Problem 1
For y≠0, dy/y=t dt, so ln|y|=t²/2+C and y=Ce^{t²/2}; y=0 is included by C=0.
Faculty use
Reward correct mathematics, explicit units, conditions, and hypotheses, interpretation, and verification. Equivalent valid reasoning should receive credit.
For y≠0, dy/y=t dt, so ln|y|=t²/2+C and y=Ce^{t²/2}; y=0 is included by C=0.
(1+y²)dy=(1+t²)dt, so y+y³/3=t+t³/3+C.
Definitions should identify the equation type, dependent and independent variables, parameters, conditions, units or domain, and the hypotheses behind rearranging a first-order equation into one-variable differentials and tracking equilibrium solutions. Use the formulas dy/g(y)=f(t)dt; ∫dy/g(y)=∫f(t)dt+C.
A strong error analysis identifies the first step that violates classification, algebra, theorem hypotheses, units, initial/boundary data, or numerical assumptions; then repairs that step and re-verifies the final result.
Application responses vary. Require defined variables, a compatible differential-equation model connected to population, finance, radioactivity, pharmacokinetics, biological growth, and geometric families, a complete computation, interpretation, and limitations.
Accept two independent checks appropriate to the task: differentiation/substitution, initial or boundary conditions, dimensional analysis, phase-line or stability evidence, residuals, step refinement, conservation/energy checks, data comparison, or a second method.
Six problems × 10 points: 5 mathematical accuracy, 2 structure/justification, 2 verification/interpretation, 1 notation and communication.