Ordinary Differential Equations Questions and Answers

Answer: C

The solution is \( N = N_{0}e^{-\lambda t} \). Setting \( N = N_{0}/2 \),

\[ e^{-\lambda t} \;=\; \tfrac{1}{2} \quad \Longrightarrow \quad \lambda t \;=\; \ln 2 \quad \Longrightarrow \quad t \;=\; \frac{\ln 2}{\lambda} \;\approx\; \frac{0.693}{\lambda}. \]

The time \( 1/\lambda \) is the mean lifetime, at which the quantity has fallen to \( 1/e \) rather than a half

Answer: B

Kirchhoff's law gives \( L\,di/dt + Ri = V \), a linear equation whose solution is

\[ i \;=\; \frac{V}{R}\left(1 - e^{-Rt/L}\right). \]

The exponent must be dimensionless, so the time constant is \( L/R \). Its reciprocal has units of inverse time and is the decay rate rather than the time constant

Answer: A

Substituting \( y = e^{mx} \) turns each derivative into a power of \( m \):

\[ m^{2} - 5m + 6 \;=\; 0 \quad \Longrightarrow \quad (m-2)(m-3) \;=\; 0. \]

So the roots are two and three, and the complementary function is \( C_{1}e^{2x} + C_{2}e^{3x} \). The negative pair would come from \( y'' + 5y' + 6y = 0 \)

Answer: D

Two solutions are needed, and the repeated root supplies only one exponential; writing it twice gives \( (C_{1}+C_{2})e^{mx} \), a single arbitrary constant in disguise. Multiplying the second by \( x \) produces a genuinely independent solution, so the general form is \( (C_{1} + C_{2}x)e^{mx} \)

Answer: C

The pair \( e^{(\alpha \pm i\beta)x} \) combines through Euler's formula into a real exponential envelope times an oscillation:

\[ y \;=\; e^{\alpha x}\left(C_{1}\cos\beta x + C_{2}\sin\beta x\right). \]

The real part sets the growth or decay and the imaginary part the frequency, so interchanging them would exchange the envelope for the oscillation

Answer: B

The auxiliary equation is

\[ m^{2} + 4 \;=\; 0 \quad \Longrightarrow \quad m \;=\; \pm 2i, \]

so \( \alpha = 0 \) and \( \beta = 2 \), giving an undamped oscillation of angular frequency two with no exponential envelope. The frequency is the square root of the coefficient, not the coefficient itself

Answer: A

Substituting \( y = Ae^{2x} \) replaces each derivative by a power of two:

\[ 4A e^{2x} - Ae^{2x} \;=\; e^{2x} \quad \Longrightarrow \quad 3A \;=\; 1 \quad \Longrightarrow \quad A \;=\; \tfrac{1}{3}. \]

Equivalently \( 1/(m^{2}-1) \) evaluated at \( m = 2 \) gives \( 1/3 \). A factor of \( x \) would be needed only if two were a root of the auxiliary equation

Answer: D

The auxiliary equation \( m^{2} - 1 = 0 \) has \( m = 1 \) as a root, so \( e^{x} \) already satisfies the homogeneous equation and \( Ae^{x} \) can never produce a non-zero right-hand side. Multiplying by \( x \) breaks the degeneracy; a factor \( x^{2} \) would be needed only for a *repeated* root

Answer: C

For a sine of angular frequency two, the operator \( d^{2}/dx^{2} \) replaces \( \sin 2x \) by \( -4\sin 2x \):

\[ \left(-4 + 1\right)A\sin 2x \;=\; \sin 2x \quad \Longrightarrow \quad A \;=\; -\tfrac{1}{3}. \]

The negative sign is essential and follows from \( -4 + 1 = -3 \). Resonance, which would bring in a factor of \( x \), happens only if the forcing frequency equals one

Answer: B

Each term then carries the same power of \( x \), since \( x^{2} \) times the second derivative of \( x^{m} \) is \( m(m-1)x^{m} \), and the equation collapses to an algebraic one in \( m \). The exponential trial belongs to the constant-coefficient case, where it is the derivatives rather than the coefficients that carry the variable