Question 3.3.3: Solve y''' + 3y" - 4y = 0....
Solve y”’ + 3y” – 4y = 0.
Step-by-Step
It should be apparent from inspection of m^3+3 m^2-4=0 that one root is m_1=1 and so m = 1 is a factor of m^3+3 m^2-4=0. By division we find
m^3+3 m^2-4=(m-1)\left(m^2+4 m+4\right)=(m-1)(m+2)^2,
and so the other roots are m_2=m_3=-2 . Thus the general solution is
y=c_1 e^x+c_2 e^{-2 x}+c_3 x e^{-2 x} .
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