Find the solution of the initial-value problem. y'' + y = 3 + 5 sin(x), y(0)...

Find the solution of the initial-value problem. y'' + y = 3 + 5 sin(x), y(0) = 5, y'(0) = 8

Expert Solution

Colby Messinger answered 2 years ago

Find the solution of the initial value problem y′′−2y′−3 y=15te2t, y(0)=2, y′(0)=0.

Find the solution of the given initial value problem: y′′′+y′=sec(t), y(0)=6, y′(0)=7, y′′(0)=−3

Homework 1.1. (a) Find the solution of the initial value problem x' = x^(3/8) , x(0)=1...

Homework 1.1. (a) Find the solution of the initial value problem x' = x^(3/8) , x(0)=1 , for all t, where x = x(t). (b) Find the numerical solution on the interval 0 ≤ t ≤ 1 in steps of h = 0.05 and compare its graph with that of the exact solution. You can do this in Excel and turn in a printout of the spreadsheet and graphs.

Find the solution of the initial value problem y′′+12y′+32y=0, y(0)=12 and y′(0)=−84.

Find the solution of the given initial value problem. 2y''+y'-4y=0 ; y(0)=0 y'(0)=1

Solve the initial value problem: y'' + y = cos(x) y(0) = 2 y'(0) = -3...

Solve the initial value problem: y'' + y = cos(x) y(0) = 2 y'(0) = -3 y' being the first derivative of y(x), y'' being the second derivative, etc.

Find the solution of the initial value problem: y'' + 4y' + 20y = -3sin(2x), y(0)...

Find the solution of the initial value problem: y'' + 4y' + 20y = -3sin(2x), y(0) = y'(0) = 0

Solve the initial value problem dy/dx = −(2x cos(x^2))y + 6(x^2)e^(− sin(x^2)) , y(0) = −5...

Solve the initial value problem dy/dx = −(2x cos(x^2))y + 6(x^2)e^(− sin(x^2)) , y(0) = −5 Solve the initial value problem dy/dt = (6t^5/(1 + t^6))y + 7(1 + t^6)^2 , y(1) = 8. Find the general solution of dy/dt = (2/t)*y + 3t^2* cos3t

find the solution of the given initial value problem. y′′−2y′−3y=3te^2t y(0)=1 y′(0)=0

(3 pts) Solve the initial value problem 25y′′−20y′+4y=0, y(5)=0, y′(5)=−e2. (3 pts) Solve the initial value...

(3 pts) Solve the initial value problem 25y′′−20y′+4y=0, y(5)=0, y′(5)=−e2. (3 pts) Solve the initial value problem y′′ − 2√2y′ + 2y = 0, y(√2) = e2, y′(√2) = 2√2e2. Consider the second order linear equation t2y′′+2ty′−2y=0, t>0. (a) (1 pt) Show that y1(t) = t−2 is a solution. (b) (3 pt) Use the variation of parameters method to obtain a second solution and a general solution.

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