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dy  xdx  0, y  x2 C. 2 x  C. 2 2 x  C. 2 x  C. 3 x  C. dy  xdx  0, y(2)  0  y  x2  2. 2 x  2. 2 2 x  2. 2 x2 2. 2 2 x  2....

dy  xdx  0, y  x2 C. 2 x  C. 2 2 x  C. 2 x  C. 3 x  C. dy  xdx  0, y(2)  0  y  x2  2. 2 x  2. 2 2 x  2. 2 x2 2. 2 2 x  2. 2 dy  2 xdx  0, y(2)  4  y  2 x. x  2. 2 2 x  4. 2 x2 2. 2 2 x  2. 2 dy  6 x dx  0  y  2 2 x  C. 3 x  C. 2 2 x  C. 2 x  C. 3 x  C.  dy  6 x 2 dx  0, y0  0  y  3 2x. 2 x. 2 2x. 3 x. x. xdy  2 ydx  0,  y  2 Cx. x2 C. 2 2 x  C. 2 3 Cx. Cx. xdy  2 ydx  0, y(1)  e  y  2 ex. x2 1  e. 2 2 2 x  2  e. 2 3 ex. ex. xdy  2 ydx  0, y(1)  1  y  2 x. x2 1  2 2. 2 x  1. 2 3 x. x. x  2dy  4x ydx  y  4 3 C x  2. 4  x 4 2  C. 2 x  C. 4 3 Cx. Cx. x 4   2 dy  4 x3 ydx, y(0)  2  y  x  2. 4 x 4 2 1.  2 x  3. 4 Cx  2. 3 Cx  2. dy    dx  0, y  0y  cos x 2 sin x  1. cos x. 2  2. sin x ctgx.  x  2. dy  dx  0  y  sin x cos x  C. sin x  C. tgx  C. ctgx  C. 1 C. cos x  dy    dx  0, y   0  y  sin x 3 cos x  0,5. sin x  3 /2. tgx  3. ctgx  1 / 3. 1 2. cos x sin 2 xdy  dx  0  y  ctgx  C. cos x  C. sin x  C. tgx  C. 1 C. cos x   sin 2 xdy  dx  0, y   1 y  4 ctgx. cos x  1  2 / 2. sin x  1  2 /2. tgx. 1  2. cos x cos 2 xdy  dx  0  y  tgx  C. ctgx  C. cos x  C. sin x  C.  1 C. cos x   cos2 xdy  dx  0, y  0y  3 tgx  3. ctgx  1 / 3. cos x  0,5. sin x  3 /2. 1 2. cos x 1  x 2 dy  dx  0  y  arcsin x  C. arccos x  C. arctgx  C. arcctgx  C. cosec x  C. 1  x 2 dy  dx  0  y  arccos x  C. arcsin x  C. arctgx  C. arcctgx  C. cosec x  C. 1  x dy  dx  0  y  2 arctgx  C. arcsin x  C. arccos x  C. arcctgx  C. cosec x  C.  1  x dy  dx  0  y  2 arcctgx  C. arctgx  C. arcsin x  C. arccos x  C. cosec x  C. 2 4 y  y  x 4 , y (1)  , y  x 3 x5 y   x2 3 x2  x sin y  cos y  1 2 y  x / 2  5 / 6 2 y   x / 2  x  e 2 x cosec x. 2 2 y  y  x 4 , y (1)  , y  x 3 x5  x 2. 3 x2  x sin y  cos y  1. 2 x / 2  5 / 6. 2  x / 2  x  e. 2 x cosec x. 2 y  y  x2 , y  x x  Cx. 3 2 x5  x 2 C 3  x2  x sin y  cos y  C 2 x / 2  C 2  x / 2  x  e  C 2 x 2 y  y  x 2 , y (1)  2, y  x x  x. 3 2 x2  x sin y  cos y  1. 2 x / 2  3 / 2. 2  x / 2  x  3,5. 2 cosec x. ,y  ,y  ,y   ,y  ,y  ,y    If f  x, y   x  xy then f kx , ky   2  k x  xy. 2 2  k x  xy . 1 2 k x  xy . 0 2 k x  xy . 3 2 k x  xy . 4 2 x 2  xy If f  x, y   then f kx , ky   y  k 1 x 2  xy . y  x 2  xy  k  . 2  y  k 0 x 2  xy . y  k 3 x 2  xy . y  k 4 x 2  xy . y x 2  xy If f  x, y   then f kx , ky   y2  k 0 x 2  xy . y2  k 1 x 2  xy . y2  x 2  xy  k  y 2 . 2     k 3 x 2  xy . y2 there is no right answer. If f  x, y   x 2  y 2 then f kx, ky   k 1 x2  y2. k 2 x2  y2. k 3 x2  y2. k 4 x2  y2. k x2  y2. 5 If f x, y   x y  xy then f kx , ky   2 2 3  k x y  xy. 2  2 k x y  xy . 2 2 2 k x y  xy . 1 2 2 k x y  xy . 4 2 2 k x y  xy . 0 2 2 If f  x, y  is a homogeneous function of degree m then f tx, ty   t f x, y . m f tx , ty   mt  f  x, y . f tx, ty   tf  x, y . m f tx, ty   t f x , y m  m m . there is no right answer. the degree of the homogeneous function f x, y   x y  xy 2 2 ? 