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Differential Equations : Higher-Order Differential Equations

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Example Questions

Example Question #1 : Higher Order Differential Equations

For the initial value problem

$x(x-1)y^{'''} -3xy^{''} + 6x^2y' - \cos{(x)}y = \sqrt{x+5}$

y(x_0) = 1, y'(x_0) = 0, y''(x_0) = 7

Which if the following intervals containing x_0 do NOT guarantees the existence of a unique solution?

Possible Answers:

(-4,-3)

(0,2)

(-5,0)

(0,1)

(1,5)

Correct answer:

(0,2)

Explanation:

Putting the equation in standard form we get that

$y^{'''} - \frac{3}{(x-1)}y^{''} + \frac{6x}{(x-1)}y' - \frac{\cos{x}}{x(x-1)}y = \frac{\sqrt{x+5}}{x(x-1)}$

We need to find where these coefficients are simultaneously continuous. This is where

$(-5,0), (0,1), (1,\infty)$ . The choice that is not a subset of these is (0,2)

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Example Question #1 : Higher Order Differential Equations

Solve the initial value problem y'' - 4y' + 8y = 0 for y(0) = 1 and y'(0) = 2

Possible Answers:

$y = 2e^{2t}\cos(2t) - 2e^{2t}\sin(2t)$

$y = e^{2t}\cos(2t)$

$y = 4e^{2t} - 2e^{-2t}$

$y = e^{2t}\cos(2t) - e^{2t}\sin(2t)$

$y = e^{2t}\sin(2t)$

Correct answer:

$y = e^{2t}\cos(2t)$

Explanation:

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

$\lambda^2 - 4\lambda + 8 = 0$ We then solve the characteristic equation and find that $(\lambda - 2)^2 + 4 = 0; \lambda = 2 \pm 2i$ (Use the quadratic formula if you'd like) This lets us know that the basis for the fundamental set of solutions to this problem (solutions to the homogeneous problem) contains $e^{2t}\sin(2t), e^{2t}\cos(2t)$ .

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, $y = c_1e^{2t}\sin(2t) + c_2e^{2t}\cos(2t)$ . Plugging in our initial condition, we find that 1 = c_2 . To plug in the second initial condition, we take the derivative and find that $y' = 2c_1e^{2t}(\sin(2t) + \cos(2t)) + 2e^{2t}(\cos(2t) - sin(2t))$ . Plugging in the second initial condition yields 2 = 2c_1 + 2 . Solving this simple system of linear equations shows us that

c_1 = 0; c_2 = 1

Leaving us with a final answer of $y = e^{2t}\cos(2t)$

(Note, it would have been very simple to find the right answer just by taking derivatives and plugging in, but this is not overly helpful for non-multiple choice questions)

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Example Question #2 : Higher Order Differential Equations

Find the general solution to y''' - y'' + y' - y = 0 .

Possible Answers:

$y = c_1e^{-t} + c_2\sin(t) + c_3\cos(t)$

$y = e^{t}(c_1 + c_2\sin(2t) + c_3\cos(2t))$

$y = c_1e^{t} + c_2\sin(2t) + c_3\cos(2t)$

$y = c_1e^{t} + c_2\sin(t) + c_3\cos(t)$

$y = c_2e^{t}\sin(2t) + c_3e^{t}\cos(2t)$

Correct answer:

$y = c_1e^{t} + c_2\sin(t) + c_3\cos(t)$

Explanation:

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

$\lambda^3 - \lambda^2 + \lambda - 1 = 0$ To factor this, in this case we may use factoring by grouping. More generally, we may use horner's scheme/synthetic division to test possible roots. Here are both methods shown.

$\lambda^3 - \lambda^2 + \lambda - 1 = (\lambda^3 + \lambda) - (\lambda^2 + 1) = \lambda(\lambda^2 + 1) - (\lambda^2 +1) = (\lambda^2 + 1)(\lambda - 1)$

Alternatively, the rational root theorem suggests that we try -1 or 1 as a root of this equation. Using horner's scheme, we see

$1 \left |\begin{matrix}1 && -1 && 1 && -1\\ && 1 && 0 && 1 \\ \hline \end{matrix}\right.$

$1 \: \, \: \: \: \, \, \: \: \: 0 \: \: \, \, \: \: \: \, \: \:1 \: \: \, \, \: \: \: \, \: \: 0$

Which tells us the the polynomial factors into $(\lambda - 1)(\lambda^2 + 1)$ and that $\lambda = 1, \pm i$ . This means that the fundamental set of solutions is e^t, sin(t), cos(t)

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, $y = c_1e^{t} + c_2\sin(t) + c_3\cos(t)$ . As this is not an initial value problem and just asks for the general solution, we are done.

