Chapter 4: Q26E (page 172)

Solve the given initial value problem. $y'' - 2 y' + y = 0;$ $y (0) = 1,$ $y' (0) = - 2$

Short Answer

Expert verified

The solution of the given initial value $y'' - 2 y' + y = 0$ is $y (t) = e^{t} - 3 {te}^{t}$ when $y (0) = 1$ and $y' (0) = - 2$ .

Step by step solution

Differentiate the value of y.

Given differential equation is $y'' - 2 y' + y = 0$

Let $y = e^{rt}$

Therefore,

$\begin{matrix} y' (t) = r e^{rt} \\ y'' (t) = r^{2} e^{rt} \end{matrix}$

Finding the general solution.

Then the auxiliary equation is $r^{2} - 2 r + 1 = 0$

$\begin{matrix} {(r - 1)}^{2} = 0 \\ r - 1 = 0 \\ r = 1 \end{matrix}$

Therefore, the general solution is $y (t) = c_{1} e^{t} + c_{2} {te}^{t}$ .

Finding the values of c1 and c2

Given initial conditions are $y (0) = 1$ and $y' (0) = - 2$

$\begin{array}{l} y (0) = c_{1} e^{0} + c_{2} \times 0 \times e^{0} \\ c_{1} = 1 \end{array}$

And $y' (t) = c_{1} e^{t} + c_{2} e^{t} + c_{2} {te}^{t}$

Then,

$\begin{array}{l} y' (0) = c_{1} e^{0} + c_{2} e^{0} + c . 0 \times e^{0} \\ c_{1} + c_{2} = - 2 \end{array}$

Substitute $c_{1}$ in the above equation

$\begin{matrix} 1 + c_{2} = - 2 \\ c_{2} = - 3 \end{matrix}$

Therefore, the solution is $y (t) = e^{t} - 3 {te}^{t}$ .

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Solve the given initial value problem. $y'' - 2 y' + y = 0;$ $y (0) = 1,$ $y' (0) = - 2$

Short Answer

Step by step solution

Differentiate the value of y.

Finding the general solution.

Finding the values of c1 and c2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Calculus

Logic and Functions

Statistics

Applied Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Solve the given initial value problem. y''-2y'+y=0;y(0)=1,y'(0)=-2

Short Answer

Step by step solution

Differentiate the value of y.

Finding the general solution.

Finding the values of c1 and c2

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Calculus

Logic and Functions

Statistics

Applied Mathematics

Mechanics Maths

Study anywhere. Anytime. Across all devices.

Solve the given initial value problem. $y'' - 2 y' + y = 0;$ $y (0) = 1,$ $y' (0) = - 2$