Answer:
The solution of this differential system is [tex]y(x) = x\cdot e^{3\cdot x}[/tex].
Step-by-step explanation:
This second-order differential equation is homogeneous and linear of the form:
[tex]y'' + p\cdot y' +q\cdot y = 0[/tex] (1)
Where:
[tex]p[/tex] - First-order constant coefficient, dimensionless.
[tex]q[/tex] - Zero-order constant coefficient, dimensionless.
Whose characteristic polynomial is:
[tex]\lambda^{2}+p\cdot \lambda + q = 0[/tex] (2)
Where [tex]\lambda[/tex] contains the roots associated with the solution of the differential equation.
If we know that [tex]p = 6[/tex] and [tex]q = 9[/tex], the roots of the characteristic equation are, respectively:
[tex]\lambda^{2}+6\cdot \lambda + 9 = 0[/tex]
[tex](\lambda +3)\cdot (\lambda + 3) = 0[/tex]
Which means that [tex]\lambda_{1} = \lambda_{2} = 3[/tex] and the solution of the differential equation is of the form:
[tex]y(x) = e^{3\cdot x}\cdot (c_{1}+c_{2}\cdot x)[/tex] (3)
Where [tex]c_{1}[/tex] and [tex]c_{2}[/tex] are integration constants.
The first derivative of the equation above is:
[tex]y'(x) = 3\cdot e^{3\cdot x}\cdot (c_{1}+c_{2}\cdot x)+c_{2}\cdot e^{3\cdot x}[/tex] (4)
Now, if we get that [tex]y(0) = 0[/tex] and [tex]y'(0) = 1[/tex], then the system of equations to the solved is:
[tex]c_{1} = 0[/tex] (3b)
[tex]c_{1}+c_{2} = 1[/tex] (4b)
The solution of this system is: [tex]c_{1} = 0[/tex], [tex]c_{2} = 1[/tex]. Therefore, the solution of this differential system is [tex]y(x) = x\cdot e^{3\cdot x}[/tex].
