DIFFERENTIAL EQUATIONS UPDATED
ACTUAL QUESTIONS AND CORRECT
ANSWERS COMPLETE STUDY GUIDE FULL
SOLUTION
●● For dy/dt = 3y^(2/3) cos(2t), find two solutions through (π/4, 0).
Answer: METHOD: Existence & Uniqueness analysis. Solution 1: y = 0
(equilibrium). Solution 2: separate and solve. 3y^(1/3) = (3/2)sin(2t) +
C; with y(π/4)=0: C = -3/2. So y = ((1/2)sin(2t) - 1/2)^3.
●● Why is the solution to dy/dt = 3y^(2/3) cos(2t) through (π/4, 0) NOT
unique?
Answer: METHOD: Check Uniqueness Theorem conditions. ∂f/∂y =
2y^(-1/3) cos(2t) is NOT continuous at y=0. Since the partial derivative
fails continuity at the point, uniqueness is not guaranteed.
●● Why IS the solution to dy/dt = 3y^(2/3) cos(2t) through (π/4, 1)
unique?
Answer: METHOD: Verify Uniqueness Theorem conditions. f(t,y) =
3y^(2/3)cos(2t) is continuous everywhere; ∂f/∂y = 2y^(-1/3)cos(2t) is
continuous at y=1. Both conditions hold, so uniqueness is guaranteed.
●● Solve dy/dx = y/x with y(1) = 2.
,Answer: METHOD: Separation of Variables. dy/y = dx/x → ln|y| = ln|x|
+ C → y = Kx. Apply y(1)=2: K=2. Solution: y = 2x.
●● For dy/dx = y/x, determine all solutions satisfying y(0) = 0.
Answer: METHOD: Existence & Uniqueness analysis. Infinitely many
solutions exist: y = Kx for any K ∈ ℝ all pass through (0,0). The IVP is
not uniquely determined because f(x,y) = y/x is undefined at x=0.
●● For dy/dx = y/x, determine all solutions satisfying y(0) = 4.
Answer: METHOD: Existence & Uniqueness analysis. NO solutions
exist. f(x,y) = y/x is undefined at x=0, so no solution can pass through
(0, 4).
●● Find general solution of dy/dt = t/y.
Answer: METHOD: Separation of Variables. y dy = t dt → y²/2 = t²/2 +
C → y² - t² = K (hyperbolas). Or y = ±√(t² + K).
●● Solve dy/dt = -2y using homogeneous linear method.
Answer: METHOD: Separation / Homogeneous Linear ODE. y = Ce^(-
2t).
●● Solve dy/dt = t²y (homogeneous linear).
Answer: METHOD: Homogeneous Linear ODE, y_h = Ce^(∫a(t)dt). y =
Ce^(t³/3).
, ●● Find general solution to dy/dt = a(t)y.
Answer: METHOD: Homogeneous Linear ODE formula. y_h =
Ce^(∫a(t)dt).
●● Solve dy/dt = 5y + 3e^t using Method of Undetermined Coefficients.
Answer: METHOD: Undetermined Coefficients. Guess y_p = αe^t. Sub
in: αe^t - 5αe^t = 3e^t → -4α = 3 → α = -3/4. General: y = Ce^(5t) -
(3/4)e^t.
●● Solve dy/dt + 4y = e^(-t) using Undetermined Coefficients.
Answer: METHOD: Undetermined Coefficients. y_h = Ce^(-4t). Guess
y_p = αe^(-t): -αe^(-t) + 4αe^(-t) = e^(-t) → α = 1/3. General: y = Ce^(-
4t) + (1/3)e^(-t).
●● Solve dy/dt + 3y = sin(2t). Why is y_p = α sin(2t) NOT viable?
Answer: METHOD: Modified Undetermined Coefficients. The
derivative introduces cos(2t), so we need BOTH sine and cosine. Guess
y_p = α sin(2t) + β cos(2t). Solving: α = 3/13, β = -2/13. General: y =
Ce^(-3t) + (3/13)sin(2t) - (2/13)cos(2t).
●● Solve dy/dt + 2y = te^t. Why is y_p = Ate^t not ideal?
