Equations in separable form MCQs With Answer

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Introduction: Equations in separable form are fundamental to B. Pharm students studying pharmacokinetics and drug dynamics. Separable differential equations allow variables to be separated and integrated, simplifying solutions for first‑order elimination, absorption models, and concentration‑time profiles. Mastery of separation, integration techniques (including partial fractions) and initial condition application helps solve one‑compartment models, half‑life calculations, and dosage problems. These MCQs focus on recognizing separability, solving for concentration functions, interpreting parameters like rate constants and half‑life, and applying results to pharmaceutical contexts. Targeted practice sharpens problem‑solving for exams and real‑world pharmacology. Now let’s test your knowledge with 50 MCQs on this topic.

Q1. Which differential equation is separable?

  • dy/dx = x + y
  • dy/dx = x*y
  • dy/dx = x^2 + y^2
  • dy/dx = x/(1 + x^2)

Correct Answer: dy/dx = x*y

Q2. What is the general solution of dy/dx = ky where k is constant?

  • y = kx + C
  • y = C*e^{kx}
  • y = (kx + C)^{1/k}
  • y = C*ln(kx)

Correct Answer: y = C*e^{kx}

Q3. For the separable equation dy/dx = g(x)h(y), the first step is:

  • Integrate g(x) and h(y) together
  • Separate as dy/h(y) = g(x) dx
  • Differentiating both sides with respect to y
  • Apply Laplace transform

Correct Answer: Separate as dy/h(y) = g(x) dx

Q4. Solve dC/dt = -0.2 C with C(0)=100. What is C(t)?

  • C(t) = 100 – 0.2t
  • C(t) = 100 e^{-0.2 t}
  • C(t) = 100/(1 + 0.2t)
  • C(t) = 100 e^{0.2 t}

Correct Answer: C(t) = 100 e^{-0.2 t}

Q5. Which step is required when integrating using partial fractions after separation?

  • Set constants equal to zero
  • Express rational function as sum of simpler fractions
  • Differentiate numerator and denominator separately
  • Multiply both sides by zero

Correct Answer: Express rational function as sum of simpler fractions

Q6. For dy/dx = x/(1+y^2), the separated integral becomes:

  • ∫ (1+y^2) dy = ∫ x dx
  • ∫ dy/(1+y^2) = ∫ x dx
  • ∫ y^2 dy = ∫ dx/x
  • ∫ dy = ∫ x(1+y^2) dx

Correct Answer: ∫ dy/(1+y^2) = ∫ x dx

Q7. The solution of dy/dx = x/(1+y^2) is:

  • arctan(y) = x^2/2 + C
  • y = tan(x^2/2 + C)
  • Both arctan(y) = x^2/2 + C and y = tan(x^2/2 + C)
  • y = arctan(x^2/2 + C)

Correct Answer: Both arctan(y) = x^2/2 + C and y = tan(x^2/2 + C)

Q8. A one‑compartment oral absorption model with first‑order absorption and elimination yields which separable DE for drug amount in central compartment A(t)?

  • dA/dt = Ka*D – Ke*A
  • dA/dt = -Ke*A
  • dA/dt = Ka*e^{-Ka t} – Ke*A
  • dA/dt = Ka*A – Ke*D

Correct Answer: dA/dt = Ka*e^{-Ka t} – Ke*A

Q9. Which equation is NOT separable?

  • dy/dx = (x^2)(y^3)
  • dy/dx = (x + y)/(xy)
  • dy/dx = (x – y)/(x + y)
  • dy/dx = (x^2 + 1)/(y^2 + 1)

Correct Answer: dy/dx = (x – y)/(x + y)

Q10. For the separable DE dy/dx = y^2 x, integrating gives which general solution?

  • y = e^{x^2/2 + C}
  • -1/y = x^2/2 + C
  • 1/y = x^2/2 + C
  • y = (x^2/2 + C)^{-1/2}

Correct Answer: -1/y = x^2/2 + C

Q11. The half‑life t1/2 for first‑order elimination (dC/dt = -kC) is:

  • t1/2 = ln(2)/k
  • t1/2 = k/ln(2)
  • t1/2 = 2/k
  • t1/2 = ln(k)/2

Correct Answer: t1/2 = ln(2)/k

Q12. Solve dN/dt = rN(1 – N/K). Is this separable?

