Differential Equations : Linear Equations

Study concepts, example questions & explanations for Differential Equations

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Example Questions

Example Question #1 : Linear Equations

Solve the initial value problem  for  and 

Possible Answers:

Correct answer:

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:

 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)

Example Question #2 : Linear Equations

Find the general solution to  .

Possible Answers:

Correct answer:

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:

 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.

 

Example Question #1 : Linear Equations

Solve the initial value problem  for  and .

Possible Answers:

 

Correct answer:

 

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:

 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)

Example Question #4 : Linear Equations

Solve the following homogeneous differential equation:

Possible Answers:

 

where  and  are  constants 

 

where  and  are  constants 

 

where  and  are  constants 

 

where  and  are  constants 

Correct answer:

 

where  and  are  constants 

Explanation:

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,

Example Question #2 : Linear Equations

Solve the General form of the differential equation:

Possible Answers:

Where  and  are arbitrary constants 

Where  and  are arbitrary constants 

Where  and  are arbitrary constants 

Where  and  are arbitrary constants 

Correct answer:

Where  and  are arbitrary constants 

Explanation:

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

Example Question #3 : Linear Equations

Solve the general homogeneous part of the following differential equation:

Possible Answers:

Where  and  are arbitrary but not meaningless constants 

Where  and  are arbitrary but not meaningless constants 

Where  and  are arbitrary but not meaningless constants 

Where  and  are arbitrary but not meaningless constants 

Correct answer:

Where  and  are arbitrary but not meaningless constants 

Explanation:

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:

Example Question #7 : Linear Equations

Solve the following homogeneous differential equation: 

Possible Answers:

where  and  are constants

where  and  are constants

where  and  are constants

where  and  are constants

Correct answer:

where  and  are constants

Explanation:

This differential equation has characteristic equation of:

 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

  in the case of a repeated root such as this,  and is the repeated root r=5.

Therefore, the solution is

 

Example Question #4 : Linear Equations

Find a general solution to the following Differential Equation

Possible Answers:

Correct answer:

Explanation:

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. 

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