Question

The differential equation of the family of curves \(y = Ae^{3x} + Be^{5x}\), where \(A\) and \(B\) are arbitrary constants, is

• A. \(\frac{d^2y}{dx^2} + 8\frac{dy}{dx} + 15y = 0\)
• B. \(\frac{d^2y}{dx^2} - \frac{dy}{dx} + y = 0\)
• C. \(\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 15y = 0\)
• D. None of the above

Hint:

For exponential solutions, the auxiliary equation roots are the exponents.

Answer 1

The Correct Option is C

To find the differential equation of the given family of curves \(y = Ae^{3x} + Be^{5x}\), we need to eliminate the arbitrary constants \(A\) and \(B\) through differentiation.

First, differentiate the equation with respect to \(x\): 

\[\frac{dy}{dx} = 3Ae^{3x} + 5Be^{5x}\]

Next, differentiate again to obtain the second derivative:

\[\frac{d^2y}{dx^2} = 9Ae^{3x} + 25Be^{5x}\]

We now have the following system of equations:

• \(y = Ae^{3x} + Be^{5x}\)
• \(\frac{dy}{dx} = 3Ae^{3x} + 5Be^{5x}\)
• \(\frac{d^2y}{dx^2} = 9Ae^{3x} + 25Be^{5x}\)

We need to eliminate \(A\) and \(B\) by expressing them in terms of the derivatives. Consider these two equations:

• Multiply the first derivative equation by 3:
• Subtract this equation from the second derivative equation:

This implies that:

\[B = \frac{\frac{d^2y}{dx^2} - 3\frac{dy}{dx}}{10e^{5x}}\]

Now, to eliminate \(A\), consider multiplying the original equation by 5 and the first derivative by 1:

• Multiply the original equation by 5:
• Multiply the first derivative equation by 1:
• Subtract the second equation from the first:

This implies that:

\[A = \frac{5y - \frac{dy}{dx}}{2e^{3x}}\]

By substituting the expressions of \(A\) and \(B\) into the relevant derivatives and simplifying, we derive the differential equation:

\[\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 15y = 0\]

As a result, the correct answer is:

\[\frac{d^2y}{dx^2} - 8\frac{dy}{dx} + 15y = 0\]