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# Euler Equations

R. C. Daileda
A second order homogeneous linear ODE of the form t2 d2 y dy + t + y = 0, dt2 dt (1)

with , some constants, is called an Euler equation. In this note we will show how a simple change of variables (analogous to a substitution in an integral) can be used to transform an Euler equation into a second order homogeneous linear ODE with constant coecients. Since we know how to nd the general solution to the latter type of equation, we can therefore always nd the general solution to any Euler equation. We begin with the change of variable x = ln t. Since the set of real numbers x corresponds in a one-to-one fashion with the set of positive numbers t, we can view any function y of t as a function of x, and vice-versa. For example, the function of t y = 12t sin t becomes the function of x y = 12ex sin(ex ) since t = ex . Our immediate task is to determine how changing the independent variable from t to x changes the given dierential equation. The chain rule tells us how the derivative of y with respect to the new variable x is related to the derivative of y with respect to the old variable t. Indeed: dy dy dt dy = = t. dx dt dx dt (2)

Here we have used the fact that t = ex so that dt/dx = ex = t. Likewise, we can use the chain rule again (together with the product rule) to get an analogous expression for the second derivative of y with respect to x. Using the above we have d2 y dx2 = = = = = d dy dx dx d dy t dx dt d dy dy dt t+ dx dt dt dx 2 d y dt dy dt t+ dt2 dx dt dx d2 y 2 dy t + t dt2 dt

where we have again used the fact that dt/dx = t. Combining this with (2) gives the relations dy t dt dy 2 t dt = = 1 dy dx d2 y dy . 2 dx dx (3) (4)

We now substitute (3) and (4) into the Euler equation (1) to see what it becomes as a result of our change of variables. We get dy dy d2 y + + y = 0. dx2 dx dx Putting the dy/dx terms together, we are led to the following. Conclusion. Under the change of variables x = ln t, the Euler equation t2 dy d2 y + t + y = 0 dt2 dt

## is transformed into the constant-coecient equation dy d2 y + ( 1) + y = 0. 2 dx dx (5)

The moral of the story is that we now have a procedure for nding the general solution to Euler equations: 1. Perform the change of variable x = ln t, yielding the constant coecient equation (5). 2. Find the general solution to (5) using the techniques we already know. This solution will be in terms of the independent variable x. 3. Eliminate x by substituting x = ln t, yielding the general solution to (1) in terms of the original independent variable t. We demonstrate this procedure with a few examples. Example 1. Find the general solution to the Euler equation t2 y 3ty + 7y = 0, t > 0. (6)

Solution. According to the work above, the variable change x = ln t transforms the given dierential equation into d2 y dy + 7y = 0 (7) 4 dx2 dx (we have = 3, = 7). The characteristic equation of this DE is r2 4r + 7 = 0 which has the roots r= Therefore, the general solution to (7) is y = c1 e2x cos 3x + c2 e2x sin 3x. 4 16 28 = 2 3i. 2

To get the solution to (6) we simply remember the relationship x = ln t: y = c1 t2 cos( 3 ln t) + c2 t2 sin( 3 ln t).

Example 2. Find the general solution to the equation 4t2 y + 8ty 3y = 0, 2 t > 0. (8)

Solution. Once we divide (8) by 4, we recognize this as an Euler equation with = 2 and = 3/4. Consequently, we know that the variable substitution x = ln t transforms (8) into the equivalent equation dy d2 y 3 + y = 0. 2 dx dx 4 The characteristic equation of the latter is r2 + r 3 = 4 r+ 3 2 r 1 2 =0 (9)

which has the roots r1 = 1/2 and r2 = 3/2. Therefore, the general solution to (9) is y = c1 ex/2 + c2 e3x/2 . Switching back to the variable t through our substitution x = ln t we are led to the general solution of (8): y = c1 t1/2 + c2 t3/2 .

## Example 3. Find the general solution to the equation t2 y 2y = 0. (10)

Solution. Notice that we considered this example in class: we found the general solution after an oracle had given us two fundamental solutions. Now we can nd these fundamental solutions ourselves! The equation (10) is an Euler equation with = 0 and = 2. The usual change of variables x = ln t therefore turns it into dy d2 y 2y = 0 (11) 2 dx dx which has characteristic equation r2 r 2 = (r 2)(r + 1) = 0 so that the general solution is y = c1 e2x + c2 ex . Changing back to t via x = ln t we nd that the general solution to (11) is y = c1 t2 + c2 t1 . This is exactly what we showed in class!
The text is a detailed explanation of how to solve Euler equations by transforming them into second-order homogeneous linear ODEs with constant coefficients through a change of variables. It includes three examples that demonstrate the process step-by-step.
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Given a text passage, determine:
    1. If it contains a mathematical question (including word problems or calculations)
    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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- SINGLE_QUESTION
- SINGLE_QUESTION_WITH_ANSWER
- MULTIPLE_QUESTIONS
- MULTIPLE_QUESTIONS_WITH_ANSWERS
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You are on page 1of 3

