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Pass your Engineering Mathematics (MATH 3321) final exam with confidence using this comprehensive question bank featuring 200 realistic problems and detailed step‑by‑step rationales – aligned with the University of Houston curriculum and updated for 2026. Master first‑order ODEs (separable, linear, exact, Bernoulli, homogeneous), applications (mixing problems, Newton’s law of cooling, radioactive decay, logistics, RC circuits), second‑order linear ODEs (characteristic equations, undetermined coefficients, variation of parameters, Euler‑Cauchy), Laplace transforms (initial value problems, step functions, Dirac delta, convolution), systems of linear ODEs (eigenvalues, phase portraits, fundamental matrices), Fourier series (sine/cosine series, convergence), PDEs (heat, wave, Laplace equations, separation of variables), and numerical methods (Euler, Runge‑Kutta, stability). Each answer explains the why – not just the correct solution.
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1. The general solution of the differential equation dydx=2x dxdy = 2 x is: A) y=x2+C y = x 2 + C B) y=2x2+C y = 2 x 2 + C C) y=x2 y = x 2 D) y=2x+C y = 2 x + C Answer: A Rationale: Integrate both sides: ∫dy=∫2xdx∫ dy =∫ 2 xdx → y=x2+C y = x 2 + C. This is the general solution.
Q2. Which of the following is a separable ODE? A) dydx=x+y dxdy = x + y B) dydx=xy dxdy = yx C) dydx=xy dxdy = xy D) Both B and C Answer: D Rationale: Separable equations can be written as g(y)dy=h(x)dx g ( y ) dy = h ( x ) dx. Option B: ydy=xdx ydy = xdx ; Option C: 1ydy=xdx y 1 dy = xdx. Option A is not separable. Q3. Solve the initial value problem: dydx=3y dxdy = 3 y , y(0)=2 y ( 0 )= 2. A) y=2e3x y = 2 e 3 x B) y=3e2x y = 3 e 2 x C) y=e3x+1 y = e 3 x + 1 D) y=2ex y = 2 ex
Q5. (Scenario) A tank initially contains 100 L of pure water. Brine with 0.2 kg/L salt flows in at 5 L/min, and the well-mixed solution flows out at 5 L/min. The salt amount S(t) S ( t ) satisfies: A) dSdt=1−0.05S dtdS = 1 −0.05 S B) dSdt=0.2−0.05S dtdS =0.2−0.05 S C) dSdt=1−0.2S dtdS = 1 −0.2 S D) dSdt=0.2+0.05S dtdS =0.2+0.05 S Answer: A Rationale: Rate in = (5 L/min)(0.2 kg/L) = 1 kg/min. Rate out = (S/100)*5 = 0.05S kg/min. So dSdt=1−0.05S dtdS = 1 −0.05 S. Q6. The equation dydx=2xy dxdy = y 2 x is: A) Linear B) Separable
C) Exact D) Homogeneous Answer: B Rationale: Rewrite as ydy=2xdx ydy = 2 xdx. Variables separate. Q7. Solve dydx=y2 dxdy = y 2 , y(0)=1 y ( 0 )= 1. A) y=11−x y = 1 − x 1 B) y=11+x y = 1 + x 1 C) y=1−x y = 1 − x D) y=ex y = ex Answer: A Rationale: Separate: y−2dy=dx y − 2 dy = dx. Integrate: −1/y=x+C−1/ y = x + C. Using y(0)=1 y ( 0 )= 1 → C=− 1 C =− 1. So −1/y=x− 1 −1/ y = x − 1 → y=1/(1−x) y =1/( 1 − x ).
C) v=ln y v =ln y D) v=ey v = ey Answer: A Rationale: The standard substitution v=y1−n v = y 1 − n reduces Bernoulli to a linear ODE in v v. Q10. The general solution of dydx=yx dxdy = xy is: A) y=Cx y = Cx B) y=C/x y = C / x C) y=Cex y = Cex D) y=ln∣ x∣+C y =ln∣ x ∣+ C Answer: A Rationale: Separate: dyy=dxx ydy = xdx. Integrate: ln∣ y∣=ln∣ x∣+Cln∣ y ∣=ln∣ x ∣+ C → y=Cx y = Cx.
Q11. The integrating factor for y′+3x2y=x2 y ′+ 3 x 2 y = x 2 is: A) ex3 ex 3 B) e3x2 e 3 x 2 C) ex2 ex 2 D) ex3/3 ex 3 / Answer: A Rationale: P(x)=3x2 P ( x )= 3 x 2. ∫Pdx=x3∫ Pdx = x 3 , so μ=ex3 μ = ex 3. Q12. Solve the homogeneous ODE y′=yx+xy y ′= xy + yx. (Hint: let v=y/x v = y / x ) A) y2=x2(ln x+C) y 2 = x 2 (ln x + C ) B) y2=x2(ln∣ x∣+C) y 2 = x 2 (ln∣ x ∣+ C ) C) y=xln x+Cx y = x ln x + Cx D) y=xln∣ x∣+Cx y = x ln∣ x ∣+ Cx
Answer: A Rationale: A(t)=A0e−kt A ( t )= A 0 e − kt. From 100→80 in 100y: 80=100e−100k 80 = 100 e − 100 k → e−100k=0.8 e − 100 k =0.8 → k=−ln (0.8)/100≈0.002231 k =−ln(0.8)/100≈0.
