Ordinary Differential Equations Questions and Answers

Answer: C

Differentiating twice returns the function itself:

\[ y' \;=\; Ae^{x} - Be^{-x}, \qquad y'' \;=\; Ae^{x} + Be^{-x} \;=\; y. \]

So \( y'' - y = 0 \). Two constants call for two differentiations and yield an equation of order two. The equation \( y'' + y = 0 \) belongs to \( \sin x \) and \( \cos x \) instead

Answer: B

Writing it as \( y\,dy = x\,dx \) puts each variable with its own differential, so

\[ \frac{y^{2}}{2} \;=\; \frac{x^{2}}{2} + C \quad \Longrightarrow \quad y^{2} - x^{2} \;=\; k. \]

The solution curves are rectangular hyperbolae. Separation is always worth trying first, since it needs no auxiliary construction at all

Answer: A

If the right side depends only on the ratio, the substitution \( y = vx \) makes the equation separable in \( v \) and \( x \). A function of \( x \) alone integrates directly; the product form is the separable case; and a function of \( y \) alone is separable too. Homogeneity is specifically about the ratio

Answer: D

Multiplying by \( \mu = e^{\int P dx} \) makes the left side an exact derivative:

\[ \frac{d}{dx}\left(\mu y\right) \;=\; \mu\left(y' + Py\right) \;=\; \mu Q, \]

so \( \mu y = \int \mu Q\,dx \). The negative exponent would undo the grouping rather than create it, and the factor must be built from \( P \), the coefficient of \( y \), not from the right-hand side

Answer: C

Here \( P = 1/x \), so

\[ \mu \;=\; e^{\int (1/x)\,dx} \;=\; e^{\ln x} \;=\; x. \]

The equation becomes \( (xy)' = x^{2} \), giving \( xy = x^{3}/3 + C \). The reciprocal \( 1/x \) would come from a coefficient of \( -1/x \), so the sign of \( P \) has to be read with care

Answer: B

Exactness means the expression is \( d\phi \) for some \( \phi \), with \( M = \phi_{x} \) and \( N = \phi_{y} \). Equality of the mixed second derivatives then requires

\[ \frac{\partial M}{\partial y} \;=\; \frac{\partial N}{\partial x}. \]

The cross pairing is what matters: comparing each with the derivative in its own variable tests nothing about a potential

Answer: A

Dividing through by \( y^{n} \) leaves \( y^{-n}y' + Py^{1-n} = Q \), and with \( v = y^{1-n} \) we get \( v' = (1-n)y^{-n}y' \), so

\[ \frac{1}{1-n}\frac{dv}{dx} + Pv \;=\; Q, \]

which is linear in \( v \). The substitution \( v = \ln y \) suits an equation with \( n = 1 \), which is separable anyway

Answer: D

The law gives \( d\theta/dt = -k(\theta - \theta_{s}) \), whose solution is an exponential decay towards the surrounding temperature. The fourth power belongs to radiative transfer by Stefan's law, which the cooling law approximates for a small excess, and proportionality to elapsed time would give a straight-line fall rather than an exponential one

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