- #1
skysurani
- 7
- 0
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
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)
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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