Ordinary Differential Equations Questions and Answers

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

Answer: A

The operator method has recipes for an exponential, a sine or cosine, a polynomial and their products. For something like \( \sec x \) or \( \ln x \) there is no such recipe, and variation of parameters builds the particular integral from the complementary function by integration instead. Repeated roots and higher order are handled by either method

Answer: D

A product of a function of each variable separates directly. The linear form takes an integrating factor built from the coefficient of \( y \). Bernoulli's form is neither, but the substitution \( v = y^{1-n} \) turns it into the linear case, after which the integrating factor applies