3. 2. 1. 0. there is no right answer.  x2  y2 the degree of the homogeneous function f  x, y   ? xy 0. 3. 2. 1. there is no right answer. x2 the degree of the homogeneous function f  x, y   tg 2 ? y 0. 3. 2. 1. there is no right answer. the degree of the homogeneous function f x, y   y  xy ? 2 2. 0. 3. 1. there is no right answer. the degree of the homogeneous function f  x, y   ln x  ln y ? 0. 2. 3. 1. there is no right answer. if DE y   f  x, y  is homogeneous then its degree must be? 0. 2. 3. 1. there is no right answer. The substitution for solving homogeneous DE y=tx. y=t/x. y=x/t. y=t. there is no right answer. If y=tx then у=?  tx+x. t+x. tx+1. 1+t. there is no right answer. If y=tx then dу=? tdx+xdt. dt+dx. tdx+1. xdt+1. there is no right answer. Choose the homogeneous DE у=sin(y/x). у=sin(x). у=sin(y). у=sin(yx). there is no right answer. Choose the homogeneous DE y2  2x2 y   xy y2  2x y  xy y  2x2 y  xy y2  2x2 y  x there is no right answer. Choose the homogeneous DE y xy  y  xtg x y y   y  xtg x y xy   1  xtg x  y y  y  xtg x there is no right answer. Choose the homogeneous DE   x  y dx  xydy  0 2 2 x  y dx  xydy  0 2 x  y dx  xydy  0 2 x  y dx  ydy  0 2 2 there is no right answer. Choose the homogeneous DE   x  2 xy dx  xydy  0 2 x  2 xy dx  ydy  0 2 x  2 y dx  xydy  0 2 x  2 xy dx  dy  0 2 there is no right answer. Integrating factor for the next DE:   1/ y x   1 / y 2   e x there is no right answer. Integrating factor for the next DE: 3x  6xy  3 y dx  2x 2 2   3xy dy  0 x   1 / y   1 / y 2   e x there is no right answer.    Integrating factor for the next DE: xy  y dx  1  xy dy  0 2 3 2    1/ y2   1 / y  x   e x there is no right answer.   Integrating factor for the next DE: y  2 x  2 dx  2 ydy  0 2   ex   1 / y   x   1 / y 2 there is no right answer. Integrating factor for the next DE: y dx  xy  1dy  0 2   1/ y   e x   x   1 / y 2 there is no right answer. Choose DE with separable variables..... Choose DE with separable variables..... Choose homogenous first order DE..... Choose homogenous first order DE..... Which of the next is liner first order DE..... Which of the next is liner first order DE..... Which of the next is Bernoulli DE..... Which of the next is Bernoulli DE..... Which of the next is Riccati DE..... Which of the next is Riccati DE..... Which of the next is Exact DE..... Which of the next is Exact DE..... Which of the next is Clairaut equation... Which of the next is Clairaut equation... Which of the next is Lagrange's equation... Which of the next is Lagrange's equation... Which of the next is the second order differential equations... Which of the next is the third order differential equations.. Which of the next is the first order differential equations y  81e 3 x The order of the next differential equations: 2 1 3 4 0 The order of the next differential equations: 3 2  1 4 0 Which of the next DE is allowed to lower its order. Which of the next