Half-life T=ln 2/k≈0.6931/0.002231≈310.7 T =ln2/ k ≈0. /0.002231≈310.7 years. Q14. The particular solution of dydx−2y=4 dxdy − 2 y = 4 that satisfies y(0)=1 y ( 0 )= 1 is: A) y=3e2x− 2 y = 3 e 2 x − 2 B) y=2e2x− 1 y = 2 e 2 x − 1 C) y=−2+3e−2x y =− 2 + 3 e − 2 x D) y=−2+3e2x y =− 2 + 3 e 2 x
Answer: D Rationale: Linear: integrating factor e−2x e − 2 x. Solve: ye−2x=∫4e−2xdx=−2e−2x+C ye − 2 x =∫ 4 e − 2 xdx =− 2 e − 2 x + C → y=−2+Ce2x y =− 2 + Ce 2 x. y(0)=1 y ( 0 )= 1 → 1=−2+C 1 = − 2 + C → C=3 C = 3. So y=−2+3e2x y =− 2 + 3 e 2 x. Q15. A solution of y′=y2− 4 y ′= y 2 − 4 is: A) y=2 y = 2 B) y=− 2 y =− 2 C) Both A and B D) y=0 y = 0 Answer: C Rationale: Equilibrium solutions when y2−4=0 y 2 − 4 = 0 → y=±2 y =±2. Both are constant solutions.
Answer: A Rationale: Separate: ydy=xdx ydy = xdx → y22=x22+C 2 y 2 = 2 x 2
exact. C: M=y,N=−x M = y , N =− x → My=1,Nx=− 1 My = 1 , Nx =− 1 → not exact. Q19. The general solution of dydx=2xy dxdy = 2 xy is: A) y=Cex2 y = Cex 2 B) y=Ce2x y = Ce 2 x C) y=Cex2/2 y = Cex 2 / D) y=Cex2 y = Cex 2 Answer: A Rationale: Separate: dyy=2xdx ydy = 2 xdx → ln∣ y∣=x2+Cln∣ y ∣= x 2 + C → y=Cex2 y = Cex 2. Q20. The ODE dydx=2x+yx−y dxdy = x − y 2 x + y is: A) Homogeneous B) Exact
Answer: A Rationale: Solution: T(t)=Tm+(T0−Tm)ekt T ( t )= Tm +( T 0 − Tm ) ekt. Here T0=100,Tm=20 T 0 = 100 , Tm = 20. So T(t)=20+80ekt T ( t )= 20 + 80 ekt. At t=10: 60=20+80e10k 60 = 20 + 80 e 10 k → 40=80e10k 40 = 80 e 10 k → e10k=0.5 e 10 k =0.5 → 10k=ln (0.5) 10 k =ln(0.5) → k= 110ln (0.5) k = 101 ln(0.5) (note k is negative). Q22. The integrating factor for y′+1xy=sin x y ′+ x 1 y =sin x is: A) x x B) ln xln x C) e1/x e 1/ x D) eln x=x e ln x = x Answer: A Rationale: μ=e∫1xdx=eln x=x μ = e ∫ x 1 dx = e ln x = x.
Q23. Solve dydx=1+y2 dxdy = 1 + y 2. A) y=tan (x+C) y =tan( x + C ) B) y=arctan (x+C) y =arctan( x + C ) C) y=tan (x)+C y =tan( x )+ C D) y=sec (x+C) y =sec( x + C ) Answer: A Rationale: Separate: dy1+y2=dx 1 + y 2 dy = dx → arctan y=x+Carctan y = x + C → y=tan (x+C) y =tan( x + C ). Q24. The solution of y′=y(1−y) y ′= y ( 1 − y ) with y(0)=0.5 y ( 0 )=0.5 is: A) Logistic function y=11+e−t y = 1 + e − t 1 B) y=11+e−t y = 1 + e − t 1 (same)
Answer: A Rationale: Exactness condition. Q26. Solve dydx=ex−y dxdy = ex − y. A) ey=ex+C ey = ex + C B) ey=ex+C ey = ex + C (same) C) y=ln (ex+C) y =ln( ex + C ) D) All of the above Answer: D Rationale: Rewrite: eydy=exdx eydy = exdx → ey=ex+C ey = ex + C. So y=ln (ex+C) y =ln( ex + C ). Q27. The population of a city grows at a rate proportional to the population. If it doubles in 40 years, the time to triple is: A) 40ln 3ln 240 ln 2 ln 3 years
B) 40ln 2ln 340 ln 3 ln 2 years C) 40ln 340 ln 3 years D) 40ln 1.5 40 ln1.5 years Answer: A Rationale: P=P0ekt P = P 0 ekt. Doubling: 2=e40k 2 = e 40 k → k=ln 240 k = 40 ln 2. Triple: 3=ekt 3 = ekt → t=ln 3k=40ln 3ln 2 t = k ln 3 = 40 ln 2 ln 3. Q28. The Bernoulli equation y′+1xy=y2 y ′+ x 1 y = y 2 can be reduced to linear by the substitution v=_____ v =_____. A) v=y− 1 v = y − 1 B) v=y2 v = y 2 C) v=y− 2 v = y − 2 D) v=ln y v =ln y
