Differential Equations - Undetermined Coefficients

Find the general solution for $y'' - 3y' - 4y = -25\cos(2t)$ .

$y = c_1e^{4t} + c_2e^{-t} + \frac{3}{2}\sin(3t) + 2\cos(3t)$

$y =\frac{3}{2}\sin(3t) + 2\cos(3t)$

$y = c_1e^{4t} + c_2e^{-t} + 2\sin(3t) + \frac{3}{2}\cos(3t)$

$y = 2\sin(3t) + \frac{3}{2}\cos(3t)$

$y = c_1e^{4t} + c_2e^{-t} + 4\cos(3t)$

Explanation

This is a higher order inhomogeneous linear differential equation. Because the inhomogeneity is a cosine, we will use variation of parameters to solve it.

First, we find the characteristic equation to solve for the homogenous solution. This gives us $\lambda^2 - 3\lambda - 4 = (\lambda -4)(\lambda + 1) = 0$ .

This tells us that the homogeneous solution is $c_1e^{4t} + c_2e^{-t}$ . As neither of these overlap with our inhomogeneity, we are safe to continue without adding a factor of t.

Thus, let us guess that $y_p = a_1\sin(2t) + a_2\cos(2t)$ . Then,

$y_p' = 4a_1\cos(2t) - 4a_2\sin(2t)$ and

$y_p'' = -4a_1\sin(2t) - 4a_2\cos(2t)$

Plugging into the original equation, we have

$\sin(2t)(-4a_1 + 6a_2 -4a_1) + \cos(2t)(-4a_2 - 6a_1 -4a_2) = -25\cos(2t)$

Which implies that and . Solving via substitution,

$a_1 = \frac{3}{4}a_2$

$-8a_2 - \frac{9}{2}a_2 = -\frac{25}{2}a_2 = -25; a_2 = 2$

$a_1 = \frac{3}{2}$

Thus, the particular solution is $y_p = \frac{3}{2}\sin(2t) + 2\cos(2t)$ and the overall solution is the particular plus the homogeneous.

$y = c_1e^{4t} + c_2e^{-t} + \frac{3}{2}\sin(3t) + 2\cos(3t)$

Find the form of a particular solution to the following differential equation that could be used in the method of undetermined coefficients:

$y'' + 3y= t^{2}e^{2t}$

The form of a particular solution is $e^{2t}(At^2+Bt+C)$ where A,B, and C are real numbers.

The form of a particular solution is $Ae^{2t}(Bt^2)$ where A and B are real numbers.

The form of a particular solution is $e^{2t}(At^3+Bt^2+Ct+D)$ where A,B, C, and D are real numbers.

The form of a particular solution is $e^{2t}(At^2+Bt)$ where A and B are real numbers.

Explanation

We first note that the differential equation has characteristic equation

since the roots of this characteristic polynomial are linearly independent of the forcing function

$t^{2}e^{2t}$ ,

we simply use undetermined coefficient combination rules to figure that the particular solution will be of form

$e^{2t}(At^2+Bt+C)$

Consider the differential equation

The particular solution used in undetermined coefficients will be of what form?

The particular solution will be of form:

$y_{p} = (At^2+Bt+C)sin(2t) + (Dt^2+Et+F)cos(2t)$

where A,B,C,D,E, and F are real numbers

The particular solution will be of form:

$y_{p} = (At^2+Bt+C)sin(2t)$

where A,B, and C are real numbers

The particular solution will be of form:

$y_{p} = (At^2)sin(2t) + (Bt^2)cos(2t)$

where A and B are real numbers

The particular solution will be of form:

$y_{p} = (At^2)sin(2t)$

where A is a real number

Explanation

We first figure that the forcing function is linearly independent to the homogeneous solution solved with the characteristic equation.

Therefore, using proper undetermined coefficients function rules, the particular solution will be of the form:

$y_{p} = (At^2+Bt+C)sin(2t) + (Dt^2+Et+F)cos(2t)$

It is important to note that when either a sine or a cosine is used, both sine and cosine must show up in the particular solution guess.

Solve for a particular solution of the differential equation using the method of undetermined coefficients.

$y''-2y' +5y = 4e^{3t}$

$y_{p} = \frac{e^{3t}}{2}$

$y_{p} = \frac{e^{3t}}{3}$

$y_{p} = 2e^{3t}$

$y_{p} = 3e^{3t}$

Explanation

We start with the assumption that the particular solution must be of the form

$y_{p} =Ae^{3t}$ .

Then we solve the first and second derivatives with this assumption, that is,

$y_p'=3Ae^{3t}$ and $y_p''=9Ae^{3t}$ .