DE is allowed to lower its order. Which of the next is linear homogeneous equations y  10 y  25 y y  81e 3 x y   10 y  25 y  x y   10 y   25 y  x  0 2 y  2 x  1 Which of the next is linear homogeneous equations y  9 y  0 y  81e3 x y   10 y  25 y  x y   10 y   25 y  x  0 2 y  2 x  1  Which of the next is linear homogeneous equations y  2 y  y  2 y  0 y   2 y   y   2 y  e x y   10 y  25 y  x y   10 y   25 y  x  0 2 y  2 x  1 Which of the next is linear homogeneous equations y IV  16 y  0 y   2 y   y   2 y  e x y   10 y  25 y  x y   10 y   25 y  x  0 2 y  2 x  1 Which of the next is linear inhomogeneous equations y   2 y   y   2 y  e x y IV  16 y  0 y  2 y  y  2 y  0 y  9 y  0 y  10 y  25 y Which of the next is linear inhomogeneous equations y   10 y  25 y  x y IV  16 y y  2 y  y  2 y  0 y  9 y  0 y  10 y  25 y Which of the next is linear inhomogeneous equations y IV  3 y   9 x 2 2 y  y  y y  2 y  y  2 y  0 y  9 y  0 y  10 y  25 y  Which of the next is linear inhomogeneous equations y   7 y   6 y  x  2e x 2 y  y  y y  2 y  y  2 y  0 y  9 y  0 y  10 y  25 y Define the type of the next DE: DE with separable variables homogenous first order DE liner first order DE Bernoulli DE Riccati DE Define the type of the next DE: DE with separable variables homogenous first order DE liner first order DE Bernoulli DE Riccati DE Define the type of the next DE: homogenous first order DE DE with separable variables liner first order DE Bernoulli DE Riccati DE Define the type of the next DE: homogenous first order DE DE with separable variables liner first order DE Bernoulli DE Riccati DE Define the type of the next DE: y  2 y  y  2 y  0 third order linear homogeneous DE  DE with separable variables liner first order DE Bernoulli DE Riccati DE Define the type of the next DE: liner first order DE DE with separable variables homogenous first order DE Bernoulli DE Riccati DE Define the type of the next DE: liner first order DE DE with separable variables homogenous first order DE Bernoulli DE Riccati DE Define the type of the next DE: Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Riccati DE Define the type of the next DE: Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Riccati DE Define the type of the next DE: Riccati DE Bernoulli DE  liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: Riccati DE Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: Exact DE Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: Exact DE Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: Clairaut equation Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: Clairaut equation Bernoulli DE liner first order DE DE with separable variables homogenous first order DE  Define the type of the next DE: Lagrange's equation Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: Lagrange's equation Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: second DE Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: third order