  • No, it is not separable
  • Yes, separate as dN/[N(1-N/K)] = r dt
  • Yes, but only linear substitution works
  • Only if r=0

Correct Answer: Yes, separate as dN/[N(1-N/K)] = r dt

Q13. When integrating ∫dy/y, the antiderivative is:

  • 1/y + C
  • ln|y| + C
  • e^{y} + C
  • y ln y – y + C

Correct Answer: ln|y| + C

Q14. For dy/dx = (2x)/(1-y^2), separation yields which integral on the y side?

  • ∫ (1-y^2) dy
  • ∫ dy/(1-y^2)
  • ∫ y^2 dy
  • ∫ dy/(1+y^2)

Correct Answer: ∫ dy/(1-y^2)

Q15. Which substitution often simplifies separable equations involving y’ = f(ax + by)?

  • u = ax + by
  • u = x*y
  • u = y/x
  • u = x – y

Correct Answer: u = ax + by

Q16. For drug concentration governed by dC/dt = -kC + R (constant infusion), is the DE separable?

  • Yes, separable into dC/(R – kC) = dt
  • No, because of additive R term
  • Yes, if R=0 only
  • No, but linear first‑order methods apply

Correct Answer: No, but linear first‑order methods apply

Q17. The separable equation dy/dx = (x^2)/(y^3) leads to general solution:

  • y = (C – x^3/3)^{1/4}
  • y^4 = x^3/3 + C
  • y = e^{x^3/3 + C}
  • y^2 = x^3 + C

Correct Answer: y^4 = x^3/3 + C

Q18. When solving separable DEs, the integration constant should be determined using:

  • Boundary conditions or initial values
  • Dimensionless analysis only
  • Random choice
  • Assuming C=0 for simplicity

Correct Answer: Boundary conditions or initial values

Q19. Which expression is a singular solution possibility for separable DEs?

  • A solution obtained by setting h(y)=0 identically
  • Integrating both sides once
  • The general integral containing constant C
  • Solution only valid for negative x

Correct Answer: A solution obtained by setting h(y)=0 identically

Q20. For dy/dx = x*sqrt(y), separate variables and integrate. Which is the correct integrated relation?

  • 2 sqrt(y) = x^2/2 + C
  • 2 sqrt(y) = x^2 + C
  • ∫ y^{-1/2} dy = ∫ x dx ⇒ 2 sqrt(y) = x^2/2 + C
  • 2 y^{3/2} = x^2 + C

Correct Answer: 2 sqrt(y) = x^2 + C

Q21. Which pharmacokinetic equation is separable and yields exponential decay?

  • One-compartment IV bolus with first-order elimination
  • Zero-order infusion with constant rate
  • Michaelis-Menten saturable elimination in nonlinear range
  • Compartmental model with transfer terms both ways

Correct Answer: One-compartment IV bolus with first-order elimination

Q22. For the separable DE dy/dx = (x^2)/(1 – y), what integral is needed on the y side?

  • ∫ (1 – y) dy
  • ∫ dy/(1 – y)
  • ∫ y/(1 – y) dy
  • ∫ dy/(1 + y)

Correct Answer: ∫ dy/(1 – y)

Q23. In separable equations, absolute values appear with ln integration because:

  • ln function requires positive arguments
  • Integration constants must be absolute
  • The derivative of ln|y| is 1/y for nonzero y
  • To complicate algebra

Correct Answer: The derivative of ln|y| is 1/y for nonzero y

Q24. Solve dC/dt = -k C^2 for initial C(0)=C0. The solution is:

  • C(t) = C0 e^{-kt}
  • C(t) = 1/(kt + 1/C0)
  • C(t) = 1/(k t^2 + C0)
  • C(t) = (1/C0) e^{kt}

