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Differential Equations : Undetermined Coefficients

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Example Questions

Example Question #1 : Undetermined Coefficients

Solve the given differential equation by undetermined coefficients.

y''-8y'+16y=24x+2

Possible Answers:

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

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

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

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

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

Correct answer:

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

Explanation:

First solve the homogeneous portion:

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

Therefore, m=4 is a repeated root thus one of the complimentary solutions y_c is,

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

Now find the remaining complimentary solution y_p .

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

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.

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

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Example Question #1 : Undetermined Coefficients

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

Possible Answers:

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

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

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

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

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

Correct answer:

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

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 -8a_1 + 6a_2 = 0 and -8a_2 - 6a_1 = -25 . 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.

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

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Example Question #1 : Undetermined Coefficients

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

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

Possible Answers:

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 $e^{2t}(At^3+Bt^2+Ct+D)$ 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.

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

Correct answer:

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

Explanation:

We first note that the differential equation has characteristic equation

r^2+3=0 ,

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

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

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Example Question #11 : Higher Order Differential Equations

Consider the differential equation

y''+2y'+2y = t^2sin(2t)