Question

Find a general solution to the given equation for t<0

y"(t)-1/ty'(t)+5/t^2y(t)=0

Answer #1

Find the general solution to the given differential
equation. 1+(1+ty)e^ty+(1+t^2e^ty) dy/dt=0

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

Find the solution of the given initial value problem.
ty′+3y=t2−t+5, y(1)=5, t>0

a) Find the general solution of the differential equation
y''-2y'+y=0
b) Use the method of variation of parameters to find the general
solution of the differential equation y''-2y'+y=2e^t/t^3

Find the general solution of the second-order nonhomogeneous
equation:
ty″ − y′ = (t^2) + t

Find the general solution of the given differential
equation.
y'' − y' − 2y = −8t + 6t2
y(t) =

Given y1(t)=t^2 and y2(t)=t^-1 satisfy the corresponding
homogeneous equation of
t^2y''−2y=2−t3, t>0
Then the general solution to the non-homogeneous equation can be
written as y(t)=c1y1(t)+c2y2(t)+yp(t)
yp(t) =

find y(t) solution of the initial value problem
y'=(2y^2 +bt^2)/(ty), y(1)=1 t>o

Use variation of parameters to find a general solution to the
differential equation given that the functions y 1 and y 2 are
linearly independent solutions to the corresponding homogeneous
equation for t>0.
ty"-(t+1)y'+y=30t^2 ; y1=e^t , y2=t+1
The general solution is y(t)= ?

Find the general solution to:
ty' -2y = t3sint

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