## Differential Equations With Polynomial Coefficients

You CAN Ace Differential Equations

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second-order

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homogeneous

polynomial coefficients

$$r(t)y'' +p(t)y' + q(t)y = 0$$

Alternate Names For Cauchy-Euler Equations

Euler-Cauchy Equations

Euler Equations

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.

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

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Technique 1 - - Substitution

A good substitution for this type of equation is $$x=\ln(t)$$. 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 and then through 3 examples.

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.

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Practice Problems

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

 Level B - Intermediate

Practice B01

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

solution

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