Linear Equations - Differential Equations
Card 0 of 8
Solve the initial value problem 
for 
and 
.
Solve the initial value problem
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:
We then solve the characteristic equation and find that
This lets us know that the basis for the fundamental set of solutions to this problem (solutions to the homogeneous problem) contains 
.
As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, 
. Plugging in our initial condition, we find that 
. To plug in the second initial condition, we take the derivative and find that 
. Plugging in the second initial condition yields 
. Solving this simple system of linear equations shows us that
Leaving us with a final answer of 
(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)
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:
As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus,
Leaving us with a final answer of
(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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Solve the initial value problem 
for 
and 
Solve the initial value problem
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:
We then solve the characteristic equation and find that
(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 
.
As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, 
. Plugging in our initial condition, we find that 
. To plug in the second initial condition, we take the derivative and find that 
. Plugging in the second initial condition yields 
. Solving this simple system of linear equations shows us that
Leaving us with a final answer of 
(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)
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:
As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus,
Leaving us with a final answer of
(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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Find the general solution to 
.
Find the general solution to
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:
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.
Alternatively, the rational root theorem suggests that we try -1 or 1 as a root of this equation. Using horner's scheme, we see
Which tells us the the polynomial factors into 
and that 
. This means that the fundamental set of solutions is 
As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus, 
. As this is not an initial value problem and just asks for the general solution, we are done.
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:
Alternatively, the rational root theorem suggests that we try -1 or 1 as a root of this equation. Using horner's scheme, we see
Which tells us the the polynomial factors into
As the given problem was homogeneous, the solution is just a linear combination of these functions. Thus,
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Solve the following homogeneous differential equation:
Solve the following homogeneous differential equation:
The ode has a characteristic equation of 
.
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.
(for real repeated roots)
Thus, the solution is,
The ode has a characteristic equation of
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.
Thus, the solution is,
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Solve the General form of the differential equation:
Solve the General form of the differential equation:
This differential equation has a characteristic equation of
, 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:
with r being the roots of the characteristic equation.
Thus, the solution is
This differential equation has a characteristic equation of
with r being the roots of the characteristic equation.
Thus, the solution is
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Solve the general homogeneous part of the following differential equation:
Solve the general homogeneous part of the following differential equation:
We start off by noting that the homogeneous equation we are trying to solve is given as
.
This differential equation thus has characteristic equation of
.
This has roots of r=3 and r=-4, therefore, the general homogeneous solution is given by:
We start off by noting that the homogeneous equation we are trying to solve is given as
This differential equation thus has characteristic equation of
This has roots of r=3 and r=-4, therefore, the general homogeneous solution is given by:
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Solve the following homogeneous differential equation:
Solve the following homogeneous differential equation:
This differential equation has characteristic equation of:
It must be noted that this characteristic equation has a double rootof r=5.
Thus the general solution to a homogeneous differential equation with a repeated root is used.
This is equation is
in the case of a repeated root such as this, 
and is the repeated root r=5.
Therefore, the solution is
This differential equation has characteristic equation of:
Thus the general solution to a homogeneous differential equation with a repeated root is used.
This is equation is
Therefore, the solution is
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Find a general solution to the following Differential Equation
Find a general solution to the following Differential Equation
Solving the auxiliary equation
Trying out candidates for roots from the Rational Root Theorem we have a root 
.
Factoring completely we have
Our general solution is
where 
are arbitrary constants.
Solving the auxiliary equation
Trying out candidates for roots from the Rational Root Theorem we have a root
Factoring completely we have
Our general solution is
where
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