Ivp Cotangent Fa1Bc3

1. **State the problem:** We need to solve the initial value problem given by the differential equation $$\frac{dy}{dt} = 6t \sin^2(y)$$ with the initial condition $$y(-2) = \frac{7\pi}{4}$$. 2. **Separate variables:** Rewrite the equation to separate variables $y$ and $t$: $$\frac{dy}{\sin^2(y)} = 6t \, dt$$. 3. **Integrate both sides:** We integrate the left side with respect to $y$ and the right side with respect to $t$: $$\int \frac{dy}{\sin^2(y)} = \int 6t \, dt$$. Recall that $\frac{1}{\sin^2(y)} = \csc^2(y)$ and $$\int \csc^2(y) \, dy = -\cot(y) + C$$. So, $$-\cot(y) = 3t^2 + C$$. 4. **Apply initial condition:** Substitute $t = -2$ and $y = \frac{7\pi}{4}$: $$-\cot\left(\frac{7\pi}{4}\right) = 3(-2)^2 + C$$. Calculate $\cot\left(\frac{7\pi}{4}\right)$: Since $\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}$, and $$\cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$, so $$\cot\left(\frac{7\pi}{4}\right) = \frac{\frac{\sqrt{2}}{2}}{-\frac{\sqrt{2}}{2}} = -1$$. Therefore, $$-(-1) = 3 \times 4 + C \implies 1 = 12 + C \implies C = 1 - 12 = -11$$. 5. **Write the implicit solution:** $$-\cot(y) = 3t^2 - 11$$ or equivalently $$\cot(y) = -3t^2 + 11$$. 6. **Solve for $y$ if desired:** $$y = \cot^{-1}(-3t^2 + 11)$$. **Final answer:** $$y(t) = \cot^{-1}(-3t^2 + 11)$$ satisfies the initial value problem.