7.3 Reduction of order
It can be hard to find y 1 y_1 y 1 y 2 y_2 y 2 y 2 y_2 y 2 y 1 y_1 y 1
For example, to solve:
( x 2 − x ) y ′ ′ − x y ′ + y = 0 (x^2 - x)y'' - xy' + y = 0 ( x 2 − x ) y ′′ − x y ′ + y = 0 We can guess that y 1 = x y_1 = x y 1 = x y 1 ′ = 1 y_1' = 1 y 1 ′ = 1 y 1 ′ ′ = 0 y_1'' = 0 y 1 ′′ = 0 y 2 y_2 y 2 x 2 − x x^2-x x 2 − x
y ′ ′ + 1 1 − x y ′ + 1 x 2 − x y = 0 y'' + \frac{1}{1 - x} y' + \frac{1}{x^2 - x}y = 0 y ′′ + 1 − x 1 y ′ + x 2 − x 1 y = 0 Then, calculate u :
u = 1 x 2 exp ( − ∫ 1 1 − x d x ) = 1 x 2 e ln ( 1 − x ) = 1 − x x 2 \begin{align}
u &= \frac{1}{x^2} \exp\left(-\int \frac{1}{1-x} \d{x} \right) \\
&= \frac{1}{x^2}\ e^{\ln(1-x)} \\
&= \frac{1-x}{x^2}
\end{align} u = x 2 1 exp ( − ∫ 1 − x 1 d x ) = x 2 1 e l n ( 1 − x ) = x 2 1 − x Finally, integrate u to obtain y 2 y_2 y 2
y 2 = 1 x 2 ∫ ( 1 x 2 − 1 x ) d x = x ( − 1 x − ln x ) \begin{align}
y_2 &= \frac{1}{x^2} \int \left( \frac{1}{x^2} - \frac{1}{x} \right)\d{x} \\
&= x \left(-\frac{1}{x}-\ln{x}\right)
\end{align} y 2 = x 2 1 ∫ ( x 2 1 − x 1 ) d x = x ( − x 1 − ln x ) Absorbing the minus sign into the unknown coefficient, the general solution is:
y = c 1 x + c 2 ( 1 + ln x ) y = c_1 x + c_2 (1 + \ln x) y = c 1 x + c 2 ( 1 + ln x ) Skill builder problems ¶ Solve the following:
Rearrange in standard form:
y ′ ′ + 2 x y ′ + y = 0 y'' + \frac{2}{x}y' + y = 0 y ′′ + x 2 y ′ + y = 0 This ODE is linear and homogeneous, so use reduction of order for y 2 y_2 y 2 u :
u = 1 y 1 2 e − ∫ p d x = x 2 cos 2 x exp ( − ∫ 2 x d x ) = x 2 cos 2 x exp ( − 2 ln ( x ) ) = x 2 cos 2 x x − 2 = sec 2 x \begin{align}
u &= \frac{1}{y_1^2}e^{-\int p \d{x}} \\
&= \frac{x^2}{\cos^2 x} \exp\left(-\int \frac{2}{x} \d{x}\right) \\
&= \frac{x^2}{\cos^2 x} \exp\left(-2\ln(x)\right) \\
&= \frac{x^2}{\cos^2 x}x^{-2} \\
&= \sec^2 x
\end{align} u = y 1 2 1 e − ∫ p d x = cos 2 x x 2 exp ( − ∫ x 2 d x ) = cos 2 x x 2 exp ( − 2 ln ( x ) ) = cos 2 x x 2 x − 2 = sec 2 x Then compute y 2 y_2 y 2
y 2 = y 1 ∫ u d x = cos x x ∫ sec 2 x d x = x − 1 cos x tan x = sin x x \begin{align}
y_{2} &= y_{1} \int u \d{x} \\
&= \frac{\cos x}{x} \int \sec^2 x \d{x} \\
&= x^{-1}\cos x \tan x \\
&= \frac{\sin x}{x}
