Solve the following differential equation ^2 x d y d x +y= x , given that y(0)=1 . Hence find y ( 4 )

Step-by-step solution

The given differential equation is ^2 x dy dx + y = x . Dividing by ^2 x , we get dy dx + y ^2 x = x ^2 x . This is a linear differential equation of the form dy dx + Py = Q , where P = ^2 x and Q = x ^2 x . The integrating factor (IF) is e^ P dx = e^ ^2 x dx = e^ x . The general solution is y IF = Q IF dx + C . Substituting the values, y e^ x = x ^2 x e^ x dx + C . Let u = x , then du = ^2 x dx . The integral becomes u e^u du = u e^u - e^u + C . Thus, y e^ x = e^ x ( x - 1) + C , which simplifies to y = x - 1 + C