Find the general solution of the differential equation d^2x/dt^2 + 7 dx/dt + 10x = 116 sin 2t - CIE - A-Level Further Maths - Question 6 - 2016 - Paper 1

To find the general solution to the differential equation, we begin by identifying the characteristic equation by setting the left-hand side equal to zero.

The characteristic equation is: $m^2 + 7m + 10 = 0$ We can solve for m using the quadratic formula: $m = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{-7 \pm \sqrt{7^2 - 4 \cdot 1 \cdot 10}}{2 \cdot 1}$ Calculating the discriminant: $7^2 - 40 = 49 - 40 = 9$ This gives us: $m = \frac{-7 \pm 3}{2}$ Thus, the roots are: $m_1 = -2 \quad \text{and} \quad m_2 = -5$

The complementary function (CF) is therefore: $CF: x_c = Ae^{-2t} + Be^{-5t}$

Next, we need to determine a particular solution (P) to the non-homogeneous equation. We can use the method of undetermined coefficients. Given that the non-homogeneous term is $116 \sin(2t)$, we assume: $P = C \sin(2t) + D \cos(2t)$

Next, we differentiate P: $P' = 2C \cos(2t) - 2D \sin(2t)$ $P'' = -4C \sin(2t) - 4D \cos(2t)$

Substituting P, P', and P'' into the original equation: $-4C \sin(2t) - 4D \cos(2t) + 7(2C \cos(2t) - 2D \sin(2t)) + 10(C \sin(2t) + D \cos(2t)) = 116 \sin(2t)$

Grouping like terms yields two equations. Solving these will give us values for C and D.

Finally, the general solution is: $x(t) = Ae^{-2t} + Be^{-5t} + C \sin(2t) + D \cos(2t)$