Correct Answer: C(t) = 1/(kt + 1/C0)

Q25. The separable DE dy/dx = f(y)g(x) fails if:

  • f(y)=0 for some y leading to constant solutions
  • g(x) is integrable
  • Both sides are functions only of x
  • You can always divide by f(y)

Correct Answer: f(y)=0 for some y leading to constant solutions

Q26. For dy/dx = (3x^2)/(2y), separating and integrating gives:

  • y^2 = x^3 + C
  • y = x^{3/2} + C
  • y^2/2 = x^3/3 + C
  • y = sqrt(3x^2/2 + C)

Correct Answer: y^2 = x^3 + C

Q27. When applying separation, you divide by h(y). What must you check?

  • That h(y) ≠ 0 for solution domain or consider constant solutions
  • That x is nonzero
  • That derivative is continuous only at x=0
  • That h(y) is linear

Correct Answer: That h(y) ≠ 0 for solution domain or consider constant solutions

Q28. Which differential equation models first‑order absorption followed by first‑order elimination and is handled by separation for the elimination part?

  • dA/dt = Ka*D(t) – Ke*A
  • dA/dt = Ka*e^{-Ka t} – Ke*A
  • dC/dt = -Ke*C (after known input function applied)
  • All of the above in appropriate contexts

Correct Answer: All of the above in appropriate contexts

Q29. The solution of dy/dx = (5x)/(y) with y(0)=2 results in:

  • y^2 = 5x^2 + 4
  • y^2 = (5/2)x^2 + 4
  • y = sqrt(5x + 2)
  • y = 2 e^{(5/2)x^2}

Correct Answer: y^2 = 5x^2 + 4

Q30. Which technique is commonly used to integrate ∫ dx/(ax + b)?

  • Partial fractions
  • u-substitution with u = ax + b
  • Integration by parts
  • Taylor expansion

Correct Answer: u-substitution with u = ax + b

Q31. True or False: The logistic equation is separable and useful in modeling saturable processes in pharmacology.

  • True
  • False
  • Only logistic with r=1 is separable
  • Only in linear regimes

Correct Answer: True

Q32. For separable DE dy/dx = (sin x)(cos y), separation gives which integrals?

  • ∫ dy/sec y = ∫ sin x dx
  • ∫ sec y dy = ∫ sin x dx
  • ∫ cos y dy = ∫ sin x dx
  • ∫ dy/cos y = ∫ sin x dx

Correct Answer: ∫ dy/cos y = ∫ sin x dx

Q33. A general separable solution often includes constant C. In pharmacokinetics, this constant typically represents:

  • Absorption rate
  • Initial concentration or initial condition
  • Half-life
  • Volume of distribution

Correct Answer: Initial concentration or initial condition

Q34. Which differential equation can be solved by separation followed by partial fractions?

  • dy/dx = (x)/(y^2 + y)
  • dy/dx = x + y
  • dy/dx = e^{x+y}
  • dy/dx = x^2 + y^2

Correct Answer: dy/dx = (x)/(y^2 + y)

Q35. For DE dC/dt = -k C/(1 + α C) (nonlinear elimination), is it separable?

  • Yes, separate as (1 + α C)/C dC = -k dt
  • No, because denominator contains C
  • Only if α=0
  • Only with integrating factor

Correct Answer: Yes, separate as (1 + α C)/C dC = -k dt

Q36. When separating variables, integrating the left side yields G(y)+C. What is the next step?

  • Differentiate both sides again
  • Solve algebraically for y if possible and apply initial condition
  • Ignore constant C
  • Set x=0

Correct Answer: Solve algebraically for y if possible and apply initial condition

Q37. For the separable DE dy/dx = (2x)/(y+1), integrating yields which implicit solution?

  • (y+1)^2 = x^2 + C
  • ln(y+1) = x^2 + C
  • y+1 = x^2 + C
  • (y+1) = sqrt(x^2 + C)

Correct Answer: (y+1)^2 = x^2 + C

Q38. Which is a correct method to check a separable solution?

