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
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Please answer the 2 questions ty.
![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](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fbb52044d-d588-4d60-8413-960f544d05a0%2F39ceca9d-bb31-4e3a-8daa-c8ab7af87a0f%2Fz0bv2v_processed.jpeg&w=3840&q=75)
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
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
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