# Euler Equations

R. C. Daileda
A second order homogeneous linear ODE of the form t2 d2 y dy + t + y = 0, dt2 dt (1)

with , some constants, is called an Euler equation. In this note we will show how a simple change of variables (analogous to a substitution in an integral) can be used to transform an Euler equation into a second order homogeneous linear ODE with constant coecients. Since we know how to nd the general solution to the latter type of equation, we can therefore always nd the general solution to any Euler equation. We begin with the change of variable x = ln t. Since the set of real numbers x corresponds in a one-to-one fashion with the set of positive numbers t, we can view any function y of t as a function of x, and vice-versa. For example, the function of t y = 12t sin t becomes the function of x y = 12ex sin(ex ) since t = ex . Our immediate task is to determine how changing the independent variable from t to x changes the given dierential equation. The chain rule tells us how the derivative of y with respect to the new variable x is related to the derivative of y with respect to the old variable t. Indeed: dy dy dt dy = = t. dx dt dx dt (2)

Here we have used the fact that t = ex so that dt/dx = ex = t. Likewise, we can use the chain rule again (together with the product rule) to get an analogous expression for the second derivative of y with respect to x. Using the above we have d2 y dx2 = = = = = d dy dx dx d dy t dx dt d dy dy dt t+ dx dt dt dx 2 d y dt dy dt t+ dt2 dx dt dx d2 y 2 dy t + t dt2 dt

where we have again used the fact that dt/dx = t. Combining this with (2) gives the relations dy t dt dy 2 t dt = = 1 dy dx d2 y dy . 2 dx dx (3) (4)

We now substitute (3) and (4) into the Euler equation (1) to see what it becomes as a result of our change of variables. We get dy dy d2 y + + y = 0. dx2 dx dx Putting the dy/dx terms together, we are led to the following. Conclusion. Under the change of variables x = ln t, the Euler equation t2 dy d2 y + t + y = 0 dt2 dt

## is transformed into the constant-coecient equation dy d2 y + ( 1) + y = 0. 2 dx dx (5)

The moral of the story is that we now have a procedure for nding the general solution to Euler equations: 1. Perform the change of variable x = ln t, yielding the constant coecient equation (5). 2. Find the general solution to (5) using the techniques we already know. This solution will be in terms of the independent variable x. 3. Eliminate x by substituting x = ln t, yielding the general solution to (1) in terms of the original independent variable t. We demonstrate this procedure with a few examples. Example 1. Find the general solution to the Euler equation t2 y 3ty + 7y = 0, t > 0. (6)

Solution. According to the work above, the variable change x = ln t transforms the given dierential equation into d2 y dy + 7y = 0 (7) 4 dx2 dx (we have = 3, = 7). The characteristic equation of this DE is r2 4r + 7 = 0 which has the roots r= Therefore, the general solution to (7) is y = c1 e2x cos 3x + c2 e2x sin 3x. 4 16 28 = 2 3i. 2

To get the solution to (6) we simply remember the relationship x = ln t: y = c1 t2 cos( 3 ln t) + c2 t2 sin( 3 ln t).

Example 2. Find the general solution to the equation 4t2 y + 8ty 3y = 0, 2 t > 0. (8)

Solution. Once we divide (8) by 4, we recognize this as an Euler equation with = 2 and = 3/4. Consequently, we know that the variable substitution x = ln t transforms (8) into the equivalent equation dy d2 y 3 + y = 0. 2 dx dx 4 The characteristic equation of the latter is r2 + r 3 = 4 r+ 3 2 r 1 2 =0 (9)

which has the roots r1 = 1/2 and r2 = 3/2. Therefore, the general solution to (9) is y = c1 ex/2 + c2 e3x/2 . Switching back to the variable t through our substitution x = ln t we are led to the general solution of (8): y = c1 t1/2 + c2 t3/2 .

## Example 3. Find the general solution to the equation t2 y 2y = 0. (10)

Solution. Notice that we considered this example in class: we found the general solution after an oracle had given us two fundamental solutions. Now we can nd these fundamental solutions ourselves! The equation (10) is an Euler equation with = 0 and = 2. The usual change of variables x = ln t therefore turns it into dy d2 y 2y = 0 (11) 2 dx dx which has characteristic equation r2 r 2 = (r 2)(r + 1) = 0 so that the general solution is y = c1 e2x + c2 ex . Changing back to t via x = ln t we nd that the general solution to (11) is y = c1 t2 + c2 t1 . This is exactly what we showed in class!
user
The text is a detailed explanation of how to solve Euler equations by transforming them into second-order homogeneous linear ODEs with constant coefficients through a change of variables. It includes three examples that demonstrate the process step-by-step.

TYPE: MULTIPLE_QUESTIONS_WITH_ANSWERS