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Now we integrate both sides of the equation we have found with the integrating factor. = | x22x + Note that the left side of the equation is the integral of the derivative of e-2xy. Therefore, up to a constant of integration, the left side reduces as follows. 1 [e-2xy] dx = e-2xy |dx = The integration on the right side of the equation requires integration by parts. √x²e-²x +36-²x dx = (-x²6-2x-x-²x-2)-( =(- + 3e-2x dx x² 2 ----( -2x X 2 ])) + c - C + C

Now we integrate both sides of the equation we have found with the integrating factor. = | x22x + Note that the left side of the equation is the integral of the derivative of e-2xy. Therefore, up to a constant of integration, the left side reduces as follows. 1 [e-2xy] dx = e-2xy |dx = The integration on the right side of the equation requires integration by parts. √x²e-²x +36-²x dx = (-x²6-2x-x-²x-2)-( =(- + 3e-2x dx x² 2 ----( -2x X 2 ])) + c - C + C

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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Now we integrate both sides of the equation we have found with the integrating factor. √ [e-²xy] dx = √x²e-²x + Note that the left side of the equation is the integral of the derivative of e-2xy. Therefore, up to a constant of integration, the left side reduces as follows. √ [e-²x] dx = e-2xy + 3e-2x dx The integration on the right side of the equation requires integration by parts. [x³²e-2x + 3e-2x dx = (-x²e-²x-xe-²x-e-²x)-([ 2 (-²---(0 4 2 = e²²x ( -. 2 -²-²-(0) 4 ])) + C + C -2x + C
Transcribed Image Text:Now we integrate both sides of the equation we have found with the integrating factor. √ [e-²xy] dx = √x²e-²x + Note that the left side of the equation is the integral of the derivative of e-2xy. Therefore, up to a constant of integration, the left side reduces as follows. √ [e-²x] dx = e-2xy + 3e-2x dx The integration on the right side of the equation requires integration by parts. [x³²e-2x + 3e-2x dx = (-x²e-²x-xe-²x-e-²x)-([ 2 (-²---(0 4 2 = e²²x ( -. 2 -²-²-(0) 4 ])) + C + C -2x + C
Using a substitution, we now have a separable differential equation. Separate the variables and integrate each side of the equation to find a solution to the equation in x and u. dx = -u²x du dx -u² du 1 dx = 1 -₁² du X -u³ In(|x|) = + Co Submit Skip (you cannot come back)
Transcribed Image Text:Using a substitution, we now have a separable differential equation. Separate the variables and integrate each side of the equation to find a solution to the equation in x and u. dx = -u²x du dx -u² du 1 dx = 1 -₁² du X -u³ In(|x|) = + Co Submit Skip (you cannot come back)
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