Section-A
(Very Short Answer Type Questions)
Note:- Answer all questions. As per the nature of the question you delimit your answer in one word, one sentence or maximum up to 30 words. Each question carries 1 (one) mark.
(i). Write down the Riccati’s equation.
(ii). Write down Euler-Largange equation.
(iii). If
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z|<1 | z | < 1 and
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left|\frac{z}{1-z}\right|<1 | z 1 − z | < 1 then write the value of
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
a
,
b
;
c
;
1
2
2F_(1)(a,b;c;(1)/(2)) 2 F_1\left(a, b ; c ; \frac{1}{2}\right) 2 F 1 ( a , b ; c ; 1 2 ) .
(iv). Write down the Legendre equation.
Section-B
(Short Answer Questions)
Note :- Answer any two questions. Each answer should be given in 200 words.
Each question carries 4 marks.
Find the general solution of the Riccati’s equation.
d
y
d
x
=
2
−
2
y
+
y
2
d
y
d
x
=
2
−
2
y
+
y
2
(dy)/(dx)=2-2y+y^(2) \frac{d y}{d x}=2-2 y+y^2 d y d x = 2 − 2 y + y 2
Whose one particular solution is
(
1
+
tan
x
)
(
1
+
tan
x
)
(1+tan x) (1+\tan x) ( 1 + tan x ) .
Prove that
d
d
x
[
2
F
1
(
a
,
b
;
c
;
x
)
]
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
d
d
x
2
F
1
(
a
,
b
;
c
;
x
)
=
a
b
c
2
F
1
(
a
+
1
,
b
+
1
;
c
+
1
;
x
)
(d)/(dx)[_(2)F_(1)(a,b;c;x)]=(ab)/(c)_(2)F_(1)(a+1,b+1;c+1;x) \frac{d}{d x}\left[{ }_2 F_1(a, b ; c ; x)\right]=\frac{a b}{c}{ }_2 F_1(a+1, b+1 ; c+1 ; x) d d x [ 2 F 1 ( a , b ; c ; x ) ] = a b c 2 F 1 ( a + 1 , b + 1 ; c + 1 ; x )
Solve the following strem Liouville problem.
y
′
′
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0
y
′
′
+
λ
y
=
0
,
y
′
(
−
π
)
=
0
,
y
′
(
π
)
=
0
y^(”)+lambda y=0,quady^(‘)(-pi)=0,quady^(‘)(pi)=0 y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(-\pi)=0, \quad y^{\prime}(\pi)=0 y ′ ′ + λ y = 0 , y ′ ( − π ) = 0 , y ′ ( π ) = 0
Prove that:
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
Section – C
(Long Answer Questions)
(
n
+
1
)
L
n
+
1
(
x
)
=
(
2
n
+
1
−
x
)
L
n
(
x
)
−
n
L
n
−
1
(
x
)
Section – C
(Long Answer Questions)
{:[(n+1)L_(n+1)(x)=(2n+1-x)L_(n)(x)-nL_(n-1)(x)],[quad” Section – C “],[” (Long Answer Questions) “]:} \begin{gathered}
(n+1) L_{n+1}(x)=(2 n+1-x) L_n(x)-n L_{n-1}(x) \\
\quad \text { Section – C } \\
\text { (Long Answer Questions) }
\end{gathered} ( n + 1 ) L n + 1 ( x ) = ( 2 n + 1 − x ) L n ( x ) − n L n − 1 ( x ) Section – C (Long Answer Questions)
Note :- Answer any one question. Each answer should be given in 800 words. Each question carries 08 marks.
(a) Find the curve with fixed boundary revolves such that its rotation about
x
x
x x x -axis generated minimal surface area.
