The particular solution of the differential equation x d y d x +y= x at y(0)=1

Step-by-step solution

The given differential equation is x dy dx + y = x . Dividing by x , we get dy dx + y x = x . This is a linear differential equation of the form dy dx + Py = Q , where P = x and Q = x . The integrating factor (IF) is e^ P dx = e^ x dx = e^ | x + x| = x + x . The general solution is y (IF) = Q (IF) dx + C . y( x + x) = x ( x + x) dx + C . y( x + x) = ( x x + ^2 x) dx + C . Using ^2 x = ^2 x - 1 , we have y( x + x) = ( x x + ^2 x - 1) dx + C . y( x + x) = x + x - x + C . Given y(0) = 1 , substituting x = 0 and y = 1