The first problem is to find the general solution of the differential equation y "+ y = 3x ^ 2. The second problem is to find the general solution of the differential equation y '- (Y-X) ^ 2 = 1 It's the Advanced Mathematics I learned in the University, but I forgot,

The first problem is to find the general solution of the differential equation y "+ y = 3x ^ 2. The second problem is to find the general solution of the differential equation y '- (Y-X) ^ 2 = 1 It's the Advanced Mathematics I learned in the University, but I forgot,

1. If the characteristic equation of the original equation is R & sup2; + 1 = 0, then the characteristic root is r = ± I
The general solution of the homogeneous equation of the original equation is y = c1cosx + c2sinx (C1, C2 are integral constants)
Let the special solution of the original equation be y = ax & sup2; + BX + C
∵y'=2Ax+B,y''=2A
Substituting into the original equation, 2A + ax & sup2; + BX + C = 3x & sup2;
==>A = 3, B = 0, 2A + C = 0 (compare coefficients of the same power)
==>A=3,B=0,C=-6
The special solution of the original equation is y = 3x & sup2; - 6
So the general solution of the original equation is y = c1cosx + c2sinx + 3x & sup2; - 6 (C1, C2 are integral constants)
2. Let u = Y-X, then y '= u' + 1
Substituting into the original equation, u '+ 1-u & sup2; = 1
==>u'-u²=0
==>du/u²=dx
==>1 / u = - x + C (C is the integral constant)
==>u=1/(C-x)
==>y-x=1/(C-x)
So the general solution of the original equation is y = x + 1 / (C-X) (C is an integral constant)