3.4 Repeated Roots; Reduction of Order

Solution of Repeated Roots

In this section, we are going to solve what if $r_1 = r_2$

$$ay'' + by' +cy = 0 \quad --(a)$$

if $r_1 = r_2$, then

$$y_1(t) = c_1 e^{\frac{-b}{2a}t}$$

To find second solution, we assume that

$$y = v(t)y_1(t) = v(t)e^{\frac{-b}{2a}t}$$ $$y' = v'(t)e^{\frac{-b}{2a}t} - \frac{b}{2a}v(t)e^{\frac{-b}{2a}t}$$ $$y'' = v''(t)e^{\frac{-b}{2a}t} - \frac{b}{a}v'(t)e^{\frac{-b}{2a}t} + \frac{b^2}{4a^2}v(t)e^{\frac{-b}{2a}t}$$

By substitude equation (a) ,cancelling the factor $e^{\frac{-b}{2a}t}$, rearranging the items, we find that

$$v'' = 0$$ $$v = c_1 + c_2 t$$

then we have

$$y = c_1 e^{\frac{-b}{2a}t} + c_2 t e^{\frac{-b}{2a}t}$$

Thus $y$ is a linear combination of the two solutions

$$y_1(t) = e^{\frac{-b}{2a}t}, \quad y_2(t) = t e^{\frac{-b}{2a}t}$$ $$W[y_1 ,y_2](t) = e^{\frac{-b}{2a}}$$

Since ,$W[y_1 ,y_2](t)$ never zero, $y_1$and $y_2$form a fundamental set of solutions.

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Reduction of Order

Suppose that we know one solution $y_1(t)$ , not everywhere zero, of

$$y'' + p(t)y' + q(t)y = 0 \quad --(b)$$

To find the second solution, let

$$y = v(t) y_1(t)$$ $$y' = v'(t)y_1(t) + v(t)y_1'(t)$$ $$y'' = v''(t)y_1(t) + v'(t)y_1'(t) + v'(t)y_1'(t) + v(t)y_1''(t)$$

Substitude $y$, $y'$and $y''$to the equation (b)

$$y_1 v'' + (2y_1' + py_1)v' + (y_1'' + p(t)y_1' + q(t)y_1)v = 0$$

Since $y_1$is a solution of equation (b).

$$y_1 v'' + (2y_1' + py_1)v' = 0$$

Here becomes a first-order differential equation for the function $v'$