Solve the given initial-value problem. y'' + y' + 2y = 0, y(0) = y'(0) = 0

Solution:

Given, the differential equation is y’’ + y’ + 2y = 0

We have to find the solution of the equation.

The differential equation can be rewritten as (D² + D + 2)y = 0

Where, D = d/dx

Auxiliary equation is m² + m + 2 = 0

Using quadratic formula,

\(x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}\)

Here, a = 1, b = 1, c = 2

\(x=\frac{-1\pm \sqrt{(1)^{2}-4(1)(2))}}{2(1)}\)

\(x=\frac{-1\pm \sqrt{1-8}}{2}\)

\(x=\frac{-1\pm \sqrt{-7}}{2}\)

Now, \(x=\frac{-1 -\sqrt{7}i}{2}\)

\(x=\frac{-1 +\sqrt{7}i}{2}\)

x = \(x=\frac{1}{2}[-1\pm \sqrt{7}i]\)

So, the general solution is of the form,

\(y(x) = C_{1}e^{m_{1}x}+C_{2}e^{m_{2}x}\)

Here, m₁ = \(x=\frac{1}{2}[-1+ \sqrt{7}i]\)

and m₂ = \(x=\frac{1}{2}[-1- \sqrt{7}i]\)

\(y(x) = C_{1}e^{(\frac{1}{2}[-1+ \sqrt{7}i])x}+C_{2}e^{(\frac{1}{2}[-1- \sqrt{7}i])x}\)

\(y(x) =C_{1}e^{x[-\frac{1}{2}-\frac{\sqrt{7}i}{2}]}+C_{2}e^{x[-\frac{1}{2}+\frac{\sqrt{7}i}{2}]}\)

\(y(x) = C_{1}[e^{-\frac{1}{2}x}e^{-\frac{\sqrt{7}i}{2}x}]+C_{2}[e^{-\frac{1}{2}x}e^{\frac{\sqrt{7}i}{2}x}]\)

\(y(x) = e^{\frac{-x}{2}}[C_{1}e^{\frac{-\sqrt{7}i}{2}x}+C_{2}e^{\frac{\sqrt{7}i}{2}x}]\)

\(y(x) = e^{\frac{-x}{2}}[C_{1}sin(\frac{\sqrt{7}x}{2})+C_{2}cos(\frac{\sqrt{7}x}{2})]\)

Given, y(0) = 0

So, \(y(0)=e^{\frac{-(0)}{2}}[C_{1}sin(\frac{\sqrt{7}(0)}{2})+C_{2}cos(\frac{\sqrt{7}(0)}{2})]=0\)

We know, e⁰ = 1

\((1)[0+C_{2}(1)]=0\)

\(C_{2}=0\)

y’(x)=\(-\frac{(C_{1}sin(\frac{\sqrt{7}x}{2})+C_{2}cos(\frac{\sqrt{7}x}{2}))e\frac{-x}{2}}{2}+(\frac{\sqrt{7}C_{1}cos(\frac{\sqrt{7}x}{2})}{2}-\frac{\sqrt{7}C_{2}sin(\frac{\sqrt{7}x}{2})}{2})e^{\frac{-x}{2}}\)

Given, y’(0) = 0

0=\(-\frac{(C_{1}sin(\frac{\sqrt{7}(0)}{2})+C_{2}cos(\frac{\sqrt{7}(0)}{2}))e\frac{-(0)}{2}}{2}+(\frac{\sqrt{7}C_{1}cos(\frac{\sqrt{7}(0)}{2})}{2}-\frac{\sqrt{7}C_{2}sin(\frac{\sqrt{7}(0)}{2})}{2})e^{\frac{-(0)}{2}}\)

0=\(-\frac{(C_{1}(0)+C_{2}(1))(1)}{2}+(\frac{\sqrt{7}C_{1}(1)}{2}-\frac{\sqrt{7}C_{2}(0)}{2})(1)\)

\(-\frac{C_{2}}{2}+(\frac{\sqrt{7}C_{1}}{2}-0)\) = 0

\(\frac{\sqrt{7}C_{1}}{2}=\frac{C_{2}}{2}\)

\(\frac{\sqrt{7}C_{1}}{2}=0\)

\(C_{1}=0\)

Therefore, the solution is y(x) = 0.

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Solve the given initial-value problem. y'' + y' + 2y = 0, y(0) = y'(0) = 0

Summary:

The solution to the given initial value problem y'' + y' + 2y = 0, y(0) = y'(0) = 0 is y(x) = 0.