Question

5.
Find a general solution of t^2x′′ − 5tx′ + 5x = 6t^3, t > 0
using the method of undetermined coefficients

Answer #1

4.
Find a general solution of t^2x′′ − 5tx′ + 5x = 6t^3, t > 0
using the method of variation of parameters.

Find the general solution of y'' − 2y' = sin(5x) using the
method of undetermined coefficients

Find the general solution of the equation using the method of
undetermined coefficients: y''-y'=5sin(2x)

Use the method of Undetermined Coefficients to find the general
solution
y'' - y' -2y = e^(2x)

Using undetermined coefficients superposition approach find the
general solution for: y'' - y = e^(x) + 2x^(2) -5

Solve the following second order differential equations:
(a) Find the general solution of y'' − 2y' = sin(3x) using the
method of undetermined coefficients.
(b) Find the general solution of y'' − 2y'− 3y = te^−t using the
method of variation of parameters.

find the solution to the following ivp
dy/dy-2ty=6t^2e^(t^2),y(0)=5

Find the arc length of a(t)=(2cosh3t,-2sinh3t,6t) for 0≤t≤5

4. Find the general
solution to the homogeneous equation, then use the method of
undetermined coefficients to find the particular solution
y’’− 2y’ + 2y =
360e−t sin3t.

Find the general solution to the system x" = 5x +
2y, y" = 2x + 8y. Note that the
system as stated above has second derivatives. Once you write it as
a first-order system, the characteristic equation will be
biquadratic (quadratic in λ2), so you can solve for
λ2 and then take square roots to get the
eigenvalues.

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