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Question

If cosx and sinx are solutions of the differential equation a0d2ydx2+a1dydx+a2y=0, where a0,a1 and a2 are all real constants, then which of the following is/are always true?

A
A cosx+Bsinx is a solution, where A and B real constants
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B
A cos(x+π4) is a solution, where A is real constant
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C
A cosxsinx is a solution, where A is real constant
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D
A cos(x+π4)+Bsin(xπ4) is a solution, where A and B are real constants
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Solution

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≠0Thus, 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)sinxwhich 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.

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