DE Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: second order DE Bernoulli DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: third order DE Bernoulli DE liner first order DE DE with separable variables homogenous first order DE  Define the type of the next DE: y   2 y   y   2 y  e x third order linear inhomogeneous DE third order linear homogeneous DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: 2 y   y   2 y  0 second order linear homogeneous DE second order linear inhomogeneous DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: y   10 y  25 y  x second order linear inhomogeneous DE second order linear homogeneous DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: y   81e 3x second order DE second order linear homogeneous DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: y  16 y  0 IV fourth order linear homogeneous DE fourth order linear inhomogeneous DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: y   10 y   25 y  x  0 2 second order linear inhomogeneous DE second order linear homogeneous DE liner first order DE DE with separable variables homogenous first order DE  Define the type of the next DE: y  3 y   9 x IV 2 fourth order linear inhomogeneous DE fourth order linear homogeneous DE liner first order DE DE with separable variables homogenous first order DE Define the type of the next DE: y  9 y  0 second order linear homogeneous DE second order linear inhomogeneous DE liner first order DE DE with separable variables homogenous first order DE Which substitution can be used for solving the next DE: y=tx y=uv y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=tx y=uv y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p  Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=1/p y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y  3 p y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx  y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y  e  p x y=tx y’=p(x) y’=p(y) y=x+p Which substitution can be used for solving the next DE: y=x+1/p y=tx y’=p(x) y’=p(y) y=x+p Which substitution can be used for solving the next DE: y  e  1 / p x y=tx y’=p(x) y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx  y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’’=p(x) y  e  1 / p x y=tx y’’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(y) y’’=p(x) y  e  1 / p x y=tx  y=x+p Which substitution can be used for solving the next DE: y=tx y=uv y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=tx y=uv y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=tx y=uv y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE:  y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y=uv y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y  p 2 y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE:  y  1 / p y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y  p y=tx y’=p(x) y’=p(y) y=p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p(x) y  e  1 / p x y=tx