Find the general solution to the following differential equations. y"+6y'+13y=0

Vikolers6 2021-12-27 Answered
Find the general solution to the following differential equations.
y6y+13y=0
You can still ask an expert for help

Want to know more about Laplace transform?

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

alexandrebaud43
Answered 2021-12-28 Author has 36 answers
Given: y6y+13y=0
y6y+13y=0
A second order linear, homogenous ODE has the form ayby+cy=0
For an equation ayby+cy=0, assume a solution of the form eλt
Rewrite the equation with y=eλt
(eλt)6(eλt)+13eλt=0
eλt(y2+6y+13)=0
y2+6y+13=0
Roots of the quadratic equation in the form ax2+bx+c=0 is
x1,2=b±b24ac2a
Here, a=1,b=6andc=13
y1,2=6±624×1×132×1
y1,2=6±36522
y1,2=32i,3+2i
For two complex roots y1y2, where y1=32i,y2=3+2i
So, the general solution takes the form: y=eαt(c1cos(βt)+c2sin(βt))
y=e3t(c1cos(2t)+c2sin(2t))
Not exactly what you’re looking for?
Ask My Question
xandir307dc
Answered 2021-12-29 Author has 35 answers
y=ekx
k2+6k+13=0
D=b24ac=624113=16
D=±4i
k1,2=3±2i
Answer: y=e3x(C1cos2x+C2sin2x)
Not exactly what you’re looking for?
Ask My Question
karton
Answered 2022-01-10 Author has 439 answers

We were given the next equation
y+6y+13y=0   (1)
The roots of the characteristic equation
λ2+6λ+13=0   (2)
are λ1=3+2i,λ2=32i
The general solution of Eq. (1) is, by the Theorem 4.3.2, given by
y(t)=c1eμtcosvt+c2eμtsinvt
where λ1,2=μ±iv. In this case, the solution is
y(t)=c1e3tcos2t+c2e3tsin2t

Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more