We are given a linear homogeneous second-order ODE
#y''+2y'tan(x)-y=0#
and that a particular solution is #y_1(x)=sin(x)#.
We know that linear homogeneous second-order ODEs have the general solution of the form #c_1y_1(x)+c_2y_2(x)#, where #c_1# and #c_2# are arbitrary constants and #y_1# and #y_2# are linear independent particular solutions. Thus, the general solution of the ODE given must be of the form #c_1sin(x)+c_2y_2(x)#. We need to solve for #y_2#.
Using reduction of order, let us first assume that #y_2=v(x)y_1(x)=v(x)sin(x)#, where #v# is some function of #x# that we need to find. Note that
#d/dx(y_2)=v'(x)sin(x)+v(x)cos(x)#
and
#d^2/dx^2(y_2)=v''(x)sin(x)+v'(x)cos(x)+v'cos(x)-v(x)sin(x)#
#\ \ \ \ \ \ \ \ \ \ " "=v''(x)sin(x)+2v'(x)cos(x)-v(x)sin(x)#
Substitute #y_2# into the ODE:
#d^2/dx^2(v(x)sin(x))+2d/dx(v(x)sin(x))tan(x)-v(x)sin(x)=0#
#v''(x)sin(x)+2v'(x)cos(x)-v(x)sin(x)+2tan(x)(v'(x)sin(x)+v(x)cos(x))-v(x)sin(x)=0#
After some simplifying, we have
#v''(x)+(2v'(x))/(sin(x)cos(x))=0#
Set #w=v'#:
#(dw)/dx+(2w)/(sin(x)cos(x))=0#
So we have successfully reduced the order of the ODE. Now, this is a simple separable first-order ODE:
#int\ (dw)/w=-int\ 2/(sin(x)cos(x))\ dx#
#ln|w|=-2ln|tan(x)|+C#
#w=C/tan^2(x)#
We defined #w=v'#, so
#v=int\ 1/tan^2(x)\ dx=C(-1/tan(x)-x)+D#
As #v# will be later multiplied by an arbitrary constant, the values of #C# and #D# are not important. So, set #C=-1, D=0# to get
#v=1/tan(x)+x#.
Then,
#y_2=v(x)y_1(x)=sin(x)(1/tan(x)+x)=xsin(x)+cos(x)#.
Thus, the general solution is
#y=c_1sin(x)+c_2(xsin(x)+cos(x))#