Then we plug in these quantities into the given equation to get:

$9Ae^{3t}-6Ae^{3t} +5Ae^{3t} = 4Ae^{3t}$ , which solves for $A = \frac{1}{2}$ .

Thus, but the method of undetermined coefficients, a particular solution to this differential equation is:

$y_p=\frac{e^{3t}}{2}$

Solve the given differential equation by undetermined coefficients.

$y=c_1e^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{7}{8}$

$y=c_1e^{4x}+c_2e^{4x}+\frac{3}{2}x+\frac{7}{8}$

$y=c_1xe^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{7}{8}$

$y=c_1e^{4x}+c_2xe^{4x}+\frac{2}{3}x+\frac{7}{8}$

$y=c_1e^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{8}{7}$

Explanation

First solve the homogeneous portion:

$\y''-8y'+16y=0 \m^2-8m+16=0 \(m-4)(m-4)=0$

Therefore, is a repeated root thus one of the complimentary solutions is,

$y_c=c_1e^{4x}+c_2xe^{4x}$

Now find the remaining complimentary solution .

$\y_p=Ax+B \y_p'=A \y_p''=0$

Now solve for and .

$\0-8A+16(Ax+B)=24x+2 \-8A+16Ax+16B=24x+2$

Where

$\16A=24 \\A=\frac{24}{16}=\frac{3}{2}$

and

$\-8A+16B=2 \\-8\left(\frac{3}{2} \right )+16B=2 \\-\frac{24}{2}+16B=2 \\-12+16B=2 \\16B=14 \\B=\frac{14}{16}=\frac{7}{8}$

Therefore,

$y_p=\frac{3}{2}x+\frac{7}{8}$

Now, combine both of the complimentary solutions together to arrive at the general solution.

$\y=y_c+y_p \y=c_1e^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{7}{8}$

Find the form of a particular solution to the following Differential Equation (Do NOT Solve)

$y^{''} + 4y = 4t^3\sin{3t}$

$(A_3t^4+A_2t^3+A_1t^2+A_0t)\cos{3t} + (B_3t^4+B_2t^3+B_1t^2+B_0t)\sin{3t}$

$(A_3t^3+A_2t^2+A_1t+A_0)\cos{3t} + (B_3t^3+B_2t^2+B_1t+B_0)\sin{3t}$

$(A_3t^4+A_2t^3+A_1t^2+A_0t)e^{-t}\cos{3t} + (B_3t^4+B_2t^3+B_1t^2+B_0t)e^{-t}\sin{3t}$

$(A_3t^4+A_2t^3+A_1t^2+A_0t)e^{-3t}\cos{t} + (B_3t^4+B_2t^3+B_1t^2+B_0t)e^{-3}\sin{t}$

None of the other answers.

Explanation

The form of a guess for a particular solution is

$(A_3t^4+A_2t^3+A_1t^2+A_0t)\cos{3t} + (B_3t^4+B_2t^3+B_1t^2+B_0t)\sin{3t}$

Solve the given differential equation by undetermined coefficients.

$y=c_1e^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{7}{8}$

$y=c_1e^{4x}+c_2e^{4x}+\frac{3}{2}x+\frac{7}{8}$

$y=c_1xe^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{7}{8}$

$y=c_1e^{4x}+c_2xe^{4x}+\frac{2}{3}x+\frac{7}{8}$

$y=c_1e^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{8}{7}$

Explanation

First solve the homogeneous portion:

$\y''-8y'+16y=0 \m^2-8m+16=0 \(m-4)(m-4)=0$

Therefore, is a repeated root thus one of the complimentary solutions is,

$y_c=c_1e^{4x}+c_2xe^{4x}$

Now find the remaining complimentary solution .

$\y_p=Ax+B \y_p'=A \y_p''=0$

Now solve for and .

$\0-8A+16(Ax+B)=24x+2 \-8A+16Ax+16B=24x+2$

Where

$\16A=24 \\A=\frac{24}{16}=\frac{3}{2}$

and

$\-8A+16B=2 \\-8\left(\frac{3}{2} \right )+16B=2 \\-\frac{24}{2}+16B=2 \\-12+16B=2 \\16B=14 \\B=\frac{14}{16}=\frac{7}{8}$

Therefore,

$y_p=\frac{3}{2}x+\frac{7}{8}$

Now, combine both of the complimentary solutions together to arrive at the general solution.

$\y=y_c+y_p \y=c_1e^{4x}+c_2xe^{4x}+\frac{3}{2}x+\frac{7}{8}$

Undetermined Coefficients

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