a)Find the differential equation of the family of curves y = e^{x} (A cosx + B sinx), where A and B are arbitrary constants.

Given that, y = e^{x} (A cosx + B sinx) ------------ (1)

Differentiating (1) with respect to x, we get

y ‘ = e^{x} (-A sinx + B cosx) + e^{x} (A cosx + B sinx)

or, y ‘ = e^{x} (-A sinx + B cosx) + y {using (1)} ------ (2)

Differentiating (2) with respect to x, we get

y “ = -e^{x} (A cosx + B sinx) + e^{x} (-A sinx + B cosx) + y’ ---- (3)

Now From Equation (1)

e^{x} (-A sinx + B cosx) = y’ – y -----------(4)

Hence eliminating A and B from equation (1), (3) and (4), we get

y” = - y + y’ – y + y’

or, y” – 2y’ +2y = 0 ---------- (5)

Equation (5) is the required differential equation.

b)By eliminating the constants a and b obtain the differential equation for which xy = ae^{x} + be^{-x} + x^{2} is a solution.

Given that, xy = ae^{x} + be^{-x} + x^{2} ---------- (1)

Differentiating (1) with respect to x, we get

xy’ + y = ae^{x }– be^{-x} + 2x ------ (2)

Differentiating (2) with respect to x, we get

xy” + y’ + y’ = ae^{x} + be^{-x} + 2

or, xy” + 2y’ = xy – x^{2} + 2

or, xy” + 2y’ - xy + x^{2} – 2 = 0 ------- (3)

Equation (3) is the required differential equation.

Thank you soooo much.

ReplyDelete