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Adamson U
Differential Equations
Practice Quiz

Question:

1
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Find the general solution of the differential equation y'' + 4y' + 4y = 0
3
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Find the general solution of the differential equation y'' - 4y = 0
5
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Find the general solution of the differential equation y'' + y = 0
7
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Find the general solution of the differential equation y'' + y' + y = 0
9
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Find the general solution of the differential equation y'' - 2y' + 2y = 0

Answers:
1
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This differential equation is a first-order linear differential equation, which means that the
highest derivative in the equation is of the first order (in this case, y'), and the coefficients of
the derivatives are constants (in this case, 1 for y' and 2 for y)
...
In this case, the integrating factor is

e^(int(2)dx) = e^(2x), and the general solution is e^(2x)y = (3/5)e^(3x) + C, where C is the
constant of integration
...


2
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The roots are r1= -2 + 2i and r2= -2 - 2i
...

This differential equation is a second-order linear differential equation, which means that the
highest derivative in the equation is of second order (in this case, y''), and the coefficients of
the products are constants (in this case, 1 for y'', 4 for y' and 4 for y)
...
In this case, the characteristic equation is r^2 + 4r + 4 = 0
...
This is a
quadratic equation whose roots are r1= -2 + 2i and r2= -2 - 2i
...
Since the roots of the characteristic
equation are complex, the general solution is described in the form of y = c1e^(-2+2i)x + c2e^(2-2i)x, where c1 and c2 are arbitrary constants
...


3
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This differential equation is a first-order non-homogeneous linear differential equation, which
means that the highest derivative in the equation is of the first order (in this case, y'), the
coefficients of the products are constants (in this case, 1 for y' and -2 for y)
...
To solve this type of differential
equation, one way is to use the method of undetermined coefficients, which consists of finding
a particular solution by guessing a form for the solution that includes arbitrary coefficients and
then solving for these coefficients by plugging the guess into the differential equation and
comparing the resulting expression with the non-homogeneous function
...
Plugging this into the differential equation, we get yp' - 2yp
= Ax^3 - 2Ax^3 = 0, which is not equal to x^2
...


Another way to guess the form of the particular solution is to note that the non-homogeneous
term is a polynomial of degree 2 and that the characteristic equation of the homogeneous part
of the differential equation is r-2 = 0, with a natural root of r=2
...

Plugging this into the differential equation, we get yp' - 2yp = Bx^2e^-2x - 2Bx^2e^-2x = x^2,
equal to the non-homogeneous function
...


To find the general solution of the differential equation, we need to add the available
resolution of the homogeneous equation (y = Ce^-2x) and the particular solution we found
above
...

It can be determined by applying initial or boundary conditions
...
Find the general solution of the differential equation y'' - 4y = 0
The characteristic equation is r^2 - 4 = 0, which gives r1 = 2 and r2 = -2
...
To find the general solution of this differential equation, one method is to
use the characteristic equation, which is obtained by setting the right-hand side of the
differential equation to zero and replacing the derivatives with a variable r
...


To find the roots of this equation, we set it to zero and solve for r: r^2 - 4 = 0, which gives us r1
= 2 and r2 = -2
...


The general solution of a second-order linear homogeneous differential equation is a linear
combination of two linearly independent solutions of the homogeneous equation, which are
represented by e^(r1x) and e^(r2x)
...


Here c1 and c2 are the arbitrary constants that can be determined by applying initial or
boundary conditions
...


5
...

This differential equation is a first-order non-homogeneous linear differential equation, which
means that the highest derivative in the equation is of the first order (in this case, y'), the
coefficients of the products are constants (in this case, 1 for y' and 1 for y)
...
To solve this type of differential
equation, one way is to use the method of undetermined coefficients, which consists of finding

a particular solution by guessing a form for the solution that includes arbitrary coefficients and
then solving for these coefficients by plugging the guess into the differential equation and
comparing the resulting expression with the non-homogeneous function
...
Plugging this into the differential equation, we get yp' + yp =
Ax^3 + Ax^3 = 2Ax^3, which is not equal to x^2
...


Another way to guess the form of the particular solution is to note that the non-homogeneous
term is a polynomial of degree 2 and that the characteristic equation of the homogeneous part
of the differential equation is r+1 = 0, with a natural root of r=-1
...
Plugging this into the
differential equation, we get yp' + yp = Bx^2e^-x + Bx^2e^-x = x^2, which equals the nonhomogeneous function
...


