Let's solve the initial value problem step by step: Given: \[ y' = 3ty^2, \quad y(0) = y_0 \] ### **Step 1: Separate Variables** \[ \frac{dy}{dt} = 3ty^2 \] Separate variables: \[ \frac{1}{y^2} dy = 3t\, dt \] ### **Step 2: Integrate Both Sides** \[ \int \frac{1}{y^2} dy = \int 3t\, dt \] \[ -\frac{1}{y} = \frac{3}{2} t^2 + C \] ### **Step 3: Solve for \( y \)** \[ \frac{1}{y} = -\frac{3}{2} t^2 - C \] \[ y = \frac{1}{-\frac{3}{2} t^2 - C} \] ### **Step 4: Use Initial Condition \( y(0) = y_0 \)** Plug in \( t = 0 \): \[ y_0 = \frac{1}{-C} \implies -C = \frac{1}{y_0} \implies C = -\frac{1}{y_0} \] ### **Step 5: Substitute Back** \[ y = \frac{1}{-\frac{3}{2} t^2 + \frac{1}{y_0}} \] Or, combine over common denominator: \[ y = \frac{1}{\frac{1}{y_0} - \frac{3}{2} t^2} \] ### **Step 6: Interval of Existence** The solution exists as long as the denominator is nonzero: \[ \frac{1}{y_0} - \frac{3}{2} t^2 \neq 0 \] \[ \frac{1}{y_0} \neq \frac{3}{2} t^2 \implies t^2 \neq \frac{2}{3 y_0} \] - **If \( y_0 > 0 \):** - The solution blows up when \( t^2 = \frac{2}{3 y_0} \), so the interval is: \[ -\sqrt{\frac{2}{3 y_0}} < t < \sqrt{\frac{2}{3 y_0}} \] - **If \( y_0 < 0 \):** - \( \frac{2}{3 y_0} < 0 \), so denominator never vanishes for real \( t \). The solution exists for all \( t \). --- ### **Final Answers** **1. General Solution:** \[ \boxed{y = \frac{1}{\frac{1}{y_0} - \frac{3}{2} t^2}} \] **2. Interval of Existence:** - For \( y_0 > 0 \): Solutions exist as long as \( |t| < \sqrt{\frac{2}{3 y_0}} \). - For \( y_0 < 0 \): Solutions are defined for all \( t \). --- Let me know if you'd like a clean summary or if you want to see a graph!