Solve the differential equation: y''+6y'+12y=0

Question
Solve the differential equation:
\(y''+6y'+12y=0\)

Answers (1)

2021-02-07
Given the Differential Equation is
\(y''+6y'+12y=0\)
This is a second-order linear differential equation with constant coefficients
Auxilary Equation is
\(m^2+6m+12=0\)
Solve This quadratic equation.
Compare with:
\(ax^2+bx+c=0\)
So roots are
\(x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}\)
Here,
\(a=1,\ b=6,\ c=12\)
Hence,
\(m=\frac{-6\pm\sqrt{36-48}}{2}\)
\(=\frac{-6+-\sqrt{-12}}{2}\)
\(=\frac{-6+-\sqrt{3i}}{2}\)
\(=-3\pm\sqrt{3i}\)
So here Roots of equations are complex
So Solution of Ordinary differential equation with root a+ib and a-ib is
\(y=e^{ax}[C_1\cos{bx}+C_2\sin{bx}]\)
Here \(a=-3\) and \(b=\sqrt3\)
So the solution is
\(y=e^{-3x}[C_1\cos\sqrt{3x}+C_2\sin\sqrt{3x}]\)
0

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