4.2 Reduction of Order

4.1 Reduce a 2nd order equation to 1st

(Eq A) (Subscript[a, 2])[x]y'' + (Subscript[a, 1])[x] y' + (Subscript[a, 0])[x]y=0

(SF) y'' + P[x] y' + Q[x]y=0

(Sol) (Subscript[y, 1])is a solution

How: Substitute y[x]=u[x](Subscript[y, 1])[x] into the equation and simplify

Why: The product rule gives

({{y, =, u[x](Subscript[y, 1])[x],  ,  ,  ,  }, {y, =, u[x](Subscript[y, 1])'[x], +,     u'[x](Subscript[y, 1])[x],  ,  }, {y, =, u[x](Subscript[y, 1])''[x], +, 2 u'[x](Subscript[y, 1])'[x], +, u''[x](Subscript[y, 1])[x]}})

The terms with the u[x] all cancel because (Subscript[y, 1]) solves the equation

Examples

y''[x] + 2 y'[x]+y[x]==0,  (Subscript[y, 1])[x]=(E^(-x))

Eq=y''[x]+2 y'[x] + y[x]==0;
y1[x_]=(E^(-x));
Sub ={
y[x]->u[x]y1[x],
y'[x]->D[u[x] y1[x],x],
y''[x]->D[u[x] y1[x],{x,2}]
}
Simplify[
Eq/.Sub
];
DSolve[Eq,y[x],x];

y''[x] + 3 y'[x]+2y[x]==0,  (Subscript[y, 1])=(E^(-2x))

Eq=y''[x]+3 y'[x] +2 y[x]==0;
y1[x_]=(E^(-2x));
Sub ={
y[x]->u[x]y1[x],
y'[x]->D[u[x] y1[x],x],
y''[x]->D[u[x] y1[x],{x,2}]
}
Simplify[
Eq/.Sub
];
DSolve[Eq,y[x],x];

(x^2)y''[x] + 3x y'[x]+y[x]==0, (Subscript[y, 1])= 1/x

Eq=(x^2)y''[x]+3 x y'[x] +y[x]==0;
y1[x_]=1/x;
Sub ={
y[x]->u[x]y1[x],
y'[x]->D[u[x] y1[x],x],
y''[x]->D[u[x] y1[x],{x,2}]
}
Simplify[
Eq/.Sub
];
DSolve[Eq,y[x],x];