Question

find the general solution of the given equation.
$y - 6y + 18y = 0$
$y(x) = \square$

Explanation:

Step1: Write the characteristic equation

For the differential equation \(y'' - 6y' + 18y = 0\), the characteristic equation is \(r^{2}-6r + 18=0\).

Step2: Solve the characteristic equation

Using the quadratic formula \(r=\frac{-b\pm\sqrt{b^{2}-4ac}}{2a}\) for \(ax^{2}+bx + c = 0\). Here \(a = 1\), \(b=-6\), \(c = 18\).

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Step3: Write the general solution

When the roots of the characteristic equation are \(r=\alpha\pm\beta i\), the general solution of the differential equation \(y''+ay'+by = 0\) is \(y(x)=e^{\alpha x}(C_{1}\cos(\beta x)+C_{2}\sin(\beta x))\). Here \(\alpha = 3\) and \(\beta=3\).

Answer:

\(y(x)=e^{3x}(C_{1}\cos(3x)+C_{2}\sin(3x))\)