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To solve y' = e" + y – 1 we first rewrite the equation in the form dy – e + y – 1, and then (1 – e2* – y)dz + dy = 0. It follows that the equation is . If we choose H(x) = e== then we obtain e"(1- e – y)dx +e *dy = 0, which is .Hence, the function is an .We solve this last problem to obtain y = ce" + e²" + A where A=-2 A=2 A--3 not exact exact full A=-1 A=3 multiplicator integrating factor A=1

To solve y' = e" + y – 1 we first rewrite the equation in the form dy – e + y – 1, and then (1 – e2* – y)dz + dy = 0. It follows that the equation is . If we choose H(x) = e== then we obtain e"(1- e – y)dx +e *dy = 0, which is .Hence, the function is an .We solve this last problem to obtain y = ce" + e²" + A where A=-2 A=2 A--3 not exact exact full A=-1 A=3 multiplicator integrating factor A=1

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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To solve y' = e +y– 1 we first rewrite the equation in the form dy - +y-1, dz and then (1 – e – y)dz + dy = 0. It follows that the equation is If we choose H(z) = e= then we obtain e "(1- e – y)dr +e*dy = 0, which is Hence, the function µ(z) is an We solve this last problem to obtain y = ce* +e + A where A=-2 A=2 A=-3 not exact exact full A=-1 A=3 multiplicator integrating factor A=1
Transcribed Image Text:To solve y' = e +y– 1 we first rewrite the equation in the form dy - +y-1, dz and then (1 – e – y)dz + dy = 0. It follows that the equation is If we choose H(z) = e= then we obtain e "(1- e – y)dr +e*dy = 0, which is Hence, the function µ(z) is an We solve this last problem to obtain y = ce* +e + A where A=-2 A=2 A=-3 not exact exact full A=-1 A=3 multiplicator integrating factor A=1
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