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I In differentlal förm M dã + N dy = 0. ne equation (3x°y+ 6 cos(3a) y ?) dæ + (3x - 4y ?)dy = 0 e differential form M dx + N dy 0 is not exact. Indeed, we have M-N, For this exercise we can find an integrating factor which is a function of y alone since My-N M is a function of y alone. Namely we have µ(y) Multiplying the original equation by the integrating factor we obtain a new equation M da + N dy = 0 where %3D M %3D N = Which is exact since M, N = are equal.

I In differentlal förm M dã + N dy = 0. ne equation (3x°y+ 6 cos(3a) y ?) dæ + (3x - 4y ?)dy = 0 e differential form M dx + N dy 0 is not exact. Indeed, we have M-N, For this exercise we can find an integrating factor which is a function of y alone since My-N M is a function of y alone. Namely we have µ(y) Multiplying the original equation by the integrating factor we obtain a new equation M da + N dy = 0 where %3D M %3D N = Which is exact since M, N = are equal.

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 this problem we consider an equation in differential form M dx + N dy = 0. The equation (3a?y+ 6 cos(3æ) y?) dæ + (3x3 – 4y-2)dy = 0 x'y+6 cos(3x) y2) dx + (3x 4y 2)dy = 0 %3D in differential form M dx + N dy = 0 is not exact. Indeed, we have M,- N, For this exercise we can find an integrating factor which is a function of y alone since M N | M is a function of y alone. Namely we have µ(y) Multiplying the original equation by the integrating factor we obtain a new equation M dx + N dy = 0 where M = N = Which is exact since M, = %3D k. N are equal.
Transcribed Image Text:In this problem we consider an equation in differential form M dx + N dy = 0. The equation (3a?y+ 6 cos(3æ) y?) dæ + (3x3 – 4y-2)dy = 0 x'y+6 cos(3x) y2) dx + (3x 4y 2)dy = 0 %3D in differential form M dx + N dy = 0 is not exact. Indeed, we have M,- N, For this exercise we can find an integrating factor which is a function of y alone since M N | M is a function of y alone. Namely we have µ(y) Multiplying the original equation by the integrating factor we obtain a new equation M dx + N dy = 0 where M = N = Which is exact since M, = %3D k. N are equal.
are equal. This problem is exact. Therefore an implicit general solution can be written in the form F(x, y) = C where F(r, y) Finally find the value of the constant C so that the initial condition y(0) = 1 is satisfied. %3D
Transcribed Image Text:are equal. This problem is exact. Therefore an implicit general solution can be written in the form F(x, y) = C where F(r, y) Finally find the value of the constant C so that the initial condition y(0) = 1 is satisfied. %3D
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