dx + x = t³ dt
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![This equation represents a first-order linear differential equation. It is given by:
\[ t \frac{dx}{dt} + x = t^3 \]
Here, \( t \) is the independent variable, and \( x \) is the dependent variable. The term \(\frac{dx}{dt}\) represents the derivative of \( x \) with respect to \( t \). The equation states that the product of \( t \) and the derivative of \( x \) with respect to \( t \), plus \( x \) itself, is equal to the cube of \( t \).
This type of equation can often be solved using an integrating factor or other methods specific to first-order linear differential equations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5025798a-663c-4f82-a82a-d0677c58328a%2Ffa496629-e161-46c8-8914-3403c1c142a6%2Fs557ay5_processed.png&w=3840&q=75)
Transcribed Image Text:This equation represents a first-order linear differential equation. It is given by:
\[ t \frac{dx}{dt} + x = t^3 \]
Here, \( t \) is the independent variable, and \( x \) is the dependent variable. The term \(\frac{dx}{dt}\) represents the derivative of \( x \) with respect to \( t \). The equation states that the product of \( t \) and the derivative of \( x \) with respect to \( t \), plus \( x \) itself, is equal to the cube of \( t \).
This type of equation can often be solved using an integrating factor or other methods specific to first-order linear differential equations.
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