A CauchyEuler Differential Equation (also called EulerCauchy Equation or just Euler Equation) is an equation with polynomial coefficients of the form \(\displaystyle{ t^2y'' +aty' + by = 0 }\).
These may seem kind of specialized, and they are, but equations of this form show up so often that special techniques for solving them have been developed. We will look at a couple of techniques on this page and direct you to other techniques that will also work on other pages.
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Technique 1   Substitution 

A good substitution for this type of equation is \(y=\ln(t)\). This substitution converts the differential equation into one with constant coefficients.
Technique 2   Trial Solution 

A second technique involves using a trial solution \(y=t^k\) where k is a constant. This video takes you through a general equation. The equations developed in this video are used in several practice problems below.
video by Houston Math Prep 

Okay, before we go on to another type of differential equation with polynomial coefficients, let's work some practice problems.
Practice
Unless otherwise instructed, solve these CauchyEuler differential equations using the techniques on this page.
\(\displaystyle{ x^2u'' + 3xu' + \frac{5u}{4} = 0 }\)
Problem Statement 

\(\displaystyle{ x^2u'' + 3xu' + \frac{5u}{4} = 0 }\)
Final Answer 

\(\displaystyle{ u = x^{1} \left[ A\cos((1/2)\ln(x)) + B\sin((1/2)\ln(x)) \right] }\)
Problem Statement 

\(\displaystyle{ x^2u'' + 3xu' + \frac{5u}{4} = 0 }\)
Solution 

Final Answer 

\(\displaystyle{ u = x^{1} \left[ A\cos((1/2)\ln(x)) + B\sin((1/2)\ln(x)) \right] }\) 
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\( 2t^2y'' + 3ty' + 5y = 0 \)
Problem Statement 

\( 2t^2y'' + 3ty' + 5y = 0 \)
Solution 

video by The Lazy Engineer 

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\( t^2y'' + 5ty' + 4y = 0 \)
Problem Statement 

\( t^2y'' + 5ty' + 4y = 0 \)
Solution 

video by The Lazy Engineer 

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\( x^2y'' + 7xy' + 8y = 0 \)
Problem Statement 

\( x^2y'' + 7xy' + 8y = 0 \)
Solution 

video by Houston Math Prep 

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\( 9x^2y'' + 3xy' + y = 0 \)
Problem Statement 

\( 9x^2y'' + 3xy' + y = 0 \)
Solution 

video by Houston Math Prep 

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\( x^2y''  9xy' + 28y = 0 \)
Problem Statement 

\( x^2y''  9xy' + 28y = 0 \)
Solution 

video by Houston Math Prep 

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\( x^2 y''  11xy'+85y = 0, y(1) = 3, y'(1) = 4 \)
Problem Statement 

\( x^2 y''  11xy'+85y = 0, y(1) = 3, y'(1) = 4 \)
Solution 

video by MIP4U 

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You CAN Ace Differential Equations
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
Set 1  basic identities  

\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) 
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) 
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) 
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) 
Set 2  squared identities  

\( \sin^2t + \cos^2t = 1\) 
\( 1 + \tan^2t = \sec^2t\) 
\( 1 + \cot^2t = \csc^2t\) 
Set 3  doubleangle formulas  

\( \sin(2t) = 2\sin(t)\cos(t)\) 
\(\displaystyle{ \cos(2t) = \cos^2(t)  \sin^2(t) }\) 
Set 4  halfangle formulas  

\(\displaystyle{ \sin^2(t) = \frac{1\cos(2t)}{2} }\) 
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) 
Trig Derivatives
\(\displaystyle{ \frac{d[\sin(t)]}{dt} = \cos(t) }\) 
\(\displaystyle{ \frac{d[\cos(t)]}{dt} = \sin(t) }\)  
\(\displaystyle{ \frac{d[\tan(t)]}{dt} = \sec^2(t) }\) 
\(\displaystyle{ \frac{d[\cot(t)]}{dt} = \csc^2(t) }\)  
\(\displaystyle{ \frac{d[\sec(t)]}{dt} = \sec(t)\tan(t) }\) 
\(\displaystyle{ \frac{d[\csc(t)]}{dt} = \csc(t)\cot(t) }\) 
Inverse Trig Derivatives
\(\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\) 
\(\displaystyle{ \frac{d[\arccos(t)]}{dt} = \frac{1}{\sqrt{1t^2}} }\)  
\(\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }\) 
\(\displaystyle{ \frac{d[\arccot(t)]}{dt} = \frac{1}{1+t^2} }\)  
\(\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
\(\displaystyle{ \frac{d[\arccsc(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 1}} }\) 
Trig Integrals
\(\int{\sin(x)~dx} = \cos(x)+C\) 
\(\int{\cos(x)~dx} = \sin(x)+C\)  
\(\int{\tan(x)~dx} = \ln\abs{\cos(x)}+C\) 
\(\int{\cot(x)~dx} = \ln\abs{\sin(x)}+C\)  
\(\int{\sec(x)~dx} = \) \( \ln\abs{\sec(x)+\tan(x)}+C\) 
\(\int{\csc(x)~dx} = \) \( \ln\abs{\csc(x)+\cot(x)}+C\) 
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Practice Instructions
Unless otherwise instructed, solve these CauchyEuler differential equations using the techniques on this page.