Question

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

Answer #1

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

Solve the differential equation:
a) y'= t^2y^3 / t^3+6
b) y'= x(e^x^2 +2) / 6y^2 ; y(0) =1

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

ﬁnd the general solution of the given differential equation
1. y''−2y'+2y=0
2. y''+6y'+13y=0
ﬁnd the solution of the given initial value problem
1. y''+4y=0, y(0) =0, y'(0) =1
2. y''−2y'+5y=0, y(π/2) =0, y'(π/2) =2
use the method of reduction of order to ﬁnd a second solution of
the given differential equation.
1. t^2 y''+3ty'+y=0, t > 0; y1(t) =t^−1

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

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

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

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

Given the differential equation
y''−2y'+y=0, y(0)=1, y'(0)=2
Apply the Laplace Transform and solve for Y(s)=L{y}
Y(s) =
Now solve the IVP by using the inverse Laplace Transform
y(t)=L^−1{Y(s)}
y(t) =

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