Problem 4 (Non-homogeneous Equations Redux). By now we are very familiar with a procedure for solving a linear, second-order initial value problem\[y'' + p(t)y' + q(t)y = f(t), \quad y(t_0) = y_0, \quad y'(t_0) = v_0.\]1. First, find the general solution to the associated homogeneous equation\[y'' + p(t)y' + q(t)y = 0.\]2. Find a particular solution to the non-homogeneous equation\[y'' + p(t)y' + q(t)y = f(t)\]and add it to the homogeneous solution.3. Plug in the initial conditions to find the solution to the IVP.There is actually another way we can proceed, and it allows us to plug the initial conditions into the homogeneous solution before finding the particular solution.(a) Suppose $ u_1(t) $ is a solution to the initial value problem\[y'' + p(t)y' + q(t)y = 0, \quad y(t_0) = y_0, \quad y'(t_0) = v_0,\]and $ u_2(t) $ is a solution to\[y'' + p(t)y' + q(t)y = f(t), \quad y(t_0) = 0, \quad y'(t_0) = 0.\]Show that $ y(t) = u_1(t) + u_2(t) $ is a solution to the original initial value problem (1).(b) Thanks to Variation of Parameters, we know that if $\{y_1, y_2\}$ is a fundamental set of solutions to the associated homogeneous equation, then\[y_p(t) = v_1(t)y_1(t) + v_2(t)y_2(t)\]is a particular solution, where\[v_1(t) = - \int_{t_0}^{t} \frac{y_2(\tau)f(\tau)}{y_1(\tau)y_2'(\tau) - y_1'(\tau)y_2(\tau)} \, d\tau,\quad v_2(t) = \int_{t_0}^{t}

NOTE: According to guideline answer of first question can be given, for other please ask in a…

Suppose u1(t) is a solution to the initial value problem y" + p(t)y' + q(t)y = 0, y(to) = Yo, y'(to) = vo, %3D %3D and u2(t) is a solution to y/" + p(t)y' + q(t)y = f(t), y(to) = 0, y (to) = 0. (2) %3D %3D %3D Show that y(t) = u1(t) + u2(t) is a solution to the original initial value problem (1). %3D

Suppose u1(t) is a solution to the initial value problem y" + p(t)y' + q(t)y = 0, y(to) = Yo, y'(to) = vo, %3D %3D and u2(t) is a solution to y/" + p(t)y' + q(t)y = f(t), y(to) = 0, y (to) = 0. (2) %3D %3D %3D Show that y(t) = u1(t) + u2(t) is a solution to the original initial value problem (1). %3D

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

See similar textbooks

Related questions

Q: Determine by substitution if y(x) =xe* is a solution of y"- 2y' +y = 0 %3D

A: Given The differential equation is y''-2y'+y=0

Q: If Y(x) = 5x is a particular solution for x'y" + 2y' = h(x) then h(x) = %3D %3D %3D A. 80x* В. 140х*…

A: Given Y(x)=5x4 is particular solution of x2y''+2y'=h(x)....(1)

Q: A form of a particular solution to = 2x + 3y-7 dt = -x - 2y +5 is dt O (a b)" O (7, 5)" O (2, y)" O…

Q: Find the general solution of 6y''+y"- 15y'=0, 3x 5x A y=c,x+cze 2 +Cze 3x 5x B y=c,+c,e ? +c.« ¯ 2…

Q: t²y' + 2ty - y³ = 0

Q: c. y' + 2y = x³ with y(1) = 2 Yo (x) = e Yp (x) = y (x) = y (x) = -2x 1½ e ²x³ - 2/ ( 10²x² -…

Q: 2. Find the general solution of x*v + 1) dx + y'(x - 1) dy = 0 x2 - y? + 2x - 2y + А. 2ln[(y + 1)/(x…

A: No option is matching the final answer. All the given four options are incorrect. The correct answer…

Q: Find the explicit solution of dy/dx=2xy² + 3x²y², y ( 1) = − 1 O 1/y=3-x²-x3 Oy=x²/2+x²³/3-3…

