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image_8e3f1d.png (Prompt)

(c) Use separation of variables to find $y = H(t)$, the particular solution to the differential equation

image_8e3f02.png (Student Submission)

(C) $\frac{dH}{dt} = \frac{1}{2}(H-1)\cos(\frac{t}{2})$

$\Rightarrow \frac{dH}{H-1} = \frac{1}{2}\cos(\frac{t}{2})dt$

$\int \frac{dH}{H-1} = \int \frac{1}{2}\cos(\frac{t}{2})dt$

$\Rightarrow \ln|H-1| = \sin(\frac{t}{2}) + C$

$\because H(0) = 4, H > 1$

$\Rightarrow H-1 = e^{\sin(\frac{t}{2}) + C}$

Scoring Overview
Estimated Grade3 / 5

Step-by-Step Breakdown

1 Separation of Variables

Evaluation: Correct. The student successfully separated the variables H and t.

2 Antiderivatives

Evaluation: Correct. Both antiderivatives are evaluated correctly.

3 Constant of Integration & IVP

Evaluation: Incorrect. Although the student included the constant of integration (+C) and correctly analyzed the absolute value, they failed to substitute the initial condition H(0) = 4 to solve for C.

4 Solve for H(t)

Evaluation: Incorrect. The student stopped prematurely and did not isolate H(t) to provide the final particular solution.

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