You CAN Ace Differential Equations
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Single Variable Calculus |
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Multi-Variable Calculus |
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Acceleration Vector |
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Differential Equations |
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Do you have a practice problem number but do not know on which page it is found? If so, enter the number below and click 'page' to go to the page on which it is found or click 'practice' to be taken to the practice problem.
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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. |
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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.
\(\displaystyle{ t^2y'' +aty' + by = 0 }\) | |
\(\displaystyle{ (1-t^2)y'' - ty' + ay = 0 }\) |
Cauchy-Euler Differential 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 on other pages.
this page |
other pages | |
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Technique 1 - - Substitution |
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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 |
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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.
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 |
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\(\displaystyle{ x^2u'' + 3xu' + \frac{5u}{4} = 0 }\)
Final Answer |
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\(\displaystyle{ u = x^{-1} \left[ A\cos((1/2)\ln(x)) + B\sin((1/2)\ln(x)) \right] }\) |
Problem Statement |
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\(\displaystyle{ x^2u'' + 3xu' + \frac{5u}{4} = 0 }\)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\(\displaystyle{ u = x^{-1} \left[ A\cos((1/2)\ln(x)) + B\sin((1/2)\ln(x)) \right] }\) |
close solution |
\( 2t^2y'' + 3ty' + 5y = 0 \)
Problem Statement |
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\( 2t^2y'' + 3ty' + 5y = 0 \)
Solution |
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video by The Lazy Engineer
close solution |
\( t^2y'' + 5ty' + 4y = 0 \)
Problem Statement |
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\( t^2y'' + 5ty' + 4y = 0 \)
Solution |
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video by The Lazy Engineer
close solution |
\( x^2y'' + 7xy' + 8y = 0 \)
Problem Statement |
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\( x^2y'' + 7xy' + 8y = 0 \)
Solution |
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video by Houston Math Prep
close solution |
\( 9x^2y'' + 3xy' + y = 0 \)
Problem Statement |
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\( 9x^2y'' + 3xy' + y = 0 \)
Solution |
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video by Houston Math Prep
close solution |
\( x^2y'' - 9xy' + 28y = 0 \)
Problem Statement |
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\( x^2y'' - 9xy' + 28y = 0 \)
Solution |
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video by Houston Math Prep
close solution |
\( x^2 y'' - 11xy'+85y = 0, y(1) = -3, y'(1) = -4 \)
Problem Statement |
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\( x^2 y'' - 11xy'+85y = 0, y(1) = -3, y'(1) = -4 \)
Solution |
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video by MIP4U
close solution |
Chebyshev's Differential Equations |
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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 |
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\( (1-x^2)u'' - xu' + v^2u = 0 \); use \( x = \cos(\theta) \)
Final Answer |
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\( u = A\cos(v\arccos(x)) + B\sin(v\arccos(x)) \) |
Problem Statement |
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\( (1-x^2)u'' - xu' + v^2u = 0 \); use \( x = \cos(\theta) \)
Solution |
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video by Dr Chris Tisdell
Final Answer |
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\( u = A\cos(v\arccos(x)) + B\sin(v\arccos(x)) \) |
close solution |