Answer: METHOD: Modified Undetermined Coefficients for
polynomial × exponential. Ate^t alone won't work because
differentiation produces both te^t and e^t terms. Guess y_p = (At +
ACTUAL QUESTIONS AND CORRECT
ANSWERS COMPLETE STUDY GUIDE FULL
SOLUTION
●● For dy/dt = 3y^(2/3) cos(2t), find two solutions through (π/4, 0).
Answer: METHOD: Existence & Uniqueness analysis. Solution 1: y = 0
(equilibrium). Solution 2: separate and solve. 3y^(1/3) = (3/2)sin(2t) +
C; with y(π/4)=0: C = -3/2. So y = ((1/2)sin(2t) - 1/2)^3.
●● Why is the solution to dy/dt = 3y^(2/3) cos(2t) through (π/4, 0) NOT
unique?
Answer: METHOD: Check Uniqueness Theorem conditions. ∂f/∂y =
2y^(-1/3) cos(2t) is NOT continuous at y=0. Since the partial derivative
fails continuity at the point, uniqueness is not guaranteed.
●● Why IS the solution to dy/dt = 3y^(2/3) cos(2t) through (π/4, 1)
unique?
Answer: METHOD: Verify Uniqueness Theorem conditions. f(t,y) =
3y^(2/3)cos(2t) is continuous everywhere; ∂f/∂y = 2y^(-1/3)cos(2t) is
continuous at y=1. Both conditions hold, so uniqueness is guaranteed.
●● Solve dy/dx = y/x with y(1) = 2.
,Answer: METHOD: Separation of Variables. dy/y = dx/x → ln|y| = ln|x|
+ C → y = Kx. Apply y(1)=2: K=2. Solution: y = 2x.
●● For dy/dx = y/x, determine all solutions satisfying y(0) = 0.
Answer: METHOD: Existence & Uniqueness analysis. Infinitely many
solutions exist: y = Kx for any K ∈ ℝ all pass through (0,0). The IVP is
not uniquely determined because f(x,y) = y/x is undefined at x=0.
●● For dy/dx = y/x, determine all solutions satisfying y(0) = 4.
Answer: METHOD: Existence & Uniqueness analysis. NO solutions
exist. f(x,y) = y/x is undefined at x=0, so no solution can pass through
(0, 4).
●● Find general solution of dy/dt = t/y.
Answer: METHOD: Separation of Variables. y dy = t dt → y²/2 = t²/2 +
C → y² - t² = K (hyperbolas). Or y = ±√(t² + K).
●● Solve dy/dt = -2y using homogeneous linear method.
Answer: METHOD: Separation / Homogeneous Linear ODE. y = Ce^(-
2t).
●● Solve dy/dt = t²y (homogeneous linear).
Answer: METHOD: Homogeneous Linear ODE, y_h = Ce^(∫a(t)dt). y =
Ce^(t³/3).
, ●● Find general solution to dy/dt = a(t)y.
Answer: METHOD: Homogeneous Linear ODE formula. y_h =
Ce^(∫a(t)dt).
●● Solve dy/dt = 5y + 3e^t using Method of Undetermined Coefficients.
Answer: METHOD: Undetermined Coefficients. Guess y_p = αe^t. Sub
in: αe^t - 5αe^t = 3e^t → -4α = 3 → α = -3/4. General: y = Ce^(5t) -
(3/4)e^t.
●● Solve dy/dt + 4y = e^(-t) using Undetermined Coefficients.
Answer: METHOD: Undetermined Coefficients. y_h = Ce^(-4t). Guess
y_p = αe^(-t): -αe^(-t) + 4αe^(-t) = e^(-t) → α = 1/3. General: y = Ce^(-
4t) + (1/3)e^(-t).
●● Solve dy/dt + 3y = sin(2t). Why is y_p = α sin(2t) NOT viable?
Answer: METHOD: Modified Undetermined Coefficients. The
derivative introduces cos(2t), so we need BOTH sine and cosine. Guess
y_p = α sin(2t) + β cos(2t). Solving: α = 3/13, β = -2/13. General: y =
Ce^(-3t) + (3/13)sin(2t) - (2/13)cos(2t).
●● Solve dy/dt + 2y = te^t. Why is y_p = Ate^t not ideal?
Answer: METHOD: Modified Undetermined Coefficients for
polynomial × exponential. Ate^t alone won't work because
differentiation produces both te^t and e^t terms. Guess y_p = (At +
