Respuesta :
Answer:
a) The general solution of the differential equation is [tex]y(t)=c_1e^{3t}+c_2te^{3t}[/tex].
b) Applying the initial conditions the solution to the differential equation is then [tex]y(t)=5e^{3t}-15e^{3t}t[/tex].
Step-by-step explanation:
We have the following second order linear homogeneous differential equation with constant coefficients [tex]y''-6y'+9y=0[/tex]
Given the differential equation [tex]ay'' + by' + cy = 0[/tex], [tex]a\neq 0[/tex], consider the quadratic polynomial [tex]ax^2+bx+c[/tex], called the characteristic polynomial. Using the quadratic formula, this polynomial always has one or two roots, call them [tex]r[/tex] and [tex]s[/tex]. The general solution of the differential equation is:
a) [tex]\ds y=Ae^{rt}+Be^{st}[/tex], if the roots [tex]r[/tex] and [tex]s[/tex] are real numbers and [tex]r\neq s[/tex].
b) [tex]\ds y=Ae^{rt}+Bte^{rt}[/tex], if [tex]r=s[/tex] is real.
c) [tex]\ds y=A\cos(\beta t)e^{\alpha t}+B\sin(\beta t)e^{\alpha t}[/tex], if the roots [tex]r[/tex] and [tex]s[/tex] are complex numbers [tex]\alpha+\beta i[/tex] and [tex]\alpha-\beta i[/tex].
Applying the above information we have that:
The characteristic polynomial is
[tex]x^2-6x+9=0[/tex]
Its roots are
[tex]x^2-6x+9=(x-3)^2=0\\x=3[/tex]
So, the general solution of the differential equation is:
[tex]y(t)=c_1e^{3t}+c_2te^{3t}[/tex]
and its derivative is:
[tex]y'(t)=\frac{d}{dt}\left(c_1e^{3t}+c_2te^{3t}\right)=c_1e^{3t}\cdot \:3+c_2\left(e^{3t}+3e^{3t}t\right)[/tex]
Now, plug in the initial conditions to get the following system of equations.
[tex]y(0)=c_1e^{3\cdot 0}+c_2\cdot 0\cdot e^{3\cdot 0}=5\\\\y'(0)=c_1e^{3\cdot 0}\cdot \:3+c_2\left(e^{3\cdot 0}+3e^{3\cdot 0}\cdot 0\right)=0[/tex]
Solving this system gives
[tex]c_1=5[/tex] and [tex]c_2=-15[/tex]
The actual solution to the differential equation is then[tex]y(t)=5e^{3t}-15e^{3t}t[/tex]
