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In order to solve with y(0) = 1 we first (2y-5)dy = (3x² - e²)dr. After y²-5y = 2³-e²+c of integration. Setting x= and where c is the have y² - 5y- (r³e² - 3) = 0. By using 5 13 (2) = +2³-e². In the quadratic formula, negative square root is chosen because of the variables. This gives both sides, we obtain y' 31² - e² 2y-5 we have Thus, we

In order to solve with y(0) = 1 we first (2y-5)dy = (3x² - e²)dr. After y²-5y = 2³-e²+c of integration. Setting x= and where c is the have y² - 5y- (r³e² - 3) = 0. By using 5 13 (2) = +2³-e². In the quadratic formula, negative square root is chosen because of the variables. This gives both sides, we obtain y' 31² - e² 2y-5 we have Thus, we

Advanced Engineering Mathematics
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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In order to solve with y(0) = 1 we first (2y-5)dy = (3x² - e²)dr. After y²-5y = 2³-e²+c of integration. Setting x= and where c is the have y² - 5y- (r³e² - 3) = 0. By using 5 13 ly(z) +2³ - e². In the quadratic formula, negative square root is chosen because of factor constant integration by parts c=-3 separate the initial conditions value x=1 integrating the variables. This gives both sides, we obtain y' 31² - e² 2y-5 disintegrate c=-3 0 eliminate c=1 y=1 we have c=0 c=3 Thus, we y=0 the quadratic formula
Transcribed Image Text:In order to solve with y(0) = 1 we first (2y-5)dy = (3x² - e²)dr. After y²-5y = 2³-e²+c of integration. Setting x= and where c is the have y² - 5y- (r³e² - 3) = 0. By using 5 13 ly(z) +2³ - e². In the quadratic formula, negative square root is chosen because of factor constant integration by parts c=-3 separate the initial conditions value x=1 integrating the variables. This gives both sides, we obtain y' 31² - e² 2y-5 disintegrate c=-3 0 eliminate c=1 y=1 we have c=0 c=3 Thus, we y=0 the quadratic formula
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