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Wednesday, February 18, 2009

Problem: Differential Equation of a Solution

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

Solution: -

Given that, y = ex (A cosx + B sinx) ------------ (1)

Differentiating (1) with respect to x, we get

y ‘ = ex (-A sinx + B cosx) + ex (A cosx + B sinx)

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

Differentiating (2) with respect to x, we get

y “ = -ex (A cosx + B sinx) + ex (-A sinx + B cosx) + y’ ---- (3)

Now From Equation (1)

ex (-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 = aex + be-x + x2 is a solution.

Solution: -

Given that, xy = aex + be-x + x2 ---------- (1)

Differentiating (1) with respect to x, we get

xy’ + y = aex – be-x + 2x ------ (2)

Differentiating (2) with respect to x, we get

xy” + y’ + y’ = aex + be-x + 2

or, xy” + 2y’ = xy – x2 + 2

or, xy” + 2y’ - xy + x2 – 2 = 0 ------- (3)

Equation (3) is the required differential equation.

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