y’=p(y) y=x+p Which substitution can be used for solving the next DE: y’=p, y’’=pp’  y  e  1 / p x y=tx y’=p, y’’=p’ y=x+p Which substitution can be used for solving the next DE: y’=p, y’’=pp’ y  e  1 / p x y=tx y’=p, y’’=p’ y=x+p Which substitution can be used for solving the next DE: y’=p, y’’=pp’ y  e  1 / p x y=tx y’=p, y’’=p’ y=x+p The characteristic equation of the next DE: r  9r  0 4 3 r  9  0 r  9r  0 3 r  9  0 3 r  5r  8r  4  0 3 2 The characteristic equation of the next DE: r  5r  8r  4  0 3 2 r  2r  2r  1  0 4 3 r  2r  r  0 4 3 2 r  2r  9r  18  0 3 2 r  9r  0 3 The characteristic equation of the next DE:  r  2r  2r  1  0 4 3 r  2r  r  0 4 3 2 r  2r  9r  18  0 3 2 r  5r  8r  4  0 3 2 r  9r  0 3 The characteristic equation of the next DE: r  9r  0 3 r3  9  0 r 4  9r 3  0 r 9  0 r 4  2r 3  2r  1  0 Characteristic equation of the next DE: r  2r  9r  18  0 3 2 r  2r  2r  1  0 4 3 r  2r  r  0 4 3 2 r  5r  8r  4  0 3 2 r  9r  0 3 The characteristic equation of the next DE: r  2r  r  0 4 3 2 r  2r  2r  1  0 4 3 r  2r  9r  18  0 3 2 r  5r  8r  4  0 3 2 r  9r  0 3 The characteristic equation of the next DE: r  16  0 4 r  9r  0 3 r  9  0 3 r  9r  0 4 3 r  9  0  y  sin x, y0   1, y (0)  0  y   sin x. sin x. cos x  1.  cos x  1. tgx. y  6 x, y1  3, y (1)  1  y  3 x. x  x. 3 x. 4 x. 2 x  1. y  81e3 x  y  9e  C1 x  C2. 3x e  C1 x  C2. 3x 3e  C1 x  C2. 3x 27 e  C1 x  C2. 3x 1 3x e  C1 x  C2. 3 y  81e3 x , y0  28, y0  10  y  9e 3x  x 1. e 3x  x9. 3e  3x  4. 3x 27 e  5 x  17 3x. 1 3x 1 29 e  x  3 3 3. y  10 y  25 y  0  y   C1e 5x  C2 xe 5 x. C1e 5x  C2e5 x. 5x 2C1e. e 5x cos5x  i sin 5x . C1e 5x  C2e10 x. y  4 y  0  y  C1 cos 2 x  C2 sin 2 x. C1e 2x  C2 xe 2 x. C1e 2x  C2e2 x. 2x 2C1e. 2ix C1e  C2e2ix. y  3 y  2 y  0  y  x C1e  C2e2 x. C1e 3x  C2e2 x. 5x 2C1e. e 5x cos x  i sin 2 x . C1e 5x  C2e10 x. y  y  2 y  0  y  x C1e  C2 e 2 x. C1e  C2e 3x 2x. 5x 2C1e. e 5x cos x  i sin 2 x . C1e 5x  C2e10 x.  y  9 y  0  y  C1 cos 3 x  C2 sin 3 x. C1e 2x  C2 xe 2 x. C1e 2x  C2e2 x. 2x 2C1e. 3ix C1e  C2e3ix. y  4 y  4 y  0  y  2 x C1e  C2 xe 2 x. C1e 3x  C2e2 x. 5x 2C1e. e 5x cos x  i sin 2 x . C1e 5x  C2e10 x. y  4 y  0  y  C1  C2e 4x. C1 cos 3 x  C2 sin 3 x. C1e 2x  C2 xe 2 x. 2x 2C1e. 3ix C1e  C2e3ix. y  4 y  13 y  0  y  e 2x C1 cos3x  C2 sin 3x . C1  C2e 4x. C1e 2x  C2 xe 2 x. 2x 2C1e. 3ix C1e  C2e3ix.  y  5 y  6 y  0  y  C1e  C2e3 x. 2x e C1 cos9 x  C2 sin 9 x . 2x C1  C2e 4x. C1e 2x  C2 xe 2 x. 2x 2C1e. y  3 y  0  y  3 x C1  C2e. e 2x C1 cos9 x  C2 sin 9 x . C1e 2x  C2 xe 2 x. 2x 2C1e. 3ix C1e  C2e3ix. y  2 y  5 y  0  y  e x C1 cos 2 x  C2 sin 2 x . C1  C2e 4x. C1e 2x  C2 xe 2 x. 2x 2C1e. 3ix C1e  C2e3ix. y  2 y  10 y  0  y  e x C1 cos3x  C2 sin 3x . C1  C2e 4x. C1e 2x  C2 xe 2 x. 2x 2C1e. 3ix C1e  C2e3ix.  y  y  2 y  0  y  2 x C1e  C2e x. e x C1 cos9 x  C2 sin 9 x. C1  C2e 4x. C1e 2x  C2 xe 2 x. 2x 2C1e. y  2 y  0  y  C1  C2e 2x. e x C1 cos9 x  C2 sin 9 x. C1e 2x  C2 xe 2 x. 2x 2C1e. 3ix C1e  C2e3ix. y  49 y  0  y  C1e 7x  