Question

find the solution of the differential equation that satisfies the given initial condition.

\\\frac{du}{dt} = \frac{2t + \sec^2(t)}{2u}, u(0) = -5\\

\\(u =\\)

Explanation:

⚡ Using what you learned: separable differential equations

Step 1: Separate variables

$$ 2u \, du = (2t + \sec^2(t)) \, dt $$

Step 2: Integrate both sides

$$ \int 2u \, du = \int (2t + \sec^2(t)) \, dt $$

$$ u^2 = t^2 + \tan(t) + C $$

Step 3: Apply the initial condition

Using \( u(0) = -5 \):

$$ (-5)^2 = 0^2 + \tan(0) + C $$

$$ 25 = C $$

So, the equation is:

$$ u^2 = t^2 + \tan(t) + 25 $$

Step 4: Solve for \( u \)

$$ u = \pm\sqrt{t^2 + \tan(t) + 25} $$

Since \( u(0) = -5 \) is negative, we choose the negative square root:

$$ u = -\sqrt{t^2 + \tan(t) + 25} $$

Answer:

\( -\sqrt{t^2 + \tan(t) + 25} \)