Question

The nonhomogeneous equation t^{2} y′′−2 y=19
t^{2}−1, t>0, has homogeneous solutions
y1(t)=t^{2}, 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)=

Answer #1

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.

Find the solution of the initial value problem
y′′+4y=t2+7et y(0)=0, y′(0)=2.
Enter an exact answer.
Enclose arguments of functions in parentheses. For example,
sin(2x).

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.

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

The function y1(t) = t is a solution to the
equation.
t2 y'' + 2ty' - 2y = 0, t > 0
Find another particular solution y2 so that
y1 and y2 form a fundamental set of
solutions. This means that, after finding a solution y2,
you also need to verify that {y1, y2} is
really a fundamental set of solutions.

The indicated functions are known linearly independent solutions
of the associated homogeneous differential equation on (0, ∞). Find
the general solution of the given nonhomogeneous equation.
x2y'' + xy' + y = sec(ln(x))
y1 = cos(ln(x)), y2 = sin(ln(x))

Determine whether the equation is exact. If it is exact, find
the solution. If it is not, enter NS.
(y/x+7x)dx+(ln(x)−2)dy=0, x>0
Enclose arguments of functions in parentheses. For example,
sin(2x).

Choose the correct answers
If y1 and y2 are two
solutions of a nonhomogeneous equation ayjj+
byj+ cy =f (x), then
their difference is a solution of the equation
ayjj+ byj+ cy =
0.
If f (x) is continuous everywhere, then there
exists a unique solution to the following initial value
problem.
f (x)yj=
y, y(0) = 0
The differential equation yjj +
t2yj −
y = 3 is linear.
There is a solution to the ODE
yjj+3yj+y...

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) =

Two solutions to the diﬀerential equation y00 + 2y0 + y = 0 are
y1(t) = e−t and y2(t) = te−t. Verify that y1(t) is a solution and
show that y1,y2 form a fundamental set of solutions by computing
the Wronskian

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