Find the differential equation of the family of curves $y = A{e^{2x}} + B{e^{ - 3x}}$, where A and B are arbitrary constants.

Question:

Find the differential equation of the family of curves $y = A{e^{2x}} + B{e^{ - 3x}}$, where A and B are arbitrary constants.

Solution:

Given, y=Ae^2x+Be^−3x......(1)

Now differentiating both sides with respect to x we get,

$\frac{{dy}}{{dx}}$=2Ae^2x−3Be^−3x ……..(2)

Multiplying equation (1) by 2 and subtracting from (2)

$\frac{{dy}}{{dx}}$ - 2y=-5Be^−3x…….(3)

Differentiating above equation, we get

$\frac{{{d^2}y}}{{d{x^2}}}$-2$\frac{{dy}}{{dx}}$=15 Be^−3x……(4)

Multiplying Equation (3) by 3 and adding with equation (4)

$\frac{{{d^2}y}}{{d{x^2}}}$-2$\frac{{dy}}{{dx}}$ + 3$\frac{{dy}}{{dx}}$-6y=15 Be^−3x -15 Be^−3x

$\frac{{{d^2}y}}{{d{x^2}}}$ + $\frac{{dy}}{{dx}}$ -6y = 0