Suppose that

y_(1)(t)

is solution of

L(y_(1))=0

and

y_(2)(t)

is solution of

L(y_(2))=b(t)!=0

, where\

L(y)=3y^('')+2y^(')+7y.

\ Choose all correct statements below.\ The function

-y_(1)+y_(2)

is solution of the nonhomogeneous equation

L(y)=b(t)

.\ The function

3y_(2)

is a solution of the nonhomogeneous equintion

L(y)=-3b(t)

\ The function

-y_(1)-y_(2)

is solution of the nonhomogeneous equation

L(y)=-b(t)

.\ The function

3y_(2)

is solution of the nonhomogeneous equation

L(y)=b(t)

.\ The function

2y_(1)-2y_(2)

is solution of the homogeneous equation

L(y)=0

.\ The function

y_(1)+y_(2)

is solution of the homogeneous equation

L(y)=0

.\ The function

2y_(1)+2y_(2)

is a solution of the nonhomogeneous equation

L(y)=2b(t)

.\ The function

3y_(1)

is solution of the homogeneous equation

L(y)=0

.

Suppose that y1(t) is solution of L(y1)=0 and y2(t) is solution of L(y2)=b(t)=0, where L(y)=3y+2y+7y. Choose all correct statements below. The function y1+y2 is solution of the nonhomogeneous equation L(y)=b(t). The function 3y2 is a solution of the nonhomogeneous equition L(y)=3b(t). The function y1y2 is solution of the nonhomogeneous equation L(y)=b(t). The function 3y2 is solution of the nonhomogeneous equation L(y)=b(t). The function 2y12y2 is solution of the homogeneous equation L(y)=0. The function y1+y2 is solution of the homogeneous equation L(y)=0. The function 2y1+2y2 is a solution of the nonhomogeneous equation L(y)=2b(t). The function 3y1 is solution of the homogeneous equation L(y)=0