Q: Find two linearly independent solutions of 2x²y" - xy' + (−3x + 1)y = 0, x > 0 = x¹¹ (1 + α₁x + a₂x²…

Q: Are y₁ (1) = x, y₂(x) = ze² a fundamental set of solution for the equation r²y" − x(x + 2)y' + (x +…

A: Given thatand are two solutions of DE

Q: 1. If y = In(V x² +1 ) , then y' = 2x (b): 2x (a) x² +1 3(x2 +1) () 3(x2 +1)-2/3 (d) 3(x2 +1)

A: y=lnx2+13⇒y=lnx2+113⇒y=13·lnx2+1 ∵lnxy=ylnx

Q: If y = u^4 − 2u^2 and x = u^2 − 2u then find d(y)/d(x)

Q: 5. Find a particular solution to the equation -2x y" + 4y' + 4y = xe¯

Q: Obtain the particular solution 1. (D³ +6D² + 12D + 8)y=0 when x = 0, y = 1, y'= -2, y" = 2 2. y" +…

Q: 3. y' = 5y + e-2*y-2 y(0) = 2 У (0) - 4. хуу' + 4x2 + у? 3D 0 о у(2) 3 —7

Q: If these two y's, y1=e^tcos(t) and y2(t)=e^tsin(t) are both solutions to y''+j(t)y'+k(t)y=0, how do…

Q: 3 log(x^4)-2 log(3x) write as a single logarithm with a coefficient of 1

Q: If y₁ and y₂ are linearly independent solutions of ty" + 3y'+te^5t y = 0 and if W(y1 y2)(1)= 3, find…

A: and are two linearly independent solution of

Q: xy' + 2y = 3x, y(1) = 5 %3D

Q: 3X 7- (0²1)y=K²" yp-e (4X-3} 8-(0-20+4)y=x+3x-5x+2 yp=Dx yo= 3.2 -2K 2. to-(D²+2}y=x²+xre+cos3X éte…

Q: What is the particular solution of y'=x²-2x+3, ify=4whenx=0?

A: What is Ordinary DIfferential Equation: A differential equation having one or more functions of one…

Q: b) y(4) + y(3) – 7y(2) – y' + 6y = 0. c) y(4) – y(3) – 7y(2) + y' + 6y = 0.

A: Solve the given differential equation Here I am giving answer of 1 st problem if you want answer of…

Q: 4. Solve Cauchy problem: y" - 5y' + 4y = 0, y(0) = 1, y'(0) = 2

A: To solve the following Cauchy problem.

Q: ==== 3y' - y + 6ety3/2 = 0, y(0) = 1/4.

Q: Obtain two linearly independent solutions valid near x = 0 x(1-x)y" - 3y' + 2y = 0

Q: Obtain two linearly independent solutions valid near x = 0 x(1 - x)y" - 3y' + 2y = 0

Q: 17. Evaluate L{y"(1) + 2y'(1) + y(1)}; y'(0) = -1, y(0) = 5. Let L {y()} = {s). A) (s°+2s+ 1)Y(s) –…

A: Here we use basic laplace transform

Concept explainers

Differential Equations

Have you ever observed the movement of the satellites and rockets or how the planets go around the sun in their orbits? How will you determine their motion? Definitely they follow some scientific laws and these kinds of problems are framed as differential equations.

Question

Problem 4 (Non-homogeneous Equations Redux). By now we are very familiar with a procedure for solving a linear, second-order initial value problem

\[
y'' + p(t)y' + q(t)y = f(t), \quad y(t_0) = y_0, \quad y'(t_0) = v_0.
\]

1. First, find the general solution to the associated homogeneous equation

\[
y'' + p(t)y' + q(t)y = 0.
\]

2. Find a particular solution to the non-homogeneous equation

\[
y'' + p(t)y' + q(t)y = f(t)
\]

and add it to the homogeneous solution.

