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solve the following promblem

Y^(,,,)-3y^(,,)+31y^(,)-37y=0

i let y = e^kx

y'= ke^kx

y''=k^2e^kx

y'''=k^3e^kx

so i got this

k^3e^kx-3(k^2e^kx)+31(ke^kx)-37(e^kx)=0

e^kx(k^3-3K2+31k-37)=0

so,

(k^3-3K2+31k-37)=0

now i have to find (K)

how should i solve for (k) from this equation k^3-3K2+31k-37=0

can i use synthetic division if yes how should i use it or which other method can i use

is this right

i solve the k by synthetic division

5 1 1 -17 -65

1 6 13 0

so the factor is (k-5) (K^2+6k+13)

then i use this equation

(-b+-squrt(b^2-(4ac)))/2a

and got k = -3 +- 2i

and my fianl answer is

y=c1e^5x+c2e^(-3x)cos(2x)+c3e^(-3x)sin(2x)

Y^(,,,)-3y^(,,)+31y^(,)-37y=0

i let y = e^kx

y'= ke^kx

y''=k^2e^kx

y'''=k^3e^kx

so i got this

k^3e^kx-3(k^2e^kx)+31(ke^kx)-37(e^kx)=0

e^kx(k^3-3K2+31k-37)=0

so,

(k^3-3K2+31k-37)=0

now i have to find (K)

how should i solve for (k) from this equation k^3-3K2+31k-37=0

can i use synthetic division if yes how should i use it or which other method can i use

is this right

i solve the k by synthetic division

5 1 1 -17 -65

__5 30 65__1 6 13 0

so the factor is (k-5) (K^2+6k+13)

then i use this equation

(-b+-squrt(b^2-(4ac)))/2a

and got k = -3 +- 2i

and my fianl answer is

y=c1e^5x+c2e^(-3x)cos(2x)+c3e^(-3x)sin(2x)

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