To find the general solution of the differential equation, we need to add the available
resolution of the homogeneous equation (y = Ce^-x) and the particular solution we found
above
...

It can be determined by applying initial or boundary conditions
...


6
...
The general solution is y
= c1sin(x) + c2cos(x)
This differential equation is a second-order linear homogeneous differential equation, which
means that the highest derivative in the equation is of second order (in this case, y''), the

coefficients of the products are constants (in this case, 1 for y'' and 1 for y) and the right side of
the equation is zero
...
In this case, the
characteristic equation is r^2 + 1 = 0
...
These are the two roots of the characteristic equation and are complex and
distinct
...
Since the roots of the characteristic equation are complex
and distinct, the general solution is represented as y = c1e^(ix) + c2e^(-ix)
...


Here c1 and c2 are the arbitrary constants that can be determined by applying initial or
boundary conditions
...


7
...


This differential equation is a first-order non-homogeneous linear differential equation, which
means that the highest derivative in the equation is of the first order (in this case, y'), the
coefficients of the products are constants (in this case, 1 for y' and -3 for y) and there is a nonzero function on the right side of the equation (in this case, x)
...


In this case, we can guess that the particular solution has the same form as x, yp = Ax, where A
is an arbitrary coefficient
...
Therefore, the particular solution of this
differential equation is yp = Ax, with A =1
...
Thus, the general answer is y = Ce^-3x + (1/4)x^2, where C is the constant of integration
...


It's worth noting that this differential equation is also known as the Riccati equation; it is a
nonlinear differential equation that appears in many physical and mathematical problems
...


8
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The roots are r1 = -1/2 + i√3/2 and r2 = -1/2 i√3/2
...
To find the general solution of this differential equation, one
method is to use the characteristic equation, which is obtained by setting the right-hand side of

the differential equation to zero and replacing the derivatives with a variable r
...


To find the roots of this equation, we set it to zero and solve for r: r^2 + r + 1 = 0, which gives us
r1 = -1/2 + i√3/2 and r2 = -1/2 - i√3/2
...


The general solution of a second-order linear homogeneous differential equation is a linear
combination of two linearly independent solutions of the homogeneous equation, which are
represented by e^(r1x) and e^(r2x)
...


Here c1 and c2 are the arbitrary constants that can be determined by applying initial or
boundary conditions
...

This differential equation is commonly used in physics and engineering to model systems like
spring-mass systems, electric circuits, and mechanical oscillations
...
Solve the differential equation y' + 2y = x^3
The solution to this differential equation is y = (1/5)x^4 + C*e^-2x, where C is the constant of
integration
...
There is a non-zero
function on the right side of the equation (in this case, x^3)
...


In this case, we can guess that the particular solution has the same form as x^3, yp = Ax^4,
where A is an arbitrary coefficient
...
Therefore, the particular
solution of this differential equation is yp = (1/5)x^4, with A =1/5
...
Thus, the
available answer is y = (1/5)x^4 + Ce^-2x, where C is the constant of integration
...


In this case, the non-homogeneous term x^3 is a polynomial of degree 3, and the characteristic
equation of the homogeneous part of the differential equation is r+2=0, with a natural root of
r=-2
...
This is a common feature of a non-homogeneous
differential equation; generally, when the non-homogeneous time is a polynomial of degree n,
the available solution will be a combination of a polynomial term of degree n and the resolution
of the homogeneous part
...
Find the general solution of the differential equation y'' - 2y' + 2y = 0
The characteristic equation is r^2 - 2r + 2 = 0
...
Thus, the general
solution is y = c1e^x + c2e^2x
This differential equation is a second-order linear homogeneous differential equation, which
means that the highest derivative in the equation is of second order (in this case, y''), the
coefficients of the products are constants (in this case, 1 for y'', -2 for y' and 2 for y) and the
right side of the equation is zero
...
In this case, the
characteristic equation is r^2 - 2r + 2 = 0
...
These are the two roots of the characteristic equation and are accurate and
distinct
...
Since the roots of the characteristic equation are real and
distinct, the general solution is defined as y = c1e^x + c2e^2x, where c1 and c2 are arbitrary
constants
...

It's essential to notice that this differential equation is known as the Cauchy-Euler equation; it
appears in many mathematical and physical problems
...
It also appears in other fields, such
as electrical and mechanical engineering, physics, and chemistry

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