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Find the solution of the second-order linear differential equation that satisfies the given initial conditions. y'' + 6y' + 9y = 0 y(0) = 0 y'(0) = 1

User Charline
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Answer:

The solution of this differential system is
y(x) = x\cdot e^(3\cdot x).

Explanation:

This second-order differential equation is homogeneous and linear of the form:


y'' + p\cdot y' +q\cdot y = 0 (1)

Where:


p - First-order constant coefficient, dimensionless.


q - Zero-order constant coefficient, dimensionless.

Whose characteristic polynomial is:


\lambda^(2)+p\cdot \lambda + q = 0 (2)

Where
\lambda contains the roots associated with the solution of the differential equation.

If we know that
p = 6 and
q = 9, the roots of the characteristic equation are, respectively:


\lambda^(2)+6\cdot \lambda + 9 = 0


(\lambda +3)\cdot (\lambda + 3) = 0

Which means that
\lambda_(1) = \lambda_(2) = 3 and the solution of the differential equation is of the form:


y(x) = e^(3\cdot x)\cdot (c_(1)+c_(2)\cdot x) (3)

Where
c_(1) and
c_(2) are integration constants.

The first derivative of the equation above is:


y'(x) = 3\cdot e^(3\cdot x)\cdot (c_(1)+c_(2)\cdot x)+c_(2)\cdot e^(3\cdot x) (4)

Now, if we get that
y(0) = 0 and
y'(0) = 1, then the system of equations to the solved is:


c_(1) = 0 (3b)


c_(1)+c_(2) = 1 (4b)

The solution of this system is:
c_(1) = 0,
c_(2) = 1. Therefore, the solution of this differential system is
y(x) = x\cdot e^(3\cdot x).

User Gotgenes
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