use the annihilator method to show y"+3y'-4y=8x+5ex, y(0)=1, y'(0)=2

use the annihilator method to show y"+3y'-4y=8x+5e^x, y(0)=1, y'(0)=2

Expert Solution

Colby Messinger answered 1 year ago

Question : y'''+4y' =0 , y'''-2y''+4y'-8y=0 , y'''-3y''+3y'-y=0 , y^4 -4y'''+6y''-4y+y=0 , y^4+6y''+9y=0 , y^6+y'''=0

Solve the IVP: y’’’ – y ’= 2sinx where: y(0)=0, y’(0)=0, y”(0)=1 Use an annihilator method,...

Solve the IVP: y’’’ – y ’= 2sinx where: y(0)=0, y’(0)=0, y”(0)=1 Use an annihilator method, please.

Again considering y'' + 4y' + 3y = 0: (a) Solve the IVP y'' + 4y'...

Again considering y'' + 4y' + 3y = 0: (a) Solve the IVP y'' + 4y' + 3y = 0; y(0) = 1, y'(0) = α where α > 0. (b) Determine the coordinates (tm,ym) of the maximum point of the solution as a function of α. (c) Determine the behavior of tm and ym as α →∞.

la place transform of 1. y´´+4y´+3y=0. y(0)=3, y´(0)=1 2. y´´+2y´+y=0. y(0)=1, y´(0)=1

Use Laplace Transforms to solve the following second-order differential equation: y"-3y'+4y=xe2x where y'(0)=1 and y(0)=2

Solve the differential equation. y''-3y'-4y=5e^4x initial conditions: y(0)=2 y'(0)=4

Solve y′′ + 4y′ + 3y = 15e^2t given y(0) = −7,y′(0) = 16 by the...

Solve y′′ + 4y′ + 3y = 15e^2t given y(0) = −7,y′(0) = 16 by the method of Laplace transforms.

Solve the ordinary differential equation analytically: y''-4y-+3y = 5cos(x) + e^(2x) y(0)=1, y'(0)=0

A) Solve the initial value problem: 8x−4y√(x^2+1) * dy/dx=0 y(0)=−8 y(x)= B) Find the function y=y(x) (for...

A) Solve the initial value problem: 8x−4y√(x^2+1) * dy/dx=0 y(0)=−8 y(x)= B) Find the function y=y(x) (for x>0 ) which satisfies the separable differential equation dy/dx=(10+16x)/xy^2 ; x>0 with the initial condition y(1)=2 y= C) Find the solution to the differential equation dy/dt=0.2(y−150) if y=30 when t=0 y=

Solve the initial value problem: y'' + 4y' + 4y = 0; y(0) = 1, y'(0)...

Solve the initial value problem: y'' + 4y' + 4y = 0; y(0) = 1, y'(0) = 0. Solve without the Laplace Transform, first, and then with the Laplace Transform.

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