Explanation Procedure used: Method of variation of parameters: The following are the steps to obtain a particular solution to a y ″ + b y ′ + c y = f . (i) Find two linearly independent solutions { y 1 ( t ) , y 2 ( t ) } to the corresponding homogenous equation and take y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) . (ii) Determine v 1 ( t ) and v 2 ( t ) by solving the below system for v ′ 1 ( t ) and v ′ 2 ( t ) and integrating. y 1 v ′ 1 + y 2 v ′ 2 = 0 y ′ 1 v ′ 1 + y ′ 2 v ′ 2 = f a This can be simplified using Cramer’s rule as, v ′ 1 ( t ) = − f ( t ) y 2 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] and v ′ 2 ( t ) = f ( t ) y 1 ( t ) a [ y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) ] (iii) Substitute v 1 ( t ) and v 2 ( t ) into the expression for y p ( t ) to obtain a particular solution. Calculation: The given differential equation is, t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t . The two independent solutions to the homogenous equation t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t are y 1 = 5 t − 1 and y 2 = e − 5 t . The general solution to the homogenous equation will be, y ( t ) = c 1 ( 5 t − 1 ) + c 2 ( e − 5 t ) . The derivatives is, y ′ 1 ( t ) = 5 and y ′ 2 ( t ) = − 5 e − 5 t . Set the particular solution y p ( t ) = v 1 ( t ) y 1 ( t ) + v 2 ( t ) y 2 ( t ) as y p ( t ) = v 1 ( t ) ( 5 t − 1 ) + v 2 ( t ) ( e − 5 t ) . Obtain the Wronskian y 1 ( t ) y ′ 2 ( t ) − y ′ 1 ( t ) y 2 ( t ) as, W = | 5 t − 1 e − 5 t 5 − 5 e − 5 t | = ( 5 t − 1 ) ( − 5 e − 5 t ) − 5 e − 5 t = − 25 t e − 5 t + 5 e − 5 t − 5 e − 5 t = − 25 t e − 5 t Comparing t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t with the general form a y ″ + b y ′ + c y = f implies a = t , b = 5 t − 1 , c = 5 and f = t 2 e − 5 t

Fundamentals of Differential Equations (9th Edition)

9th Edition

ISBN: 9780321977069

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

See similar textbooks

Concept explainers

Optimization

Optimization comes from the same root as "optimal". "Optimal" means the highest. When you do the optimization process, that is when you are "making it best" to maximize everything and to achieve optimal results, a set of parameters is the base for the se…

Integration

Integration means to sum the things. In mathematics, it is the branch of Calculus which is used to find the area under the curve. The operation subtraction is the inverse of addition, division is the inverse of multiplication. In the same way, integratio…

Application of Integration

In mathematics, the process of integration is used to compute complex area related problems. With the application of integration, solving area related problems, whether they are a curve, or a curve between lines, can be done easily.

Volume

In mathematics, we describe the term volume as a quantity that can express the total space that an object occupies at any point in time. Usually, volumes can only be calculated for 3-dimensional objects. By 3-dimensional or 3D objects, we mean objects th…

Area

Area refers to the amount of space a figure encloses and the number of square units that cover a shape. It is two-dimensional and is measured in square units.

Videos

Question

Chapter 4.6, Problem 25E

To determine

The general solution to the differential equation ty″+(5t−1)y′−5y=t2e−5t, where y1=5t−1 and y2=e−5t are linearly independent solutions to the corresponding homogenous equation for t>0.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 4.6, Problem 24E

Chapter 4.7, Problem 1E

Students have asked these similar questions

>tt 1:32 > trend.1m 1m (sales > summary(trend.1m) - tt) #3###23 (i) #### Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 2107.220 57.997 36.332e-16 *** tt -43.500 3.067 -14.18 7.72e-15 *** > trend = ts (fitted (trend.1m), start-start (sales), freq-frequency (sales)) sales trend ###23%23 (ii) #### as.numeric((1:32 %% 4) > X > q1 > q2 > q3 > 94 = = = = - as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) as.numeric((1:32 %% 4) == 1) 2) == == 3) == 0) > season.lm = 1m (resid (trend.1m) 0+q1 + q2 + q3 + q4) #3##23%23 (iii) #### > summary(season.1m) Coefficients: Estimate Std. Error t value Pr(>|t|) q1 -38.41 43.27 -0.888 0.38232 92 18.80 43.27 0.435 0.66719 q3 -134.78 43.27 -3.115 0.00422 ** 94 154.38 43.27 3.568 0.00132 ** > season = ts (fitted (season.lm), start=start (sales), freq=frequency (sales)) > Y X season %23%23%23%23 (iv) #### >ar (Y, aic=FALSE, order.max=1) #23%23%23%23 (v) #### Coefficients: 1 0.5704 Order selected 1 sigma 2 estimated as 9431 > ar(Y, aic=FALSE,…

Refer to page 52 for solving the heat equation using separation of variables. Instructions: • • • Write the heat equation in its standard form and apply boundary and initial conditions. Use the method of separation of variables to derive the solution. Clearly show the derivation of eigenfunctions and coefficients. Provide a detailed solution, step- by-step. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS31Z9qo Hazb9tC440AZF/view?usp=sharing]

Refer to page 20 for orthogonalizing a set of vectors using the Gram-Schmidt process. Instructions: • Apply the Gram-Schmidt procedure to the given set of vectors, showing all projections and subtractions step-by-step. • Normalize the resulting orthogonal vectors if required. • Verify orthogonality by computing dot products between the vectors. Link: [https://drive.google.com/file/d/1wKSrun-GlxirS3IZ9qoHazb9tC440AZF/view?usp=sharing]

Chapter 4 Solutions

Fundamentals of Differential Equations (9th Edition)

Ch. 4.1 - Verify that for b = 0 and Fext(t) = 0, equation...Ch. 4.1 - If Fext(t) = 0, equation (3) becomes my + by + ky...Ch. 4.1 - Show that if Fext(t) = 0, m = 1, k = 9, and b = 6,...Ch. 4.1 - Verify that y = sin 3t + 2 cos 3t is a solution to...Ch. 4.1 - Prob. 5E Ch. 4.1 - An external force F(t) = 2 cos 2t is applied to a...Ch. 4.1 - Prob. 7E Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - In Problems 79, find a synchronous solution of the...Ch. 4.1 - Prob. 10E

Knowledge Booster

$Advanced Math$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, advanced-math and related others by exploring similar questions and additional content below.

The general solution to the differential equation t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t , where y 1 = 5 t − 1 and y 2 = e − 5 t are linearly independent solutions to the corresponding homogenous equation for t &gt; 0 .

Concept explainers

Videos

The general solution to the differential equation ty″+(5t−1)y′−5y=t2e−5t, where y1=5t−1 and y2=e−5t are linearly independent solutions to the corresponding homogenous equation for t>0.

Want to see the full answer?

Chapter 4 Solutions

The general solution to the differential equation t y ″ + ( 5 t − 1 ) y ′ − 5 y = t 2 e − 5 t , where y 1 = 5 t − 1 and y 2 = e − 5 t are linearly independent solutions to the corresponding homogenous equation for t > 0 .