  • Plug the solution into the original DE to verify
  • Differentiate randomly
  • Assume constant C=1 and check
  • Only check initial value, not DE

Correct Answer: Plug the solution into the original DE to verify

Q39. For dA/dt = -k1*A + k2*B where B(t) known, is the equation separable for A?

  • No, because of coupling, but if B(t) is known it becomes linear nonhomogeneous
  • Yes, always separable
  • Yes, separate as dA/(k1 A – k2 B) = dt
  • Only if k2=0

Correct Answer: No, because of coupling, but if B(t) is known it becomes linear nonhomogeneous

Q40. Solve dy/dx = (1)/(x y). Separating and integrating gives:

  • y^2 = 2 ln|x| + C
  • ln|y| = ln|x| + C
  • y = ln(x) + C
  • y = sqrt(ln|x| + C)

Correct Answer: y^2 = 2 ln|x| + C

Q41. In solving dy/dx = (x^2 + 1)/(y^2 + 1), which function appears after integrating both sides?

  • arctan(y) and arctan(x)
  • ln|y| and ln|x|
  • sin(y) and sin(x)
  • sqrt(y) and sqrt(x)

Correct Answer: arctan(y) and arctan(x)

Q42. For separable DE dy/dx = y^m x^n, the general method uses:

  • Separation to integrate y^{-m} dy = x^n dx
  • Laplace transforms
  • Fourier series
  • Matrix exponentials

Correct Answer: Separation to integrate y^{-m} dy = x^n dx

Q43. The drug elimination DE dC/dt = -kC gives area under curve (AUC) for bolus dose D and Vd as:

  • AUC = D/(k Vd)
  • AUC = D*Vd/k
  • AUC = D/(Vd)
  • AUC = k*D/Vd

Correct Answer: AUC = D/(k Vd)

Q44. For separable equation dy/dx = (cos x)/(1 + y^2), integrating yields:

  • arctan(y) = sin x + C
  • ln|y| = sin x + C
  • y = tan(sin x + C)
  • Both arctan(y) = sin x + C and y = tan(sin x + C)

Correct Answer: Both arctan(y) = sin x + C and y = tan(sin x + C)

Q45. When confronted with dy/dx = F(ax)/G(by), a useful strategy is:

  • Rescale variables to reduce parameters, u = ax, v = by
  • Ignore coefficients a and b
  • Use Laplace transforms
  • Assume solution is polynomial

Correct Answer: Rescale variables to reduce parameters, u = ax, v = by

Q46. For dy/dx = (y – 1)(x^2), the constant solution y=1 is:

  • A singular (equilibrium) solution
  • Impossible because derivative nonzero
  • Only valid at x=0
  • Not relevant

Correct Answer: A singular (equilibrium) solution

Q47. After separation, ∫ f(y) dy = ∫ g(x) dx + C. If f and g are continuous, solutions are:

  • Implicit relations that may be solved explicitly for y(x)
  • Only numerical approximations
  • Always linear
  • Always periodic

Correct Answer: Implicit relations that may be solved explicitly for y(x)

Q48. For dC/dt = Ka*D0 e^{-Ka t} – Ke*C, treating input as known yields which type of equation for C?

  • Separable equation for C directly
  • Linear nonhomogeneous first‑order ODE solvable by integrating factor
  • Nonseparable algebraic equation
  • Second‑order differential equation

Correct Answer: Linear nonhomogeneous first‑order ODE solvable by integrating factor

Q49. If separable integration yields ln|y| = x^3 + C, then y(x) equals:

  • y = e^{x^3} + C
  • y = C e^{x^3}
  • y = ln(C e^{x^3})
  • y = x^3 e^{C}

Correct Answer: y = C e^{x^3}

Q50. Best practice when solving separable DEs in pharmaceutical problems includes:

  • Keeping track of units, applying initial/boundary conditions, and checking solutions against physical constraints
  • Dropping constants to simplify algebra
  • Only solving numerically without attempting analytic form
  • Assuming linearity for all cases

Correct Answer: Keeping track of units, applying initial/boundary conditions, and checking solutions against physical constraints

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