What is the General Solution of the Differential Equation y''-6y'+10y = 0?

Feb 7, 2017

We have;

$y ' ' - 6 y ' + 10 y = 0$

This is a Second Order Homogeneous Differential Equation which we solve as follows:

We look at the Auxiliary Equation, which is the quadratic equation with the coefficients of the derivatives, i.e.

${m}^{2} - 6 m + 10 = 0$

This quadratic does not factorise to I will solve by completing the square (you could equally use the quadratic formula)

${\left(m - 3\right)}^{2} - {3}^{2} + 10 = 0$
$\therefore {\left(m - 3\right)}^{2} = - 1$
$\therefore m - 3 = \pm i$
$\therefore m - 3 = 3 \pm i$

Because this has two distinct complex solutions $p \pm q i$, the solution to the DE is;

$y = {e}^{p t} \left(A \cos q t + B \sin q t\right)$

Where $A , B$ are arbitrary constants. and so the General Solution is;

$y = {e}^{3 t} \left(A \cos t + B \sin t\right)$

Feb 7, 2017

$y = \left({K}_{1} \cos t + {K}_{2} \sin t\right) {e}^{3 t}$

Explanation:

The general solution for this kind of differential equation (homogeneous linear with constant coefficients) is

$y = C {e}^{\lambda t}$

substituting into the differential equation we have

$C \left({\lambda}^{2} - 6 \lambda + 10\right) {e}^{\lambda t} = 0$

but $C {e}^{\lambda t} \ne 0$ so the feasible $\lambda ' s$ obey the condition

${\lambda}^{2} - 6 \lambda + 10 = 0$ so ${\lambda}_{1} = 3 - i$ and ${\lambda}_{2} = 3 + i$ then the solution is a linear combination of both

$y = {C}_{1} {e}^{\left(3 - i\right) t} + {C}_{2} {e}^{\left(3 + i\right) t}$ or

$y = \left({C}_{1} {e}^{- i t} + {C}_{2} {e}^{i t}\right) {e}^{3 t}$
with ${C}_{2} = \setminus t i l \mathrm{de} {C}_{1}$ where $t i l \mathrm{de} \left(\left(\cdot\right)\right)$ means that ${C}_{1}$ and ${C}_{2}$ are conjugate.

Using de Moivre's identity

${e}^{i t} = \cos t + i \sin t$ and substituting we have

$y = \left(\left({C}_{2} + {C}_{1}\right) \cos t + i \left({C}_{2} - {C}_{1}\right) \sin t\right) {e}^{3 t}$ or

$y = \left({K}_{1} \cos t + {K}_{2} \sin t\right) {e}^{3 t}$

Here ${K}_{1} = {C}_{1} + \setminus t i l \mathrm{de} {C}_{1}$ and ${K}_{2} = i \left({C}_{1} - t i l \mathrm{de} {C}_{1}\right)$