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Example Question #3 : Higher Order Differential Equations

Solve the initial value problem y'' - 4y' - 5y = 0 for y(0) = 0 and y'(0) = 6 .

Possible Answers:

$y = -e^{5t} + e^{-t}$

$y = e^{5t} - e^{-t}$

$y = e^{3t} + 3e^{-3t}$

$y = 5e^{5t} - e^{-t}$

$y = -e^{3t} - 3e^{-3t}$

Correct answer:

$y = e^{5t} - e^{-t}$

Explanation:

This is a linear higher order differential equation. First, we need the characteristic equation, which is just obtained by turning the derivative orders into powers to get the following:

$\lambda^2 - 4\lambda - 5 = 0$ We then solve the characteristic equation and find that $(\lambda - 5)(\lambda + 1) = 0 ; \lambda = 5, -1$ This lets us know that the basis for the fundamental set of solutions to this problem (solutions to the homogeneous problem) contains $e^{5t}, e^{-t}$ .

As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, $y = c_1e^{5t} + c_2e^{-t}$ . Plugging in our initial condition, we find that 0 = c_1 + c_2 . To plug in the second initial condition, we take the derivative and find that $y' = 5c_1e^{5t} - c_2e^{-t}$ . Plugging in the second initial condition yields 6 = 5c_1 - c_2 . Solving this simple system of linear equations shows us that

c_1 = 1; c_2 = -1

Leaving us with a final answer of $y = e^{5t} - e^{-t}$

(Note, it would have been very simple to find the right answer just by taking derivatives and plugging in, but this is not overly helpful for non-multiple choice questions)

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Example Question #1 : Higher Order Differential Equations

Solve the following homogeneous differential equation:

y''-4y'+4y=0

Possible Answers:

$y(t) = c_{1}e^{2t}+c_2e^{2t}$

where c_1 and c_2 are constants

$y(t) = c_{1}e^{2t}+c_2te^{2t}$

where c_1 and c_2 are constants

$y(t) = c_{1}e^{4t}+c_2te^{4t}$

where c_1 and c_2 are constants

$y(t) = c_1te^{2t}$

where c_1 and c_2 are constants

Correct answer:

$y(t) = c_{1}e^{2t}+c_2te^{2t}$

where c_1 and c_2 are constants

Explanation:

The ode has a characteristic equation of r^2 -4r+4=0 .

This yields the double root of r=2. Then the roots are plugged into the general solution to a homogeneous differential equation with a repeated root.

$y= c_1e^{rt}+c_2te^{rt}$ (for real repeated roots)

Thus, the solution is,

$y(t) = c_{1}e^{2t}+c_2te^{2t}$

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Example Question #1 : Higher Order Differential Equations

Solve the General form of the differential equation:

y''-5y'+6y = 0

Possible Answers:

$y(t)= c_1e^{2t}+c_2te^{3t}$

Where c_1 and c_2 are arbitrary constants

$y(t)= c_1e^{-2t}+c_2e^{3t}$

Where c_1 and c_2 are arbitrary constants

$y(t)= c_1e^{2t}+c_2e^{3t}$

Where c_1 and c_2 are arbitrary constants

$y(t)= c_1e^{-2t}+c_2e^{-3t}$

Where c_1 and c_2 are arbitrary constants

Correct answer:

$y(t)= c_1e^{2t}+c_2e^{3t}$

Where c_1 and c_2 are arbitrary constants

Explanation:

This differential equation has a characteristic equation of

r^2 -5r+6=0 , which yields the roots for r=2 and r=3. Once the roots or established to be real and non-repeated, the general solution for homogeneous linear ODEs is used. this equation is given as:

$y(t)= c_1e^{r_1t}+c_2e^{r_2t}$

with r being the roots of the characteristic equation.

