page 173 number 27
This is one of the situations where starting the problem abstractly is beneficial.
$Ay^{\prime\prime}+By^\prime+Cy=0$
$y_2=y_1v$
$y_2^\prime=y_1^\prime v+y_1 v^\prime$
$y_2^{\prime\prime}=y_1^{\prime\prime}v+2y_1^\prime v^\prime+y_1 v^{\prime\prime}$
Terms that survive:
$A(2y_1^\prime v^\prime+y_1 v^{\prime\prime})+By_1 v^\prime=0$
Which is:
$Ay_1 v^{\prime\prime}+(2Ay_1^\prime+By_1)v^\prime=0$
Let $w=v^\prime$ to obtain:
$Ay_1 w^\prime+(2Ay_1^\prime+By_1)w=0$
So then:
$\int \frac{1}{w} dw=\int \left[-(2\frac{y_1^\prime}{y_1}+\frac{B}{A})\right] dx$
Note at this point I just enter the coefficients $A$ and $B$:
$\ln w=-2\ln(y_1)-\int \left[\frac{-1}{x}\right] dx$
$v^\prime=e^{\ln(y_1)^{-2}+\ln(x)+C}=C_2\cdot x(y_1)^{-2}$
I originally forgot the constant here. You should remember it.
$v=C_2\int(\frac{x}{\sin^2 x^2})dx=C_2\frac{1}{2}\int(\csc^2(x^2))dx^2$
$=-C_2\frac{1}{2}(\cot x^2+C)=C_3\cot x^2+C_4$
I made the error of forgetting the minus sign. You should not make this error. That is:
$\frac{d}{dx}\cot u=-\csc^2 u \frac{du}{dx}$
Though in this problem the error doesn't matter. We pick $C_3=1$ and $C_4=0$ and get
$y_2=\cot x^2 \sin x^2 =\cos x^2$
