Finding the Differential Equation from its General Solution

The question asks us to identify the differential equation that has the general solution \(y = ae^x + be^{-x}\). This general solution contains two arbitrary constants, \(a\) and \(b\), which suggests that the corresponding differential equation is a second-order linear homogeneous differential equation with constant coefficients.

Understanding the General Solution Structure

For a second-order linear homogeneous differential equation of the form \(Ay'' + By' + Cy = 0\), the general solution depends on the roots of its characteristic equation \(Am^2 + Bm + C = 0\).

If the characteristic equation has distinct real roots \(m_1\) and \(m_2\), the general solution is given by \(y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\).

Identifying the Roots from the Given General Solution

The given general solution is \(y = ae^x + be^{-x}\). Comparing this with the standard form \(y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\), we can identify the roots of the characteristic equation:

• The coefficient of \(x\) in the first term \(e^x\) is 1. So, one root is \(m_1 = 1\).
• The coefficient of \(x\) in the second term \(e^{-x}\) is -1. So, the other root is \(m_2 = -1\).

Thus, the roots of the characteristic equation are \(m=1\) and \(m=-1\).

Constructing the Characteristic Equation

If the roots of a quadratic equation are \(m_1\) and \(m_2\), the equation can be written in the form \((m - m_1)(m - m_2) = 0\).

Using the roots \(m_1 = 1\) and \(m_2 = -1\), the characteristic equation is:

\((m - 1)(m - (-1)) = 0\)

\((m - 1)(m + 1) = 0\)

Expanding this equation, we get:

\(m^2 - 1 = 0\)

Forming the Differential Equation

Now we convert the characteristic equation \(m^2 - 1 = 0\) back into a second-order linear homogeneous differential equation. In this conversion:

• \(m^2\) corresponds to the second derivative term, \(\frac{d^2y}{dx^2}\) or \(y''\).
• \(m\) corresponds to the first derivative term, \(\frac{dy}{dx}\) or \(y'\) (if present).
• A constant term corresponds to the \(y\) term.

So, the characteristic equation \(m^2 - 1 = 0\) corresponds to the differential equation:

\(\frac{d^2y}{dx^2} - y = 0\)

Comparing with the Given Options

Let's compare the derived differential equation with the provided options:

From the comparison, the differential equation \(\frac{d^2y}{dx^2}-y=0\) matches the structure that yields the general solution \(y = ae^x + be^{-x}\).

Conclusion

The differential equation that has the general solution \(y = ae^x + be^{-x}\) is \(\frac{d^2y}{dx^2}-y=0\).

Revision Table: Differential Equations and Solutions

Additional Information on Second-Order DEs

A second-order linear homogeneous differential equation with constant coefficients has the form \(\frac{d^2y}{dx^2} + P\frac{dy}{dx} + Qy = 0\), where \(P\) and \(Q\) are constants. To solve this, we assume a solution of the form \(y = e^{mx}\). Substituting this into the differential equation gives the characteristic equation \(m^2 + Pm + Q = 0\).

The nature of the roots of this quadratic equation determines the form of the general solution:

• Distinct Real Roots: If \(m^2 + Pm + Q = 0\) has two distinct real roots \(m_1\) and \(m_2\), the general solution is \(y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\).
• Repeated Real Roots: If the equation has a single real root \(m\) (with multiplicity 2), the general solution is \(y = (c_1 + c_2 x) e^{mx}\).
• Complex Conjugate Roots: If the roots are complex conjugates, \(m = \alpha \pm i\beta\), the general solution is \(y = e^{\alpha x} (c_1 \cos(\beta x) + c_2 \sin(\beta x))\).

In our specific problem, the roots \(m=1\) and \(m=-1\) are distinct real roots, which corresponds to the case \(y = c_1 e^{m_1 x} + c_2 e^{m_2 x}\).