Question

The number of linearly independent solutions of the differential equation $\frac{d^4y}{dx^4} - \frac{d^3y}{dx^3} - 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} - 2y = 0$ of the form $e^{ax}$ ($a$ is a real number) is}

• A. 1
• B. 2
• C. 3
• D. 4

Hint:

Be careful with the wording "of the form $e^{ax}$".
Even though $m = 1$ is a triple root and contributes three linearly independent terms to the general solution, only $e^x$ is of the form $e^{ax}$.
The other terms ($x e^x, x^2 e^x$) do not fit this form.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
This question asks for the number of linearly independent solutions of a fourth-order linear homogeneous differential equation that have the specific exponential form $e^{ax}$, where the parameter $a$ must be a real number.

Step 2: Key Formula or Approach:
For a linear differential equation with constant coefficients, substituting $y = e^{mx}$ leads to a characteristic polynomial equation.
Each distinct real root $m_i$ of this characteristic equation yields a linearly independent solution of the form $e^{m_i x}$.

Step 3: Detailed Explanation:

• Write the characteristic equation of the given differential equation:
\[ m^4 - m^3 - 3m^2 + 5m - 2 = 0 \]
• Let's find the roots of this polynomial. Test $m = 1$:
\[ (1)^4 - (1)^3 - 3(1)^2 + 5(1) - 2 = 1 - 1 - 3 + 5 - 2 = 0 \] Since $m=1$ is a root, $(m-1)$ is a factor.

• Perform division to factor the polynomial:
\[ m^4 - m^3 - 3m^2 + 5m - 2 = (m - 1)(m^3 - 3m + 2) \]
• Factor the cubic term $q(m) = m^3 - 3m + 2$. Test $m=1$:
\[ (1)^3 - 3(1) + 2 = 1 - 3 + 2 = 0 \] So $(m-1)$ is a factor of $q(m)$ as well:
\[ m^3 - 3m + 2 = (m - 1)(m^2 + m - 2) \]
• Factor the quadratic term:
\[ m^2 + m - 2 = (m + 2)(m - 1) \]
• Putting all the factors together gives the complete factored characteristic equation:
\[ (m - 1)^3 (m + 2) = 0 \]
• The roots are:
$m = 1$ (multiplicity 3)
$m = -2$ (multiplicity 1)

• The general solution to the differential equation is:
\[ y(x) = (c_1 + c_2 x + c_3 x^2) e^x + c_4 e^{-2x} \]
• The individual, distinct solutions of the form $e^{ax}$ with real $a$ are $e^x$ (for $a=1$) and $e^{-2x}$ (for $a=-2$).

• Other independent solutions like $x e^x$ and $x^2 e^x$ are not of the form $e^{ax}$ because of the algebraic multipliers.

• Therefore, there are exactly 2 linearly independent solutions of the form $e^{ax}$ with real $a$.

Step 4: Final Answer:
The number of linearly independent solutions of the specified form is 2, which corresponds to option (B).