need to add the solution of the associated homogeonous equation to obtain the complete solution. Luckily, that isn't too hard since what is the solution of the followin?

\frac{{dx}}{{dt}} + x = 0 \Leftrightarrow \frac{{dx}}{{dt}} = - x

If you don't see it immediately, integrate.

To find the coefficients a and b of your particular solution for x, find dx/dt and substitute in the equation to identify coefficients.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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