Question

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

Answer 1

Answer:

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

Step-by-step 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 $y(x) = x\cdot e^{3\cdot x}$.