C2 xe 7 x. e x C1 cos9 x  C2 sin 9 x. C1  C2e 4x. 2x 2C1e. 3ix C1e  C2e3ix. y  4 y  5 y  0  y  e 2x C1 cos x  C2 sin x. C1  C2e 4x. C1e 2x  C2 xe 2 x. 2x 2C1e.  3ix C1e  C2e3ix. roots of the corresponding characteristic equation 0, 1, 6 0, 1, 7 0, 7 0, 6 0, -7 roots of the corresponding characteristic equation 0, -3, 3 0, 9 0, 1 0, -9, 9 0, 5 , roots of the corresponding characteristic equation 0, 1 3, 1 3, -1 0, -1 2, 1 , roots of the corresponding characteristic equation -1, 1 0, 1 -2, -1 1, 2 -2, 2 , roots of the corresponding characteristic equation 1, -3 1, 3 0, 3 -5, 3 -3, 3 , roots of the corresponding characteristic equation -2, -1, 0  0, 1, 2 1, 3, 2 1, 3 0, 1 , roots of the corresponding characteristic equation 0, 1, 12 1, -13, 12 1, 2, 3 1, 2 0, 3 , roots of the corresponding characteristic equation -2,-1, 1, 2 4, 2 -3, 3 -3, 0, 3 1, -5, 4 , roots of the corresponding characteristic equation -1, 1, 3 -2, 1 -3, 3 -3, 0, 3 1, -5, 4 , roots of the corresponding characteristic equation 1, -i, i -1, 1 0 -1 i , the form of particular solution y*  Ax 2  Bx y *  Ax 2e6 x y*   Ax  B e x  y *   Ax  B e  x y *  Ae  x , the form of particular solution y*  A cos 2 x  B sin 2 x    y*  Ax 2  Bx cos x  Cx2  Dx sin x y*   Ax  B  cos 2 x  Cx  D sin 2 x y *  Ax 2e6 x y *   Ax  B e x , the form of particular solution y *  Ax 2e6 x y*   Ax  B e x y *   Ax  B e x x y  Ae * y  Ae * 3x , the form of particular solution y *   Ax  B e  x y *  Ax 2e6 x y*   Ax  B e x x y *  Ae y*  Ae3 x , the form of particular solution  y *  Ae  x y *   Ax  B e  x y  Ax e * 2 6x y*   Ax  B e x y*  Ae3 x , the form of particular solution     y*  Ax 2  Bx cos x  Cx2  Dx sin x y*  A cos x  B sin x y *   Ax  B  cos 2 x  Cx  D sin 2 x y *  Ax 2e6 x y*  Ae3 x , the form of particular solution y *  Ae 3x y *   Ax  B e  x y *  Ax 2e6 x y *   Ax  B e x y *  Ae  x , the form of particular solution y*  Ax 2e x x y *  Ae y   Ax  B e * x  y *  Ax 2e6 x y*  Ae3 x , the form of particular solution y*  Ax  B y*  Ax 2  Bx y *  Ax 2e6 x y*   Ax  B e x x y *  Ae , the form of particular solution y *  Ax 2 e 2 x y*  Ax 2  Bx y *  Ax 2e6 x y*   Ax  B e x y *   Ax  B e  x the corresponding characteristic equation k 2  6k  5  0 k  2k  3  0 2 k  9  0 2 k  8k  15  0 2 k  k  2  0 2 the corresponding characteristic equation k 2  2k  3  0 k 2  6k  5  0  k 9  0 2 k  8k  15  0 2 k  k  2  0 2 the corresponding characteristic equation k 9  0 2 k  6k  5  0 2 k  2k  3  0 2 k  8k  15  0 2 k  k  2  0 2 the corresponding characteristic equation k  2k  3  0 2 k  6k  5  0 2 k  2k  3  0 2 k  9  0 2 k  8k  15  0 2 the corresponding characteristic equation k 1  0 2 k  6k  5  0 2 k  9  0 2 k  8k  15  0 2 k  k  2  0 2 the corresponding characteristic equation k 2  8k  15  0 k 2  6k  5  0  k  2k  3  0 2 k  9  0 2 k  k  2  0 2 the corresponding characteristic equation k2  k  0 k  6k  5  0 2 k  2k  3  0 2 k  9  0 2 k  8k  15  0 2 the corresponding characteristic equation k 1  0 2 k  k  