Reduction of order

Let

$$L(y) = y^{\prime\prime} + a_{1}(x)y^{\prime} + a_{0}(x)y = f(x)$$

Suppose that a solution $y_{1}(x)$ of $L(y) = 0$ is known. Then substitute $y = u(x)y_{1}(x)$ into $L(y) = f(x)$.

$$(u^{\prime\prime}y_{1} + 2u^{\prime}y_{1}^{\prime} + uy_{1}^{\pri... ...ime}) + a_{1}(x)(u^{\prime}y_{1} + uy_{1}^{\prime}) + a_{0}(x)(uy_{1}) = f(x).$$

or

$$y_{1}u^{\prime\prime} + (2y_{1}^{\prime} + a_{1}y_{1})u^{\prime} + (y_{1}^{\prime\prime} + a_{1}y_{1}^{\prime} + a_{0}y_{1})u = f(x)$$

Since $L(y_{1}) = 0$, we note that the coefficient of $u$ is 0. Now let $u^{\prime} = w$. Then we have a 1st order linear differential equation in $w$.

$$y_{1}w^{\prime} + (2y_{1}^{\prime} + a_{1}y_{1})w = f(x).$$

Example 2..5   Using the $y_{1} = e^{x}$ is a solution of $y^{\prime\prime} - y = 0$, find the general solution of

$$y^{\prime\prime} - y = xe^{x}.$$

SOLUTION Let $y = uy_{1} = ue^{x}$. Then $y^{\prime} = u^{\prime}e^{x} + ue^{x}, y^{\prime\prime} = u^{\prime\prime}e^{x} + 2u^{\prime}e^{x} + ue^{x}$. Substitute these into $y^{\prime\prime} - y = xe^{x}$. Then

$$u^{\prime\prime}e^{x} + 2u^{\prime}e^{x} = xe^{x}.$$

or

$$u^{\prime\prime} + 2u^{\prime} = x.$$

Now let $u^{\prime} = w$. Then we have the following linear differential equation in $w$.

$$w^{\prime} + 2w = x$$

$\mu = e^{\int 2dx} = e^{2x}$ and $d(e^{2x}w) = xe^{2x}dx$. Integrate this with respect to $x$ to get

$$e^{2x}w = \frac{1}{2}xe^{2x}-\frac{1}{4}e^{2x} + c_1$$

Thus,

$$w = c_{1}e^{-2x} + \frac{x}{2} - \frac{1}{4}$$

Now integrate with respect to $x$

$$u = \int w dx = -\frac{c_{1}}{2}e^{-2x} + \frac{x^{2}}{4} - \frac{x}{4} + c_{2}$$

Since $y = ue^{x}$, the general solution is

$$y = -\frac{c_{1}}{2}e^{-x} + c_{2}e^{x} + (\frac{x^{2}}{4} - \frac{x}{4})e^{x}\ensuremath{\ \blacksquare}$$

Example 2..6   Using $L(x^{2}) \equiv 0$, Reduce the order of the following differential equation

$$L(y)= y^{\prime\prime\prime} + 2x^{2}y^{\prime\prime} -3xy^{\prime} + 2y = 0$$

SOLUTION Let $y = ux^{2}$. Then $y^{\prime} = u^{\prime}x^{2} + 2ux$, $y^{\prime\prime} = u^{\prime\prime}x^{2} + 4u^{\prime}x + 2u,$ $y^{\prime\prime\prime} = u^{\prime\prime\prime}x^{2} + 6u^{\prime\prime}x + 6u^{\prime}$. Thus

$$L(y) = L(ux^{2}) = x^{2}u^{\prime\prime\prime} + (2x^{4} + 6x)u^{\prime\prime} + (5x^{3} + 6)u^{\prime} = 0$$

Now let $u^{\prime} = w$. Then

$$x^{2}w^{\prime\prime} + (2x^{4} + 6x)w^{\prime} + (5x^{3} + 6)w = 0\ensuremath{\ \blacksquare}$$