Let's solve the given differential equation step-by-step: ### **Given:** \[ \frac{dy}{dx} = x^3 y^2 + 49x^3 \] with initial condition \( y(0) = 7 \). --- ### **Step 1: Factor the right side** \[ \frac{dy}{dx} = x^3 (y^2 + 49) \] --- ### **Step 2: Separate the variables** \[ \frac{1}{y^2 + 49} \frac{dy}{dx} = x^3 \] \[ \frac{1}{y^2 + 49} dy = x^3 dx \] --- ### **Step 3: Integrate both sides** \[ \int \frac{1}{y^2 + 49} dy = \int x^3 dx \] Recall: \( \int \frac{1}{a^2 + y^2} dy = \frac{1}{a} \arctan\left(\frac{y}{a}\right) \), with \(a = 7\): \[ \frac{1}{7} \arctan\left(\frac{y}{7}\right) = \frac{x^4}{4} + C \] --- ### **Step 4: Solve for \(C\) using initial condition** Given \( y(0) = 7 \): \[ \frac{1}{7} \arctan\left(\frac{7}{7}\right) = \frac{0^4}{4} + C \] \[ \frac{1}{7} \arctan(1) = C \] \[ \arctan(1) = \frac{\pi}{4} \] \[ C = \frac{1}{7} \cdot \frac{\pi}{4} = \frac{\pi}{28} \] --- ### **Step 5: Write the general solution** \[ \frac{1}{7} \arctan\left(\frac{y}{7}\right) = \frac{x^4}{4} + \frac{\pi}{28} \] Multiply both sides by 7: \[ \arctan\left(\frac{y}{7}\right) = \frac{7x^4}{4} + \frac{\pi}{4} \] Take tangent of both sides: \[ \frac{y}{7} = \tan\left(\frac{7x^4}{4} + \frac{\pi}{4}\right) \] So, \[ y(x) = 7 \tan\left(\frac{7x^4}{4} + \frac{\pi}{4}\right) \] --- ### **Final Answer:** \[ \boxed{ y(x) = 7 \tan\left(\frac{7x^4}{4} + \frac{\pi}{4}\right) } \]