0 2 k  6k  5  0 2 k  2k  3  0 2 k  9  0 2 the corresponding characteristic equation k 2  3k  2  0 k  k  0 2 k  6k  5  0 2 k  2k  3  0 2 k  8k  15  0 2 the corresponding characteristic equation k k 20 2 k2  k  0  k  2k  3  0 2 k  9  0 2 k  8k  15  0 2 the corresponding characteristic equation k 2  10k  16  0 k  6k  5  0 2 k  2k  3  0 2 k  9  0 2 k  8k  15  0 2 the corresponding characteristic equation k 2  9k  14  0 k  k  0 2 k  2k  3  0 2 k  9  0 2 k  8k  15  0 2 the corresponding characteristic equation k  6k  8  0 2 k  k  0 2 k  6k  5  0 2 k  2k  3  0 2 k  9  0 2 the corresponding characteristic equation k 2  6k  7  0 k 2  6k  5  0  k  2k  3  0 2 k  9  0 2 k  8k  15  0 2 the corresponding characteristic equation k  7k  0 2 k  k  0 2 k  6k  5  0 2 k  9  0 2 k  8k  15  0 2 y=? y  5  6e x  e 6 x y  C1  C2 x  C3e x y  C1e x  C2 xe x  C3e3 x y  C1  C2e x  C3e12x y  C1e x  C2 cos x  C3 sin x y=? 3 x y  C1  C2 x  C3 x 2  C4 e  C5 e 3x x x 2 x y  C1e  C2 xe  C3 x e  C4 e x y  C1e2 x  C2e x  C3e x  C4e2 x y  C1e3 x  C2e x  C3e x  C4e3 x y  C1  C2e x  C3e12x , y=? y  C1  C2 x  C3e x  y  5  6e x  e 6 x y  C1e x  C2 xe x  C3e3 x y  C1  C2e x  C3e12x y  C1e x  C2 cos x  C3 sin x , y=? y  C1e x  C2 xe x  C3 x2e x  C4e x y  C1  C2 x  C3 x 2  C4 e 3 x  C5e3 x y  C1e2 x  C2e x  C3e x  C4e2 x y  C1e3 x  C2e x  C3e x  C4e3 x y  C1  C2e x  C3e12x , y=? y  C1e x  C2 xe x  C3e3 x y  5  6e x  e 6 x y  C1  C2 x  C3e x y  C1  C2e x  C3e12x y  C1e x  C2 cos x  C3 sin x , y=? y  C1  C2e x  C3e2 x y  C1  C2 x  C3e x y  C1e x  C2 xe x  C3e3 x y  C1  C2e x  C3e12x y  C1e x  C2 cos x  C3 sin x  , y=? y  C1  C2 e x  C3e12 x y  5  6e x  e 6 x y  C1  C2 x  C3e x y  C1e x  C2 xe x  C3e3 x y  C1e x  C2 cos x  C3 sin x , y=? y  C1e2 x  C2e x  C3e x  C4e2 x y  C1  C2 x  C3 x 2  C4 e 3 x  C5e3 x y  C1e x  C2 xe x  C3 x2e x  C4e x y  C1e3 x  C2e x  C3e x  C4e3 x y  C1  C2e x  C3e12x , y=? y  C1e3 x  C2e x  C3e x  C4e3 x y  C1  C2 x  C3 x 2  C4 e 3 x  C5e3 x y  C1e x  C2 xe x  C3 x2e x  C4e x y  C1e2 x  C2e x  C3e x  C4e2 x y  C1  C2e x  C3e12x , y=? y  C1e x  C2 cos x  C3 sin x y  5  6e x  e 6x y  C1  C2 x  C3e x  y  C1e x  C2 xe x  C3e3 x y  C1  C2e x  C3e12x Roots of the corresponding characteristic equation 1, 5 -1, 3 -3, 3 -3, 1 -1, 1 Roots of the corresponding characteristic equation -1, 3 1, 5 -3, 3 -3, 1 -1, 1 Roots of the corresponding characteristic equation -3, 3 1, 5 -1, 3 -3, 1 -1, 1 Roots of the corresponding characteristic equation -3, 1 1, 5 -1, 3 -3, 3 -1, 1 Roots of the corresponding characteristic equation -1, 1 1, 5  -1, 3 -3, 3 -3, 1 Roots of the corresponding characteristic equation 3, 5 1, 5 -1, 3 -3, 3 -3, 1 Roots of the corresponding characteristic equation -1, 0 1, 5 -3, 3 -3, 1 -1, 1 Roots of the corresponding characteristic equation -1, 1 1, 5 -1, 3 -3, 3 -3, 1 Roots of the corresponding characteristic equation 1, 2 1, 5 -1, 3 -3, 3 -1, 1 Roots of the corresponding characteristic equation -2, 1 1, 5 -1, 3 -3, 1  -1, 1