The equation dy + v y² + ¢y+x = 0, dt where y, 4, and x are functions of t, is called the gen- eralized Riccati equation. In general, the equation is not integrable by quadratures. However, suppose that one so- lution, say y = yı, is known. (a) Show that the substitution y = yi + z reduces the generalized Riccati equation to dz + (2yı + ¢) z + y z? = 0, dt which is an instance of Bernoulli's equation (see Ex- ercise 22). (b) Use the fact that y = 1/t is a particular solution of dy 1 y ,2 | dt t2 to find the equation's general solution.
The equation dy + v y² + ¢y+x = 0, dt where y, 4, and x are functions of t, is called the gen- eralized Riccati equation. In general, the equation is not integrable by quadratures. However, suppose that one so- lution, say y = yı, is known. (a) Show that the substitution y = yi + z reduces the generalized Riccati equation to dz + (2yı + ¢) z + y z? = 0, dt which is an instance of Bernoulli's equation (see Ex- ercise 22). (b) Use the fact that y = 1/t is a particular solution of dy 1 y ,2 | dt t2 to find the equation's general solution.
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
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Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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Equations and Inequations
Equations and inequalities describe the relationship between two mathematical expressions.
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A linear function can just be a constant, or it can be the constant multiplied with the variable like x or y. If the variables are of the form, x2, x1/2 or y2 it is not linear. The exponent over the variables should always be 1.
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Please help to answer 27b step-by-step in pic1. Pic2 is the solution - please especially explain the step with the orange arrow.
![(b) Since y, = 1/t is a solution, we set z = y+1/t.
Then y = z – 1/t, and y² = z² – 2z/t +1/t²,
Now it is a matter of unravelling the changes of
variable. First
%3D
so
1
z' = y'
2
1
z(t) :
2
%3D
t2
w(t)
t + 2C†³
t + Bt3 '
1
+y² -
y
1
Where we have set B = 2C. Then
t2
3z
+z?.
t2
1
y(t) = z(t)
2
1
= --
t + Bt3
This is a Bernoulli equation with n = 2. Thus
we set w = 1/z. Differentiating, we get
This looks a little better if we use partial frac-
%3D
tions to write
1
w'
2
1
y(t) :
t + Bt3
2Bt
1
+
= -
1+ Bt2
1
2Bt
3
= -
= -
tz
t
1+ Bt2 °
3w
1.
This is a linear equation, and t-3 is an integrat-
ing factor.
3w
-t-3
[w] = -t-3
1-³w(t)
1
-2
+ C
=
1
w(t) = ; +Cr³](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F1b1372ae-e668-4971-89ee-29da7ac7c466%2F92c92b7e-c494-4caa-bcce-586cc8db49a3%2F3ijx2gs_processed.jpeg&w=3840&q=75)
Transcribed Image Text:(b) Since y, = 1/t is a solution, we set z = y+1/t.
Then y = z – 1/t, and y² = z² – 2z/t +1/t²,
Now it is a matter of unravelling the changes of
variable. First
%3D
so
1
z' = y'
2
1
z(t) :
2
%3D
t2
w(t)
t + 2C†³
t + Bt3 '
1
+y² -
y
1
Where we have set B = 2C. Then
t2
3z
+z?.
t2
1
y(t) = z(t)
2
1
= --
t + Bt3
This is a Bernoulli equation with n = 2. Thus
we set w = 1/z. Differentiating, we get
This looks a little better if we use partial frac-
%3D
tions to write
1
w'
2
1
y(t) :
t + Bt3
2Bt
1
+
= -
1+ Bt2
1
2Bt
3
= -
= -
tz
t
1+ Bt2 °
3w
1.
This is a linear equation, and t-3 is an integrat-
ing factor.
3w
-t-3
[w] = -t-3
1-³w(t)
1
-2
+ C
=
1
w(t) = ; +Cr³
Transcribed Image Text:27. The equation
dy
+ ý y² + ¢y+ x = 0,
dt
where y, 4, and x are functions of t, is called the gen-
eralized Riccati equation. In general, the equation is not
integrable by quadratures. However, suppose that one so-
lution, say y = y1, is known.
(a) Show that the substitution y
generalized Riccati equation to
yi + z reduces the
dz
+ (2yı½ + ¢) z + y z? = 0,
dt
which is an instance of Bernoulli's equation (see Ex-
ercise 22).
(b) Use the fact that y = 1/t is a particular solution of
dy
1
+ y?
t
|
dt
t2
to find the equation's general solution.
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