The correct options are
A A
cos(x+π4) is a solution, where A is real constant
B A
cosx+Bsinx is a solution, where A and B real constants
D A
cos(x+π4)+Bsin(x−π4) is a solution, where A and B are real constants
(a) Let f(x)=cosx and g(x)=sinxConsider the Wronskian of f(x) and g(x),W=∣∣∣f(x)g(x)f′(x)g′(x)∣∣∣
=∣∣∣cosxsinx−sinxcosx∣∣∣
=cos2x+sin2x=1≠0
Thus, the functions are linearly independent. So, the general solution of given differential equation is given by y=Acosx+Bsinx, where A and B are real constants.
[∵ if y1 and y2 are linearly independent solutions of the differential equation ay"+by′+c=0, then the general solution is
y=c1y1+C2y2, where c1 and c2 are constants]
Hence, option (a) is true.
(b) Let y=Acos(x+π4)
=A(cosx⋅cosπ4−sinx⋅sinπ4)
=A√2(cosx−sinx)
=A√2cosx+(−A√2)sinx
which is in the form of general solution.
Hence, option (b) is true.
(c) Let y=Acosxsinx, Which cannot be expressed in the form of general solution.
(d) Let y=Acos(x+π4)+Bsin(x−π4)
=A(cosx⋅1√2−sinx⋅1√2)+B(sinx⋅1√2−cosx⋅1√2)
=cosx(A√2−B√2)+sinx(B√2−A√2) which is in the form of general solution.
Hence, option (d) is true.