3. Plug in the initial conditions to find the solution to the IVP.

There is actually another way we can proceed, and it allows us to plug the initial conditions into the homogeneous solution before finding the particular solution.

(a) Suppose $ u_1(t) $ is a solution to the initial value problem

\[
y'' + p(t)y' + q(t)y = 0, \quad y(t_0) = y_0, \quad y'(t_0) = v_0,
\]

and $ u_2(t) $ is a solution to

\[
y'' + p(t)y' + q(t)y = f(t), \quad y(t_0) = 0, \quad y'(t_0) = 0.
\]

Show that $ y(t) = u_1(t) + u_2(t) $ is a solution to the original initial value problem (1).

(b) Thanks to Variation of Parameters, we know that if $\{y_1, y_2\}$ is a fundamental set of solutions to the associated homogeneous equation, then

\[
y_p(t) = v_1(t)y_1(t) + v_2(t)y_2(t)
\]

is a particular solution, where

\[
v_1(t) = - \int_{t_0}^{t} \frac{y_2(\tau)f(\tau)}{y_1(\tau)y_2'(\tau) - y_1'(\tau)y_2(\tau)} \, d\tau,
\quad v_2(t) = \int_{t_0}^{t}

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 2 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

$Differential Equation$

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.

Similar questions

y" + 6y' + 9y = 7+ e-4x
Obtain general solution of the D.E. y + (1/3)y = e'y A) 1/x = e (c - 3x) B 1/y³ = e* (c - 3x) 1/y = e* (c - 3y) 1/x = e* (c - 3y)
Consider the following IVP: y" + 3y + 2y = 12x²; y(0) = a, y'(0) = b Find the solution. A y(x) = c₁₂e-²x + c₂e (B © 1 c₂e-2x+6x³-18x²+2 y(x) = c₁₂e-²x + c₂ex + 3x² - 18x+21 -3x y(x) = c₂₁e² + c₂e -2x + 2x² - 6x +4 y(x) = c₂e²x + c₂e² -2x+6x²-18x+21
3. Find the general solution of the following DE. y' = 3 (y + 1) O y = ce³x - 1 3y = ce³x - 1 y = 3ce³x - 1 O y = ce³x - 3/
11. Find " if х* + у? %3D ху. dx А. у'— х-2у 2х-у B. y'= У+2х X-2y C. y'= Y-2x 2y-x 2y+x D. y'= х+2у
Find the solution to 2y'' + y' − 3y = 0, y(0) = a, y'(0) = b.
The general solution to D. 2y(4) – y'" – 2y" +y' = ( is given by y(x) = cje* + c2xe*+ Cze + c4e* cje¯* + C2e C3 + C4e* cje* + c2x e* + Cze¯* + c4ez* cje-* + c2x e¯*+ Cze" + c4eż* cje* + czez* + C3 + C4xe* C1 + C2e* + Cze * + C4e* C1 + c2e* + Cze¯* +
2y³ + y_dy y* + y2 + 1/ 6y /9 + e-6y -
Solve y′′+ 6y′+ 9y = u_3(t), y(0) = 1, y′(0) = 2
Given an IVP d"y + an-1(x) dn-ly dy +...+ a1 (x) dx + ao (x)y = g(x) drr-1 у (хо) — Уо, у (tо) — Ул, -.., у" "(2о) — Ул-1 If the coefficients an (x), ..., ao (x) and the right hand side of the equation g(x) are continuous on an interval I and if an (x) +0 on I then the IVP has a unique solution for the point xo e I that exists on the whole interval I. Consider the IVP on the whole real line d'y - 36) dx4 dy + y= sin(x) 1 (x² dz3 x2 + 36 dx у(3) — 916, у (3) — 14, у"(3) — 8, у" (3) — 8, The Fundamental Existence Theorem for Linear Differential Equations guarantees the existence of a unique solution on the interval
2t. у" -у - бу 3D56(t-3)+ e 2; у(0)30, у'(0)-5 y(t) =O %3D
Identify if the series converges or diverges, and if it converges, solve for its sum.