Differential equations from context; separation of variables practice questions.

A quantity satisfies dy/dx = ky. Determine the general solution of this differential equation.

• y = Ae^(kx)
• y = kx + C
• y = A e^(x/k)
• y = ½kx² + C

Answer: y = Ae^(kx)

Separating: (1/y)dy = k dx gives ln|y| = kx + c, so y = Ae^(kx) — exponential growth/decay.

Solve dy/dx = 2xy given that y = 3 when x = 0. Determine y as a function of x.

• y = 3e^(x²)
• y = 3e^(2x)
• y = x² + 3
• y = 3x² + 1

Answer: y = 3e^(x²)

(1/y)dy = 2x dx gives ln|y| = x² + c; y(0) = 3 gives y = 3e^(x²).

A tank drains so that dV/dt = −0.4V, where V is volume in litres and t in minutes. If V = 500 initially, determine the volume (to 3 s.f.) after 5 minutes.

• 67.7 L
• 406 L
• 270 L
• 100 L

Answer: 67.7 L

V = 500e^(−0.4t); at t = 5, V = 500e^(−2) = 500 × 0.1353 ≈ 67.7 L.

Solve the differential equation dy/dx = y/x for x > 0, given y = 4 when x = 2. Determine y in terms of x.

• y = 2x
• y = x + 2
• y = 4/x
• y = x²

Answer: y = 2x

(1/y)dy = (1/x)dx gives ln|y| = ln|x| + c, so y = Ax; y(2) = 4 gives A = 2, hence y = 2x.

A cup of water warms in a room at 25 °C according to dT/dt = −k(T − 25). Which statement about the constant k is correct?

• k > 0, so that T increases toward 25 °C
• k < 0, so the object heats without bound
• k = 0, so temperature is constant
• k depends on the sign of (T − 25)

Answer: k > 0, so that T increases toward 25 °C

For an object below ambient (T < 25), dT/dt must be positive; with (T − 25) negative this requires k > 0, giving T → 25 °C.

The amount of drug in the body satisfies dA/dt = −0.25A (t in hours). Determine the time (to 3 s.f.) for the amount to fall to 10% of its initial value.

• 9.21 h
• 2.77 h
• 4.00 h
• 0.400 h

Answer: 9.21 h

A = A₀e^(−0.25t); 0.1 = e^(−0.25t) gives t = ln(0.1)/(−0.25) = 2.3026/0.25 ≈ 9.21 h.