You CAN Ace Differential Equations

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On this page, we discuss differential equations with polynomial coefficients of the form \(\displaystyle{ r(t^n)y^{(n)}(t) + a_{n-1}t^{n-1}y^{n-1}(t) + \cdots + a_0y(t) = g(t) }\), where \(r(t^n)\) is an nth-order polynomial.

General Form: \(r(t)y'' + p(t)y' + q(t)y = 0\)

Classification: second-order, linear, homogeneous

\(r(t)\), \(p(t)\) and \(q(t)\) are polynomials

For now, we will stick with these two special types of second-order homogeneous equations.

Cauchy-Euler Equations

\(\displaystyle{ t^2y'' +aty' + by = 0 }\)

Chebyshev's Equations

\(\displaystyle{ (1-t^2)y'' - ty' + ay = 0 }\)

Cauchy-Euler Differential Equation

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 on other pages.

this page

 

other pages

substitution

 

reduction of order

trial solution

 

integrating factors

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.

Instructions - - 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

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

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

620 solution video

video by Dr Chris Tisdell

Final Answer

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

close solution

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

Problem Statement

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

Solution

2228 solution video

close solution

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

Problem Statement

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

Solution

2229 solution video

close solution

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

Problem Statement

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

Solution

2230 solution video

close solution

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

Problem Statement

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

Solution

2231 solution video

close solution

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

Problem Statement

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

Solution

2232 solution video

close solution

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

Problem Statement

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

Solution

2291 solution video

video by MIP4U

close solution

Chebyshev's Differential Equations

Chebyshev's equations are of the form \(\displaystyle{ (1-t^2)y'' - ty' + ay = 0 }\) where a is a real constant. Again, these seem specialized but they occur so often, they are worth discussing separately. These equations can be solved by using the substitution \( t = \cos( \theta )\). This changes the differential equation into a form that can be solved more easily. See the practice problems for examples.

Instructions - - Unless otherwise instructed, solve these Chebyshev differential equations using the techniques on this page.

\( (1-x^2)u'' - xu' + v^2u = 0 \); use \( x = \cos(\theta) \)

Problem Statement

\( (1-x^2)u'' - xu' + v^2u = 0 \); use \( x = \cos(\theta) \)

Final Answer

\( u = A\cos(v\arccos(x)) + B\sin(v\arccos(x)) \)

Problem Statement

\( (1-x^2)u'' - xu' + v^2u = 0 \); use \( x = \cos(\theta) \)

Solution

621 solution video

video by Dr Chris Tisdell

Final Answer

\( u = A\cos(v\arccos(x)) + B\sin(v\arccos(x)) \)

close solution
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