The differential equation of the family of curves y=e2x(acosx+bsinx), where a and b are arbitrary constants, is given by
y2–4y1+5y=0
2y2–y1+5y=0
y2+4y1–5y=0
y2–2y1+5y=0
Explanation for the correct answer:
Given curve,
y=e2x(acosx+bsinx)y1=dydxy2=dy1dx
On Differentiating, we get
dydx=e2x·2(acosx+bsinx)+e2x(-asinx+bcosx)y1=2y+e2x(-asinx+bcosx)...(1)dy1dx=2dydx+2e2x(-asinx+bcosx)+e2x(-acosx-bsinx)
y2=2y1+2e2x-asinx+bcosx-yy2=-y+2y1-2y+2y1[From(1)]y2=-y+4y1-4y
⇒y2-4y1+5y=0
Hence, the correct option is (A)