Question

Which of the following is the general solution of `("d"^2y)/("d"x^2) - 2 ("d"y)/("d"x) + y` = 0?

पर्याय

• y = (Ax + B)e^x

• y = (Ax + B)e^–x

• y = Ae^x + Be^–x

• y = Acosx + Bsinx

Answer 1

y = (Ax + B)e^x 

Explanation:

The given differential equation is `("d"^2y)/("d"x^2) - 2 ("d"y)/("d"x) + y` = 0

Since the above equation is of second order and first degree

∴ `"D"^2y - 2"D"y + y` = 0

Where D = `"d"/("d"x)`

⇒ `("D"^2 - 2"D" + 1)y` = 0

∴ Auxiliary equation is m² – 2m + 1 = 0

⇒ (m – 1)² = 0

⇒ m = 1, 1

If the roots of Auxiliary equation are real and equal say (m)

Then CF = `("c"_1 + "c"_2) . "e"^(mx)`

∴ CF = `("A"x + "B")"e"^x`

So y = `("A"x + "B")."e"^x`