• :00Days
  • :00Hours
  • :00Mins
  • 00Seconds
A new era for learning is coming soonSign up for free
Log In Start studying!

Select your language

Suggested languages for you:
Answers without the blur. Sign up and see all textbooks for free! Illustration


Linear Algebra With Applications
Found in: Page 442
Linear Algebra With Applications

Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

Answers without the blur.

Just sign up for free and you're in.


Short Answer

Solve the differential equation f"(t)+2f'(t)+f(t)=0 and find all the real solutions of the differential equation.

The solution is f(t)=e-tC1+C2t .

See the step by step solution

Step by Step Solution

Step1: Definition of characteristic polynomial

Consider the linear differential operator


The characteristic polynomial of is defined as


Step2: Determination of the solution

The characteristic polynomial of the operator as follows.


Then the characteristic polynomial is as follows.


Solve the characteristic polynomial and find the roots as follows.

λ2+2λ+1=0 λ2+λ+λ+1=0λλ+1+1λ+1=0 λ+1λ+1=0

Simplify further as follows.

λ+1=0 λ=-1λ+1=0 λ=-1

Therefore, the roots of the characteristic polynomials are -1 and -1 .

Step3: Conclusion of the solution

Since, the roots of the characteristic equation is different real numbers.

The exponential functions e-t and te-t form a basis of the kernel of T .

Hence, they form a basis of the solution space of the homogenous differential equation is T(f) = 0 .

Thus, the general solution of the differential equation f"t+2f't=0 is ft+e-tC1+C2t .

Recommended explanations on Math Textbooks

94% of StudySmarter users get better grades.

Sign up for free
94% of StudySmarter users get better grades.