Question

Solve the initial value problem: y''−2y'+y=e^t/(1+t^2), y(0) = 1, y'(0) = 0.

Answer #1

For the initial value problem
• Solve the initial value problem.
y' = 1/2−t+2y withy(0)=1

Solve the differential equation with initial value
y''−2y'+y=e^t/(1+t^2), y(0) = 1, y'(0) = 0.

For 2y' = -tan(t)(y^2-1) find general solution (solve for y(t))
and solve initial value problem y(0) = -1/3

solve the initial value problem y''-2y'+5y=u(t-2) y(0)=0
y'(0)=0

Solve the initial value problem below for the Cauchy-Euler
equation
t^2y"(t)+10ty'(t)+20y(t)=0, y(1)=0, y'(1)=2
y(t)=

for the given initial value problem: (2-t)y' + 2y
=(2-t)3(ln(t)) ; y(1) = -2
solve the initial value problem

Solve the initial value problem xy′ +2y = e^x2 , y(1) = −2

Use Laplace transform to solve the following initial value
problem: y '' − 2y '+ 2y = e −t , y(0) = 0 and y ' (0) =
1
differential eq

1. Solve the following initial value problem using Laplace
transforms.
d^2y/dt^2+ y = g(t) with y(0)=0 and dy/dt(0) = 1 where g(t) = t/2
for 0<t<6 and g(t) = 3 for t>6

Use Laplace transforms to solve the given initial value
problem.
y"-2y'+5y=1+t y(0)=0 y’(0)=4

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