Consider the family of curves \(y = Ae^{x} + Be^{3x}\), where A and B are arbitrary constants, and the following statements:
(A) The differential equation of this family is of order 2.
(B) The differential equation of this family is \(\displaystyle\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 3y = 0\).
(C) \(y = 5e^{3x}\) is a solution of the differential equation of this family.
(D) \(y = e^{x} + 2\) is a solution of the differential equation of this family.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → (A), (B) and (C) only
The family has two arbitrary constants, so its differential equation is of order 2. Differentiating twice:
\( \displaystyle \frac{dy}{dx} = Ae^{x} + 3Be^{3x} \)
\( \displaystyle \frac{d^2y}{dx^2} = Ae^{x} + 9Be^{3x} \)
Substituting:
\( \displaystyle \frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 3y = (1 - 4 + 3)Ae^{x} + (9 - 12 + 3)Be^{3x} = 0 \)
Every member of the family is a solution; \(y = 5e^{3x}\) is the member with A = 0, B = 5.
For \(y = e^{x} + 2\): \(\frac{dy}{dx} = \frac{d^2y}{dx^2} = e^{x}\), so
\(e^{x} - 4e^{x} + 3(e^{x} + 2) = 6\)
So the equation is \(\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 3y = 0\), of order 2; \(5e^{3x}\) satisfies it and \(e^{x} + 2\) does not.
Checking the statements
(A) The differential equation of this family is of order 2. — Correct. Two arbitrary constants, A and B.
(B) The differential equation of this family is \(\frac{d^2y}{dx^2} - 4\frac{dy}{dx} + 3y = 0\). — Correct. The coefficients of \(Ae^{x}\) and \(Be^{3x}\) both become 0.
(C) \(y = 5e^{3x}\) is a solution of the differential equation of this family. — Correct. \(45e^{3x} - 60e^{3x} + 15e^{3x} = 0\).
(D) \(y = e^{x} + 2\) is a solution of the differential equation of this family. — Incorrect. The left side becomes 6, not 0.