The differential equation, representing the curve y = eˣ(a cosx + b sinx) where a and b are arbitrary constants, is
Answer
The correct answer is C: d²y/dx² – 2(dy/dx) + 2y = 0. Differentiate y = eˣ(a cosx + b sinx): dy/dx = eˣ(a cosx + b sinx) + eˣ(–a sinx + b cosx) = y + eˣ(–a sinx + b cosx). So dy/dx – y = eˣ(–a sinx + b cosx).
A.d²y/dx² + 2y = 0
B.d²y/dx² + 2(dy/dx) + 2y = 0
C.d²y/dx² – 2(dy/dx) + 2y = 0✓ Correct
D.d²y/dx² + y = 0
Explanation
Differentiate y = eˣ(a cosx + b sinx): dy/dx = eˣ(a cosx + b sinx) + eˣ(–a sinx + b cosx) = y + eˣ(–a sinx + b cosx). So dy/dx – y = eˣ(–a sinx + b cosx). Differentiate again: d²y/dx² – dy/dx = eˣ(–a sinx + b cosx) + eˣ(–a cosx – b sinx) = (dy/dx – y) – y. So d²y/dx² – dy/dx = dy/dx – 2y, giving d²y/dx² – 2(dy/dx) + 2y = 0.
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