Reduction of Order

Exercise 2.2

1. Solve the following differential equations using the reduction of order.

(a) $\ y^{\prime\prime} - 3y^{\prime} + 2y = 0,\ y_{1} = e^{2x}$
(b) $\ x^{2}y^{\prime\prime} - 3xy^{\prime} + 4y = 0, \ y_{1} = x^{2}$
(c) $\ y^{\prime\prime} - y = e^{-x}, \ y_{1} = e^{-x}$
(d) $\ y^{\prime\prime} + y = \sec{x}, \ y_{1} = \cos{x}$

2. Suppose that one of the solutions of $y^{\prime\prime} + a_{1}(x)y^{\prime} + a_{0}(x)y = 0$ is $y_{1}(x)$，then the other solution $y_{2}(x)$ by using the reduction of order is given by

$$y_{2}(x) = y_{1}(x)\int \frac{e^{-\int a_{1}(x)dx}}{y_{1}^{2}} dx$$

Show that $\{y_{1},y_{2}\}$ are linearly independent.

Answer
1. (a) Let $y = uy_{1} = u e^{2x}$. Then

$$2(y = u e^{2x})$$

$$-3(y^{\prime} = 2ue^{2x} + u^{\prime} e^{2x})$$

$$\underline{+ y^{\prime\prime} } = \underline{ 4u e^{2x} + 4u^{\prime}e^{2x} + u^{\prime\prime} e^{2x} }$$

$$= u^{\prime}e^{2x} + u^{\prime\prime}e^{2x}$$ 0

Thus, we have

$$u^{\prime\prime} + u^{\prime} = 0$$

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

$$w^{\prime} + w = 0$$

This is a first order linear differential equation or a separable equation. Using the integrating factor, we have $\mu = e^{\int dx} = e^{x}$. Multiply $\mu$ to both sides of equation, we have $(e^{x} w)^{\prime} = 0$. Solving this, we have $e^{x} w = c_{1}$ and

$$w = c_{1}e^{-x}$$

Since $w = u^{\prime}$, we have

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

Lastly noting $y = ue^{2x}$ and we obtain the general solution.

$$y = ue^{2x} = c_{1}e^{x} + c_{2}e^{2x} \ \$$

(b) Let $y = uy_{1} = u x^{2}$. Then

$$4(y = u x^{2})$$

$$-3x(y^{\prime} = 2 u x + u^{\prime} x^{2})$$

$$\underline{x^{2} y^{\prime\prime} } = \underline{ 2u + 4u^{\prime}x + u^{\prime\prime} x^{2} }$$

$$= u^{\prime}x^{3} + u^{\prime\prime}x^{4}$$ 0

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

$$x^{4}w^{\prime} + x^{3}w = 0$$

This is a first order linear differential equation or a separable differential equation. Here we use the separation of variable.

$$\frac{w^{\prime}}{w} = - \frac{1}{x}$$

Then $\log{\vert w\vert} = - \log{\vert x\vert} + c$. Thus, $w = c_{1}/x$. Note that $w = u^{\prime}$. Then

$$u = \int \frac{c_{1}}{x} dx = c_{1}\log{x} + c_{2}$$

Note also that $y = uex^{2}$. Then we have the general solution.

$$y = ux^{2} = c_{1}x^{2}\log{x} + c_{2}x^{2} \ \$$

(c) Let $y = uy_{1} = u e^{-x}$. Then

$$-(y = u e^{-x})$$

$$y^{\prime} = - u e^{-x} + u^{\prime} e^{-x}$$

$$\underline{+ y^{\prime\prime} } = \underline{ u e^{-x} - 2u^{\prime}e^{-x} + u^{\prime\prime} e^{-x} }$$

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

Thus,

$$u^{\prime\prime} - 2u^{\prime} = 1$$

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

$$w^{\prime} - 2 w = 1$$

This is a first order linear differential equation. So, we find the integrating factor. $\mu = e^{\int -2 dx} = e^{-2x}$. Multiply $\mu$ to both sides of equation. Then the left-hand side is the derivative of the product of $\mu$ and $w$.