\end{align} y 2 = y 1 ∫ u d x = x cos x ∫ sec 2 x d x = x − 1 cos x tan x = x sin x Hence, the general solution is:
y = c 1 cos x x + c 2 sin x x y = c_{1}\frac{\cos x}{x} + c_{2}\frac{\sin x}{x} y = c 1 x cos x + c 2 x sin x By inspection y 1 = x y_{1} = x y 1 = x y 1 ′ = 1 y_{1}' = 1 y 1 ′ = 1 y 1 ′ ′ = 0 y_{1}'' = 0 y 1 ′′ = 0 y 2 y_{2} y 2
y ′ ′ − 2 x 1 − x 2 y ′ + 2 1 − x 2 y = 0 y'' - \frac{2x}{1-x^2}y' + \frac{2}{1-x^2}y = 0 y ′′ − 1 − x 2 2 x y ′ + 1 − x 2 2 y = 0 Now compute u
u = 1 y 2 2 e − ∫ p d x = 1 x 2 exp ( ∫ 2 x 1 − x 2 d x ) = 1 x 2 exp [ − ln ( 1 − x 2 ) ] = 1 x 2 ( 1 − x 2 ) \begin{align}
u &= \frac{1}{y_{2}^2}e^{-\int p \d{x}} \\
&= \frac{1}{x^2} \exp\left(\int \frac{2x}{1-x^2} \d{x}\right) \\
&= \frac{1}{x^2} \exp\left[-\ln(1 - x^2)\right] \\
&= \frac{1}{x^2(1 - x^2)}
\end{align} u = y 2 2 1 e − ∫ p d x = x 2 1 exp ( ∫ 1 − x 2 2 x d x ) = x 2 1 exp [ − ln ( 1 − x 2 ) ] = x 2 ( 1 − x 2 ) 1 Then y 2 y_2 y 2
y 2 = y 1 ∫ u d x = x ∫ 1 x 2 ( 1 − x 2 ) d x \begin{align}
y_{2} &= y_{1} \int u \d{x} \\
&= x \int \frac{1}{x^2(1 - x^2)} \d{x}
\end{align} y 2 = y 1 ∫ u d x = x ∫ x 2 ( 1 − x 2 ) 1 d x Use partial-fraction decomposition to evaluate the integral:
1 x 2 ( 1 − x ) ( 1 + x ) = A x + B x 2 + C 1 + x + D 1 − x \frac{1}{x^2(1 - x)(1 + x)} = \frac{A}{x} + \frac{B}{x^2} +
\frac{C}{1 + x} + \frac{D}{1 - x} x 2 ( 1 − x ) ( 1 + x ) 1 = x A + x 2 B + 1 + x C + 1 − x D Use the coverup method to get B = 1 B = 1 B = 1 C = 1 / 2 C = 1/2 C = 1/2 D = 1 / 2 D = 1/2 D = 1/2 A = 0 A = 0 A = 0
y 2 = x ∫ [ 1 x 2 + 1 2 ( 1 + x ) + 1 2 ( 1 − x ) ] d x = x [ − 1 x + 1 2 ln ( 1 + x ) − 1 2 ln ( 1 − x ) ] = − 1 − x 2 ln ( 1 − x 1 + x ) \begin{align}
y_2 &= x \int \left[ \frac{1}{x^2} + \frac{1}{2(1 + x)}
+ \frac{1}{2(1 - x)} \right] \d{x} \\
&= x \left[ -\frac{1}{x} + \frac{1}{2}\ln(1 + x)
- \frac{1}{2}\ln(1 - x) \right] \\
&= -1 - \frac{x}{2} \ln \left( \frac{1 - x}{1 + x} \right)
\end{align} y 2 = x ∫ [ x 2 1 + 2 ( 1 + x ) 1 + 2 ( 1 − x ) 1 ] d x = x [ − x 1 + 2 1 ln ( 1 + x ) − 2 1 ln ( 1 − x ) ] = − 1 − 2 x ln ( 1 + x 1 − x ) Hence, the general solution is:
y = c 1 x + c 2 [ 1 + x 2 ln ( 1 − x 1 + x ) ] y = c_{1}x + c_{2} \left[ 1 + \frac{x}{2}
\ln \left( \frac{1 - x}{1 + x} \right) \right] y = c 1 x + c 2 [ 1 + 2 x ln ( 1 + x 1 − x ) ]