Question

Solve y'' + 2y' + 4y = 1 subject to y(0) = y'(0) = 0 (differential equations)

Answer #1

Solve the given system of differential equations by
elimination.
x'-2x-y = 1
x+y'-4y=0

Solve the following differential equations using inspection:
1) y”+4y=12
2) y””+4y”+4y=-20
3) (D^4 -4D^2)y=24
4) y”-y=x-1
5) D(D-3)y=4
6) (D^2+2D-8)(D+3)y=0
7) y”-y’-2y=18xe^(2x)

Solve the following differential equations
y''-4y'+4y=(x+1)e2x (Use Wronskian)
y''+(y')2+1=0 (non linear second order equation)

differential equations solve
(2xy+6x)dx+(x^2+4y^3)dy, y(0)=1

(differential equations): solve for x(t) and y(t)
2x' + x - (5y' +4y)=0
3x'-2x-(4y'-y)=0
note: Prime denotes d/dt

solve differential equation
y^(4) +2y''' +2y''=0

Differential Equations: Use the Laplace transform to solve the
given initial value problem:
y′′ −2y′ +2y=cost;
y(0)=1,
y′(0)=0

x^2y'' − 3xy'+ 4y = 0 ; y(1)=5 y'(1)=3
differential equation using the Cauchy-Euler method

x^2y'' − 3xy'+ 4y = 0 ; y(1)=5 y'(1)=3
differential equation using the Cauchy-Euler method

(a) Solve the differential equations.
4y´´ + 12y´ + 9y= 0
(b) Find the general solution of the ODE
y´´- 5y´+4y=e^4t
(c) Solve the following initial value problem:
y´´+ y =0, y(0)=2, y´(0)=2

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