The order of a differential equation is the order of
Answer: A
Order counts derivatives; degree counts powers. So \( (y'')^{3} + y = 0 \) is of order two and degree three, while \( y''' + (y')^{5} = 0 \) is of order three and degree one, the degree being read from the *highest* derivative present. Confusing the two is the commonest slip in classifying an equation
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The equation \( \left(\dfrac{d^{2}y}{dx^{2}}\right)^{3} + \dfrac{dy}{dx} = 0 \) has order and degree
Answer: D
The highest derivative present is the second, so the order is two. That derivative appears cubed, so the degree is three. The first derivative's own power is irrelevant to the degree, which is read only from the highest-order term after the equation has been cleared of radicals and fractions in the derivatives
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Eliminating the arbitrary constants from \( y = Ae^{x} + Be^{-x} \) gives
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
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The equation \( \dfrac{dy}{dx} = \dfrac{x}{y} \) is solved by
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
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A first-order equation is homogeneous when it can be written as \( dy/dx = \)
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
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For the linear equation \( \dfrac{dy}{dx} + Py = Q \) with \( P \) a function of \( x \), the integrating factor is
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
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The integrating factor for \( \dfrac{dy}{dx} + \dfrac{y}{x} = x \) is
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
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The condition for \( M\,dx + N\,dy = 0 \) to be exact is
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
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Bernoulli's equation \( \dfrac{dy}{dx} + Py = Qy^{n} \) is made linear by the substitution
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
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Newton's law of cooling leads to a differential equation stating that the rate of temperature fall is proportional to
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
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