\( \int _ { 0 } ^ { x } y ( t ) d t = x _ { + } ^ { 3 } \int _ { 0 } ^ { x } \sin ( x , t ) y ( t ) d t \)

To solve the equation \[ \int _ { 0 } ^ { x } y ( t ) d t = x _ { + } ^ { 3 } \int _ { 0 } ^ { x } \sin ( x , t ) y ( t ) d t, \] we differentiate both sides with respect to \(x\): \[ y(x) = 3x_{+}^2 \int _ { 0 } ^ { x } \sin(x, t) y(t) \, dt + x_{+}^3 \left( \sin(x, x) y(x) + \int _ { 0 } ^ { x } \frac{\partial}{\partial x} \sin(x, t) y(t) \, dt \right). \] This leads to the solution: \[ y(x) = \frac{3x_{+}^2 \int _ { 0 } ^ { x } \sin(x, t) y(t) \, dt + x_{+}^3 \int _ { 0 } ^ { x } \frac{\partial}{\partial x} \sin(x, t) y(t) \, dt}{1 - x_{+}^3 \sin(x, x)}. \] This equation defines \(y(x)\) in terms of its own integral, requiring further analysis or specific forms for \(\sin(x, t)\) and \(y(t)\) to find an explicit solution.