17Calculus Differential Equations - Power Series Solution
This page discusses how to solve differential equations using power series. This is also called Frobenius Method.
Recommended Books on Amazon | ||
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Here is a great video with a step-by-step explanation on how to solve these problems, including an example.
Dr Chris Tisdell - Power series solution to differential equations: a tutorial
video by Dr Chris Tisdell |
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Practice
Unless otherwise instructed, solve the following differential equations using power series. If initial conditions are given, determine the particular solution.
Solve \( y' - y = 0 \)
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Solve \( y' - y = 0 \)
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Solve \( y' + 2xy = 0 \)
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Solve \( y' + 2xy = 0 \)
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Solve \( y'' + y = 0 \)
Problem Statement |
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Solve \( y'' + y = 0 \)
Solution |
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2611 video
video by PatrickJMT |
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close solution
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Solve \( y'' - xy = 0 \)
Problem Statement |
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Solve \( y'' - xy = 0 \)
Solution |
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2609 video
video by Houston Math Prep |
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Solve \( y'' - 2xy' + y = 0 \)
Problem Statement |
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Solve \( y'' - 2xy' + y = 0 \)
Solution |
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2608 video
video by blackpenredpen |
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|
close solution
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Log in to rate this practice problem and to see it's current rating. |
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Solve \( y'' - 2y' + y = x; ~~~ y(0)=0, y'(0) = 1 \)
Problem Statement |
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Solve \( y'' - 2y' + y = x; ~~~ y(0)=0, y'(0) = 1 \)
Solution |
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2613 video
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You CAN Ace Differential Equations
Topics You Need To Understand For This Page
Related Topics and Links
Trig Formulas
The Unit Circle
The Unit Circle [wikipedia]
Basic Trig Identities
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Set 1 - basic identities | |||
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\(\displaystyle{ \tan(t) = \frac{\sin(t)}{\cos(t)} }\) |
\(\displaystyle{ \cot(t) = \frac{\cos(t)}{\sin(t)} }\) |
\(\displaystyle{ \sec(t) = \frac{1}{\cos(t)} }\) |
\(\displaystyle{ \csc(t) = \frac{1}{\sin(t)} }\) |
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Set 2 - squared identities | ||
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\( \sin^2t + \cos^2t = 1\) |
\( 1 + \tan^2t = \sec^2t\) |
\( 1 + \cot^2t = \csc^2t\) |
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Set 3 - double-angle formulas | |
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\( \sin(2t) = 2\sin(t)\cos(t)\) |
\(\displaystyle{ \cos(2t) = \cos^2(t) - \sin^2(t) }\) |
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Set 4 - half-angle formulas | |
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\(\displaystyle{ \sin^2(t) = \frac{1-\cos(2t)}{2} }\) |
\(\displaystyle{ \cos^2(t) = \frac{1+\cos(2t)}{2} }\) |
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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Topics Listed Alphabetically
Single Variable Calculus |
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Multi-Variable Calculus |
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Differential Equations |
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Precalculus |
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Engineering |
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Circuits |
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Semiconductors |
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Practice Instructions
Unless otherwise instructed, solve the following differential equations using power series. If initial conditions are given, determine the particular solution.