$$(e^{-2x} w)^{\prime} = e^{-2x}$$

Solve this equation, we have

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

Thus,

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

Note that $w = u^{\prime}$.

$$u = \int (-\frac{1}{2} + c e^{2x}) dx = -\frac{x}{2} + c_{1}e^{2x} + c_{2}$$

Note also that $y = ue^{-x}$. Then we have the general solution.

$$y = ue^{-x} = c_{1}e^{x} + c_{2}e^{-x} - \frac{xe^{-x}}{2} \ \$$

(d) Let $y = uy_{1} = u \cos{x}$. Then

$$+(y = u \cos{x})$$

$$y^{\prime} = u^{\prime} \cos{x} - u \sin{x}$$

$$\underline{+ y^{\prime\prime} } = \underline{ u^{\prime\prime} \cos{x} - 2u^{\prime} \sin{x} - u \cos{x} }$$

$$\sec{x} = u^{\prime\prime} \cos{x} - 2u^{\prime} \sin{x}$$

Thus,

$$u^{\prime\prime} - 2u^{\prime} \tan{x} = \sec^{2}{x}$$

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

$$w^{\prime} - 2\tan{x} w = \sec^{2}{x}$$

This is a linear differential equation. We find the integrating factor. $\mu = e^{\int -2\tan{x} dx} = e^{2 \log{\cos{x}}} = \cos^{2}{x}$. We multiply $\mu$ to both sides of equation. Then we get $(\cos^{2}{x} w)^{\prime} = 1$. Solve this equation, we have

$$\cos^{2}{x} w = \int dx = x + c_{1}$$

Thus,

$$w = (x+c_{1})\sec^{2}{x}$$

Note that $w = u^{\prime}$.

$$u = \int (x + c_{1}) \sec^{2}{x} dx \ \ \left(\begin{array}{ll} u = x... ...sec^{2}{x} dx\\ du = dx & v = \int sec^{2}{x} dx = \tan{x} \end{array}\right )$$

$$= (x + c)\tan{x} - \int \tan{x} dx = (x + c_{1})\tan{x} + \log{\vert\cos{x}\vert} + c_{2}$$

Note also that $y = u \cos{x}$. Then we have the general solution.

$$y = u\cos{x} = c_{1}\sin{x} + c_{2}\cos{x} + \cos{x}\log{\vert\cos{x}\vert} + x\sin{x} \ \$$

2. Let $y_{2} = uy_{1}$. Then

$$+a_{0}(x)(y_{2} = u y_{1})$$

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

$$\underline{+ y_{2}^{\prime\prime} } = \underline{ u^{\prime\prime} y_{1} + 2u^{\prime}y_{1}^{\prime}+ u y^{\prime\prime} }$$

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

Thus,

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

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

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

This is a first order linear differential equation. So, we find the integrating factor.

$$\mu = e^{\int (\frac{2y_{1}^{\prime}}{y_{1}} + a_{1}(x))dx } = e^{2\log{y_{1}} + \int a_{1}(x) dx} = y_{1}^{2}e^{\int a_{1}(x) dx}$$

Multiply $\mu$ to both sides. Then the left-hand side is the derivative of the product of $\mu$ and $w$.

$$(y_{1}^{2}e^{\int a_{1}(x) dx} w)^{\prime} = 0$$

Integrate both sides, we have .

$$y_{1}^{2}e^{\int a_{1}(x) dx} w = c$$

Thus,

$$w = \frac{c e^{- \int a_{1}(x) dx}}{y_{1}^{2}}$$

Note that $w = u^{\prime}$. Then

$$u = \int \frac{c e^{\int a_{1}(x) dx}}{y_{1}^{2}}$$

$$= \int \frac{c e^{\int a_{1}(x) dx}}{y_{1}^{2}}$$

Note also that $y_{2} = uy_{1}$. Then

$$y_{2} = y_{1}\int \frac{e^{\int a_{1}(x) dx}}{y_{1}^{2}} \ \$$