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y(x² + y² - 1)dx + x(x² + y 2 + 1)dy = 0

y(x² + y² - 1)dx + x(x² + y 2 + 1)dy = 0

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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Integrating factors found by inspection. Pls use guide the second picture as solving
Example Problem No. 4 y(x* - y")dx + x(x* + y )dy -0 Step 1: Let's expand the terms by distributing the factors. xtydx - yidx + xidy + xyady =0 Step 2: To make it easy for us to identify the differentials, let us group them according to their degrees. xtydx - y'dx + x'dy xy'dy =0 xtydx + x'dy - y'dx + xy'dy =0 (x*ydx + x'dy) + (ydx + xy'dy) = 0 Let us factor out common terms from this equation. x* (ydx + xdy) - y (ydx - xdy) 0 ydx + xdy is also equal to d(xy). However, ydx - xdy is almost simlar to d(xy) except that it lacks y on the denominator's side. So let us divide the whole equation by y to have a denominator of y for the second term. x* (ydx + xdy) - y (ydx - xdy) = 0 ) ydx – xdy . (ydx + xdy) – y Since ydx + xdy is also d(xy) and (ydx - xdyy is also d(wy), the equation will become: (vdx + xdy) – ydx – xdy y =0 Step 3: Idently the integrating factor. For this problem, we wil be needing to identify the integrating factor since we need x and y to appear in the first term because of the d(xy). It could be xy, or xy?, or xy, or xty or the like. They may also appear in the denominator side like , or or or the like. For the second term, they must appear in traction with the x in the numerator and y in the denominator due to the presence of diwy) Since we needed xly for the second term, we can just divide the equation by xy and the equation will become: d(xy) - () » -0 And now, the equation can already be differentiated. 3/4 Salay) - xy - Since it may be hard for some to integrate the second term, we will let u = xly and du = d(xly) d(u) =C xy - xy- zy --c 1 xy +T=C But u= xly xy + 3() xy + C ху Step 4: Simplify. 3x'y + y = 3Cx 3xty + y Cx (fnal answer) :::
Transcribed Image Text:Example Problem No. 4 y(x* - y")dx + x(x* + y )dy -0 Step 1: Let's expand the terms by distributing the factors. xtydx - yidx + xidy + xyady =0 Step 2: To make it easy for us to identify the differentials, let us group them according to their degrees. xtydx - y'dx + x'dy xy'dy =0 xtydx + x'dy - y'dx + xy'dy =0 (x*ydx + x'dy) + (ydx + xy'dy) = 0 Let us factor out common terms from this equation. x* (ydx + xdy) - y (ydx - xdy) 0 ydx + xdy is also equal to d(xy). However, ydx - xdy is almost simlar to d(xy) except that it lacks y on the denominator's side. So let us divide the whole equation by y to have a denominator of y for the second term. x* (ydx + xdy) - y (ydx - xdy) = 0 ) ydx – xdy . (ydx + xdy) – y Since ydx + xdy is also d(xy) and (ydx - xdyy is also d(wy), the equation will become: (vdx + xdy) – ydx – xdy y =0 Step 3: Idently the integrating factor. For this problem, we wil be needing to identify the integrating factor since we need x and y to appear in the first term because of the d(xy). It could be xy, or xy?, or xy, or xty or the like. They may also appear in the denominator side like , or or or the like. For the second term, they must appear in traction with the x in the numerator and y in the denominator due to the presence of diwy) Since we needed xly for the second term, we can just divide the equation by xy and the equation will become: d(xy) - () » -0 And now, the equation can already be differentiated. 3/4 Salay) - xy - Since it may be hard for some to integrate the second term, we will let u = xly and du = d(xly) d(u) =C xy - xy- zy --c 1 xy +T=C But u= xly xy + 3() xy + C ху Step 4: Simplify. 3x'y + y = 3Cx 3xty + y Cx (fnal answer) :::
У(x2+ у2 - 1)dx + x(x2 + у 2 +1)dy %3D0 х(x2 + у 2+1)dy 3D0
Transcribed Image Text:У(x2+ у2 - 1)dx + x(x2 + у 2 +1)dy %3D0 х(x2 + у 2+1)dy 3D0
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