Question

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) =

Answer #1

In this problem verify that the given functions y1 and y2
satisfy the corresponding homogeneous equation. Then find a
particular solution of the nonhomogeneous equation.
x^2y′′−3xy′+4y=31x^2lnx, x>0, y1(x)=x^2, y2(x)=x^2lnx. Enter an
exact answer.

Consider the differential equation:
66t^2y''+12t(t-11)y'-12(t-11)y=5t^3, . You can verify that y1 = 5t
and y2 = 4te^(-2t/11)satisfy the corresponding homogeneous
equation.
The Wronskian W between y1 and y2 is W(t) =
(-40/11)t^2e^((-2t)/11)
Apply variation of parameters to find a particular solution.
yp = ?????

Given that y1 = t, y2 = t 2 are solutions to the homogeneous
version of the nonhomogeneous DE below, verify that they form a
fundamental set of solutions. Then, use variation of parameters to
find the general solution y(t).
(t^2)y'' - 2ty' + 2y = 4t^2 t > 0

Use variation of parameters to find a general solution to the
differential equation given that the functions y1 and y2 are
linearly independent solutions to the corresponding homogeneous
equation for t>0.
y1=et y2=t+1
ty''-(t+1)y'+y=2t2

Consider the differential equation
L[y] = y′′ + p(t)y′ + q(t)y = f(t) + g(t), and suppose L[yf] =
f(t) and L[yg] = g(t).
Explain why yp = yf + yg is a solution to L[y] = f + g.
Suppose y and y ̃ are both solutions to L[y] = f + g, and
suppose
{y1, y2} is a fundamental set of solutions to the homogeneous
equation L[y] = 0. Explain why
y = C1y1 + C2y2 + yf...

The nonhomogeneous equation t2 y′′−2 y=29 t2−1, t>0, has
homogeneous solutions y1(t)=t2, y2(t)=t−1. Find the particular
solution to the nonhomogeneous equation that does not involve any
terms from the homogeneous solution.

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)= ?

The nonhomogeneous equation t2 y′′−2 y=19
t2−1, t>0, has homogeneous solutions
y1(t)=t2, y2(t)=t−1. Find the particular
solution to the nonhomogeneous equation that does not involve any
terms from the homogeneous solution.
Enter an exact answer.
Enclose arguments of functions in parentheses. For example,
sin(2x).
y(t)=

1. For each of these problems, (i) verify by direct substitution
that y1 and y2 are both solutions of the ODE, and (ii) find the
particular solution in the form y(x) = c1y1(x) + c2y2(x) that
satisfies the given initial conditions. (a) y''+5y'-6y=0, y1(x) =
e^−6x , y2(x) = e^x , y(0)=2, y'(0)=1

1) find a solution for a given differential equation
y1'=3y1-4y2+20cost ->y1 is not y*1 & y2 is not y*2
y2'=y1-2y2
y1(0)=0,y2(0)=8
2)by setting y1=(theta) and y2=y1', convert the following 2nd
order differential equation into a first order system of
differential equations(y'=Ay+g)
(theta)''+4(theta)'+10(theta)=0

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