No. 10 - Consider the Riccati equation dy dx where = · P(x) + yQ(x) + y²R(x), (1) P(x) = 6x22x5+4x8, Q(x) = x² - 4x5, and R(x) = x². Find values of the constants C and y > 0 so that y=v=CxY is a solution of (1). Substituting y = v = Cx into equation (1) gives 2 Cуx-1=6x² - 2x5 + 4x8 + CС x² + 1 - 4 Cx5 +7 + C²x²+2y. Consider first the powers of x involving Y. Since y > 0, all the powers involving on the RHS are strictly greater than y- 1. The smallest power of x on the LHS must match the smallest power of x on the RHS, which is therefore 2. So the only possible value for y is y = 3, and hence by comparing coefficients, C′ = 2. γ It is straightforward to check that with these values for C and Y, equation (1) is satisfied. No. 610-0 Consider the Riccati equation dy dx where = P(x) + yQ(x) + y² R(x), (1) P(x) = 6 x² - 2x5 + 4x8, Q(x) = x²-4x5, and R(x) = x². A particular solution of (1) is y = v = * Writing Y 1 = v+ Z solution of (1). 2x³. find a first-order differential equation for z such that y = v + How come Substituting y = 2 x³ + 1/2 into equation (1) gives dz 6 x² dx = z2 ?? 1 is a אן 6 x² − 2 x³ + 4x³ + (2 ׳ + 1/z)(x² − 4x³) + (2 x ³ + ±) ² (x²) Z Expanding, 6 x² - dz dx z2 = 2 6 x² + x2 x2 z2 Z This can be simplified to zx² + dz dx =
No. 10 - Consider the Riccati equation dy dx where = · P(x) + yQ(x) + y²R(x), (1) P(x) = 6x22x5+4x8, Q(x) = x² - 4x5, and R(x) = x². Find values of the constants C and y > 0 so that y=v=CxY is a solution of (1). Substituting y = v = Cx into equation (1) gives 2 Cуx-1=6x² - 2x5 + 4x8 + CС x² + 1 - 4 Cx5 +7 + C²x²+2y. Consider first the powers of x involving Y. Since y > 0, all the powers involving on the RHS are strictly greater than y- 1. The smallest power of x on the LHS must match the smallest power of x on the RHS, which is therefore 2. So the only possible value for y is y = 3, and hence by comparing coefficients, C′ = 2. γ It is straightforward to check that with these values for C and Y, equation (1) is satisfied. No. 610-0 Consider the Riccati equation dy dx where = P(x) + yQ(x) + y² R(x), (1) P(x) = 6 x² - 2x5 + 4x8, Q(x) = x²-4x5, and R(x) = x². A particular solution of (1) is y = v = * Writing Y 1 = v+ Z solution of (1). 2x³. find a first-order differential equation for z such that y = v + How come Substituting y = 2 x³ + 1/2 into equation (1) gives dz 6 x² dx = z2 ?? 1 is a אן 6 x² − 2 x³ + 4x³ + (2 ׳ + 1/z)(x² − 4x³) + (2 x ³ + ±) ² (x²) Z Expanding, 6 x² - dz dx z2 = 2 6 x² + x2 x2 z2 Z This can be simplified to zx² + dz dx =
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:No. 10
-
Consider the Riccati equation
dy
dx
where
=
· P(x) + yQ(x) + y²R(x), (1)
P(x) = 6x22x5+4x8,
Q(x) = x² - 4x5, and
R(x) = x².
Find values of the constants C and y > 0 so that
y=v=CxY
is a solution of (1).
Substituting y = v = Cx into equation (1) gives
2
Cуx-1=6x² - 2x5 + 4x8 + CС x² + 1 - 4 Cx5 +7 + C²x²+2y.
Consider first the powers of x involving Y. Since y > 0, all the powers involving
on the
RHS are strictly greater than y- 1. The smallest power of x on the LHS must match the
smallest power of x on the RHS, which is therefore 2.
So the only possible value for y is y = 3, and hence by comparing coefficients, C′ = 2.
γ
It is straightforward to check that with these values for C and Y, equation (1) is satisfied.
Transcribed Image Text:No. 610-0
Consider the Riccati equation
dy
dx
where
=
P(x) + yQ(x) + y² R(x), (1)
P(x) = 6 x² - 2x5 + 4x8,
Q(x) = x²-4x5, and
R(x) = x².
A particular solution of (1) is y = v =
*
Writing
Y
1
=
v+
Z
solution of (1).
2x³.
find a first-order differential equation for z such that y = v +
How come
Substituting y = 2 x³ + 1/2 into equation (1) gives
dz
6 x²
dx
=
z2
??
1
is a
אן
6 x² − 2 x³ + 4x³ + (2 ׳ + 1/z)(x² − 4x³) + (2 x ³ + ±) ² (x²)
Z
Expanding,
6 x²
-
dz
dx
z2
=
2
6 x² +
x2
x2
z2
Z
This can be simplified to
zx² +
dz
dx
=
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