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Question

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

A
Acosx+Bsinx is a solution, where A and B are real costants
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B
Acos(x+π4) is a solution, where A is real constant
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C
Acosxsinx is a solution, where A is real constant
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D
Acos(x+π4)+Bsin(xπ4) is a solution, where A and B are real constants
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Solution

## The correct options are A Acosx+Bsinx is a solution, where A and B are real costants B Acos(x+π4) is a solution, where A is real constant D Acos(x+π4)+Bsin(x−π4) is a solution, where A and B are real constantsGiven differential equation is a0d2ydx2+a1dydx+a2y=0Also given that, cosx and sinx are solutions to the given differential equation.⟹Acosx+Bsinx is a solution to the given differential equation as it is homogenous, where A and B are real constants.Acos(x+Π4)=A√2(−sinx+cosx)=C1sinx+C2cosx is also a solution where A=−BConsider, Acos(x+Π4)+Bsin(x−Π4)=A√2(cosx−sinx)+B√2(cosx−sinx)=(A√2+B√2)cosx−(A√2+B√2)sinxwhich is of the formC1cosx+C2sinx.Hence, options A, B, and D are true.

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