The equation in differential form M dx + Ñ dy = 0 is not exact. Indeed, we have M₁ - Ñ , For this exercise we can find an integrating factor which is a function of x alone since M, - Ñ , I Ñ = N = can be considered as a function of x alone. Namely we have μ(x) Multiplying the original equation by the integrating factor we obtain a new equation M dx + N dy = 0 where M Which is exact since My N₂ (3y + 2xe-³¹) dx + (1 − 2ye¯³¹)dy = 0 = = = = are equal. This problem is exact. Therefore an implicit general solution can be written in the form F(x, y) = C where F(x, y) = Finally find the value of the constant C so that the initial condition y(0) = 1. C =
The equation in differential form M dx + Ñ dy = 0 is not exact. Indeed, we have M₁ - Ñ , For this exercise we can find an integrating factor which is a function of x alone since M, - Ñ , I Ñ = N = can be considered as a function of x alone. Namely we have μ(x) Multiplying the original equation by the integrating factor we obtain a new equation M dx + N dy = 0 where M Which is exact since My N₂ (3y + 2xe-³¹) dx + (1 − 2ye¯³¹)dy = 0 = = = = are equal. This problem is exact. Therefore an implicit general solution can be written in the form F(x, y) = C where F(x, y) = Finally find the value of the constant C so that the initial condition y(0) = 1. C =
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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Transcribed Image Text:The equation
in differential form M dx + Ñ dy = 0 is not exact. Indeed, we have
M₁ - Ñ ,
For this exercise we can find an integrating factor which is a function of x alone since
M₁ - Ñ ,
I
Ñ
=
N =
can be considered as a function of x alone.
Namely we have μ(x)
Multiplying the original equation by the integrating factor we obtain a new equation M dx + N dy = 0 where
M
Which is exact since
My
N₂
(3y + 2xe-³¹) dx + (1 − 2ye¯³¹)dy = 0
=
=
=
=
are equal.
This problem is exact. Therefore an implicit general solution can be written in the form F(x, y) = C where
F(x, y) =
Finally find the value of the constant C so that the initial condition y(0) = 1.
C =
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