## 17Calculus Differential Equations - Cauchy-Euler Equation

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A Cauchy-Euler Differential Equation (also called Euler-Cauchy 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.

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.

### Houston Math Prep - Cauchy-Euler Differential Equations (2nd Order) [12mins-56secs]

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 Cauchy-Euler differential equations using the techniques on this page.

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

Problem Statement

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

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

Problem Statement

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

Solution

### 620 video

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

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$$2t^2y'' + 3ty' + 5y = 0$$

Problem Statement

$$2t^2y'' + 3ty' + 5y = 0$$

Solution

### 2228 video

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$$t^2y'' + 5ty' + 4y = 0$$

Problem Statement

$$t^2y'' + 5ty' + 4y = 0$$

Solution

### 2229 video

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$$x^2y'' + 7xy' + 8y = 0$$

Problem Statement

$$x^2y'' + 7xy' + 8y = 0$$

Solution

### 2230 video

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$$9x^2y'' + 3xy' + y = 0$$

Problem Statement

$$9x^2y'' + 3xy' + y = 0$$

Solution

### 2231 video

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$$x^2y'' - 9xy' + 28y = 0$$

Problem Statement

$$x^2y'' - 9xy' + 28y = 0$$

Solution

### 2232 video

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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

### 2291 video

video by MIP4U

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You CAN Ace Differential Equations

### Trig Formulas

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 - double-angle formulas

$$\sin(2t) = 2\sin(t)\cos(t)$$

$$\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }$$

Set 4 - half-angle 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

 $$\displaystyle{ \frac{d[\arcsin(t)]}{dt} = \frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arccos(t)]}{dt} = -\frac{1}{\sqrt{1-t^2}} }$$ $$\displaystyle{ \frac{d[\arctan(t)]}{dt} = \frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arccot(t)]}{dt} = -\frac{1}{1+t^2} }$$ $$\displaystyle{ \frac{d[\arcsec(t)]}{dt} = \frac{1}{\abs{t}\sqrt{t^2 -1}} }$$ $$\displaystyle{ \frac{d[\arccsc(t)]}{dt} = -\frac{1}{\abs{t}\sqrt{t^2 -1}} }$$

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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Single Variable Calculus

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Differential Equations

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Engineering

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