Chapter 6: Q4E (page 341)

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.
$y^{'''} - 3 y^{''} + 3 y^{'} - y = e^{x}$

Short Answer

Expert verified

The particular solution is $y_{p} = \frac{x^{3} e^{x}}{6}$

Step by step solution

Definition

Variation of parameters, also known as variation of constants, is a general method to solve inhomogeneous linear ordinary differential equations.

Find complementary solution

The given equation is: $y^{'''} - 3 y^{''} + 3 y^{'} - y = e^{x}$

The auxiliary equation is $m^{3} - 3 m^{2} + 3 m - 1 = 0$

Solving we get $m = 1, 1, 1$

Therefore, the complimentary solution is given by $y_{c} = c_{1} e^{x} + c_{2} x e^{x} + c_{3} x^{2} e^{x}$

Calculate Wornkians

Compare with. $C F = c_{1} y_{1} (x) + c_{2} y_{2} (x) + c_{3} y_{3} (x)$

$y_{1} (x) = e^{x}, y_{2} (x) = x e^{x} and y_{3} (x) = x^{2} e^{x}$

$W [\begin{matrix} e^{x} & x e^{x} & x^{2} e^{x} \end{matrix}] = [\begin{matrix} e^{x} & x e^{x} & x^{2} e^{x} \\ e^{x} & e^{x} (x + 1) & e^{x} (x^{2} + 2 x) \\ e^{x} & e^{x} (x + 2) & e^{x} (x^{2} + 4 x + 2) \end{matrix}] = 2 e^{3 x}$

Calculate Wornkians

The value of wronkians $W_{1}, W_{2}, W_{3}$ is:

$W_{1} = (- 1)^{3 - 1} W [\begin{matrix} x e^{x} & x^{2} e^{x} \end{matrix}] = e^{2 x} (x^{3} + 2 x^{2} - x^{3} - x^{2}) = e^{2 x} x^{2}$

The particular solution is thus given by:

$y_{p} = e^{x} \int \frac{x^{2} e^{2 x} (e^{x})}{2 e^{3 x}} d x + x e^{x} \int \frac{- 2 x e^{2 x} (e^{x})}{2 e^{3 x}} d x + x^{2} e^{x} \int \frac{e^{2 x} \cdot e^{x}}{2 e^{3 x}} d x y_{p} = e^{x} \frac{x^{3}}{6} - \frac{x^{3}}{2} e^{x} + \frac{x^{3}}{2} e^{x} = \frac{x^{3} e^{x}}{6}$

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.

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.
$y^{'''} - 3 y^{''} + 3 y^{'} - y = e^{x}$

Short Answer

Step by step solution

Definition

Find complementary solution

Calculate Wornkians

Calculate Wornkians

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Statistics

Decision Maths

Geometry

Calculus

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.y'''-3y''+3y'-y=ex

Short Answer

Step by step solution

Definition

Find complementary solution

Calculate Wornkians

Calculate Wornkians

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Logic and Functions

Statistics

Decision Maths

Geometry

Calculus

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

In Problems 1-6, use the method of variation of parameters to determine a particular solution to the given equation.
$y^{'''} - 3 y^{''} + 3 y^{'} - y = e^{x}$