Thus, the solution is

$y(t)= c_1e^{2t}+c_2e^{3t}$

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Example Question #1 : Higher Order Differential Equations

Solve the general homogeneous part of the following differential equation:

y'' +y'-12y = cos(3t)

Possible Answers:

$y(t)= c_1te^{3t}+c_2te^{-4t}$

Where c_1 and c_2 are arbitrary but not meaningless constants

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

Where c_1 and c_2 are arbitrary but not meaningless constants

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

Where c_1 and c_2 are arbitrary but not meaningless constants

$y(t)= c_1te^{3t}+c_2e^{4t}$

Where c_1 and c_2 are arbitrary but not meaningless constants

Correct answer:

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

Where c_1 and c_2 are arbitrary but not meaningless constants

Explanation:

We start off by noting that the homogeneous equation we are trying to solve is given as

y'' +y'-12y = 0 .

This differential equation thus has characteristic equation of

r^2+r-12=0 .

This has roots of r=3 and r=-4, therefore, the general homogeneous solution is given by:

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

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Example Question #1 : Higher Order Differential Equations

Solve the following homogeneous differential equation:

y''-10y'+25y = 0

Possible Answers:

$y(t)= c_1e^{4t}+c_2te^{7t}$

where c_1 and c_2 are constants

$y(t)= c_1e^{4t}+c_2e^{5t}$

where c_1 and c_2 are constants

$y(t)= c_1e^{-5t}+c_2te^{-5t}$

where c_1 and c_2 are constants

$y(t)= c_1e^{5t}+c_2te^{5t}$

where c_1 and c_2 are constants

Correct answer:

$y(t)= c_1e^{5t}+c_2te^{5t}$

where c_1 and c_2 are constants

Explanation:

This differential equation has characteristic equation of:

r^2 - 10r + 25 It must be noted that this characteristic equation has a double root of r=5.

Thus the general solution to a homogeneous differential equation with a repeated root is used.

This is equation is

$y(t)= c_1e^{r_1t}+c_2te^{r_2t}$ in the case of a repeated root such as this, r_1 = r_2 and is the repeated root r=5.

Therefore, the solution is

$y(t)= c_1e^{5t}+c_2te^{5t}$

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Example Question #21 : Differential Equations

Find a general solution to the following Differential Equation

$y^{'''} + 2y^{''} - 11y' - 12y = 0$

Possible Answers:

$y(t) = C_1e^{-3x} + C_2e^{4x} + C_3e^{-x}$

$y(t) = C_1e^{-3x} + C_2e^{4x} + C_3e^{x}$

$y(t) = C_1e^{3x} + C_2e^{-4x} + C_3e^{x}$

$y(t) = C_1e^{3x} + C_2e^{-4x} + C_3e^{-x}$

$y(t) = C_1e^{2x} + C_2e^{-6x} + C_3e^{-x}$

Correct answer:

$y(t) = C_1e^{3x} + C_2e^{-4x} + C_3e^{-x}$

Explanation:

Solving the auxiliary equation

r^3+2r^2-11r-12 = 0

Trying out candidates for roots from the Rational Root Theorem we have a root r = -1 .

Factoring completely we have

(r+1)(r^2+r-12) = (r+4)(r-3)(r+1)=0

Our general solution is

$y(t) = C_1e^{3x} + C_2e^{-4x} + C_3e^{-x}$

where C_1, C_2, C_3 are arbitrary constants.

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Example Question #2 : Higher Order Differential Equations

Solve the given differential equation by undetermined coefficients.

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

Possible Answers:

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

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

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

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

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

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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Differential Equations : Higher-Order Differential Equations

Study concepts, example questions & explanations for Differential Equations

All Differential Equations Resources

Example Questions

Example Question #1 : Higher Order Differential Equations

Example Question #1 : Higher Order Differential Equations

Example Question #2 : Higher Order Differential Equations

Example Question #3 : Higher Order Differential Equations

Example Question #1 : Higher Order Differential Equations

Example Question #1 : Higher Order Differential Equations

Example Question #1 : Higher Order Differential Equations

Example Question #1 : Higher Order Differential Equations

Example Question #21 : Differential Equations

Example Question #2 : Higher Order Differential Equations

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