Let y=y(x) be a continuous solution curve of the differential equation d y d x + 1 x+1 y = f(x) , where f(x) = cases x, & 0 x 1 0, & x > 1 cases If y(0) = 0 , then the value of y(4) is :

Step-by-step solution

The given differential equation is a first-order linear differential equation. The integrating factor (IF) for all x 0 is: IF = e^ 1 x+1 d x = e^ (x+1) = x+1 Case 1: For 0 x 1 , f(x) = x . The equation becomes d y d x + 1 x+1 y = x . Multiplying by the IF: d d x [y(x+1)] = x(x+1) = x^2 + x Integrating both sides: y(x+1) = x^3 3 + x^2 2 + C₁ Using the initial condition y(0) = 0 : 0 = 0 + 0 + C₁ C₁ = 0 So, for 0 x 1 , y(x+1) = x^3 3 + x^2 2 . At x = 1 , we have: y(1) 2 = 1 3 + 1 2 = 5 6 y(1) = 5 12 Case 2: For x > 1