(b) Solve in series :
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0
x
(
1
−
x
)
d
2
y
d
x
2
+
(
1
+
5
x
)
d
y
d
x
−
4
y
=
0
x(1-x)(d^(2)y)/(dx^(2))+(1+5x)(dy)/(dx)-4y=0 x(1-x) \frac{d^2 y}{d x^2}+(1+5 x) \frac{d y}{d x}-4 y=0 x ( 1 − x ) d 2 y d x 2 + ( 1 + 5 x ) d y d x − 4 y = 0
(a) Prove that :
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
f
1
(
a
,
b
;
λ
;
z
t
)
d
t
B
(
λ
,
c
−
λ
)
2
F
1
(
a
,
b
;
c
;
z
)
=
∫
0
1
t
λ
−
1
(
1
−
t
)
c
−
λ
−
1
2
f
1
(
a
,
b
;
λ
;
z
t
)
d
t
B(lambda,c-lambda)2F_(1)(a,b;c;z)=int_(0)^(1)t^(lambda-1)(1-t)^(c-lambda-1)2f_(1)(a,b;lambda;zt)dt B(\lambda, c-\lambda) 2 F_1(a, b ; c ; z)=\int_0^1 t^{\lambda-1}(1-t)^{c-\lambda-1} 2 f_1(a, b ; \lambda ; z t) d t B ( λ , c − λ ) 2 F 1 ( a , b ; c ; z ) = ∫ 0 1 t λ − 1 ( 1 − t ) c − λ − 1 2 f 1 ( a , b ; λ ; z t ) d t
(b) Prove that:
(
2
n
+
1
)
(
x
2
−
1
)
P
n
′
=
n
(
n
+
1
)
(
P
n
+
1
−
P
n
−
1
)
(
2
n
+
1
)
x
2
−
1
P
n
′
=
n
(
n
+
1
)
P
n
+
1
−
P
n
−
1
(2n+1)(x^(2)-1)P_(n)^(‘)=n(n+1)(P_(n+1)-P_(n-1)) (2 n+1)\left(x^2-1\right) P_n^{\prime}=n(n+1)\left(P_{n+1}-P_{n-1}\right) ( 2 n + 1 ) ( x 2 − 1 ) P n ′ = n ( n + 1 ) ( P n + 1 − P n − 1 )
and hence deduce that
∫
−
1
1
(
x
2
−
1
)
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
∫
−
1
1
x
2
−
1
P
n
+
1
(
x
)
P
n
′
(
x
)
d
x
=
2
n
(
n
+
1
)
(
2
n
+
1
)
(
2
n
+
3
)
int_(-1)^(1)(x^(2)-1)P_(n+1)(x)P_(n)^(‘)(x)dx=(2n(n+1))/((2n+1)(2n+3)) \int_{-1}^1\left(x^2-1\right) P_{n+1}(x) P_n^{\prime}(x) d x=\frac{2 n(n+1)}{(2 n+1)(2 n+3)} ∫ − 1 1 ( x 2 − 1 ) P n + 1 ( x ) P n ′ ( x ) d x = 2 n ( n + 1 ) ( 2 n + 1 ) ( 2 n + 3 )
Answer:
Question:-01(a)
Write down the Riccati’s equation.
Answer:
Riccati’s Equation:
The Riccati equation is a first-order, nonlinear differential equation of the form:
d
y
d
x
=
a
(
x
)
y
2
+
b
(
x
)
y
+
c
(
x
)
,
d
y
d
x
=
a
(
x
)
y
2
+
b
(
x
)
y
+
c
(
x
)
,
(dy)/(dx)=a(x)y^(2)+b(x)y+c(x), \frac{dy}{dx} = a(x) y^2 + b(x) y + c(x), d y d x = a ( x ) y 2 + b ( x ) y + c ( x ) ,
where:
y
=
y
(
x
)
y
=
y
(
x
)
y=y(x) y = y(x) y = y ( x ) is the unknown function,
a
(
x
)
,
b
(
x
)
,
c
(
x
)
a
(
x
)
,
b
(
x
)
,
c
(
x
)
a(x),b(x),c(x) a(x), b(x), c(x) a ( x ) , b ( x ) , c ( x ) are given functions of
x
x
x x x .
Special Cases:
If
a
(
x
)
=
0
a
(
x
)
=
0
a(x)=0 a(x) = 0 a ( x ) = 0 , the Riccati equation reduces to a linear first-order differential equation:
d
y
d
x
=
b
(
x
)
y
+
c
(
x
)
.
d
y
d
x
=
b
(
x
)
y
+
c
(
x
)
.
(dy)/(dx)=b(x)y+c(x). \frac{dy}{dx} = b(x) y + c(x). d y d x = b ( x ) y + c ( x ) .
If
b
(
x
)
=
0
b
(
x
)
=
0
b(x)=0 b(x) = 0 b ( x ) = 0 , it becomes:
d
y
d
x
=
a
(
x
)
y
2
+
c
(
x
)
,
d
y
d
x
=
a
(
x
)
y
2
+
c
(
x
)
,
(dy)/(dx)=a(x)y^(2)+c(x), \frac{dy}{dx} = a(x) y^2 + c(x), d y d x = a ( x ) y 2 + c ( x ) ,
which is a Bernoulli equation when
c
(
x
)
=
0
c
(
x
)
=
0
c(x)=0 c(x) = 0 c ( x ) = 0 .
Question:-01(b)
Write down Euler-Lagrange equation.
Answer:
Euler-Lagrange Equation
The Euler-Lagrange equation is a fundamental equation in the calculus of variations. It provides the necessary condition for a functional to have an extremum.
Statement:
Given a functional of the form:
J
[
y
]
=
∫
x
1
x
2
L
(
x
,
y
(
x
)
,
y
′
(
x
)
)
d
x
,
J
[
y
]
=
∫
x
1
x
2
L
(
x
,
y
(
x
)
,
y
′
(
x
)
)
d
x
,
J[y]=int_(x_(1))^(x_(2))L(x,y(x),y^(‘)(x))dx, J[y] = \int_{x_1}^{x_2} L(x, y(x), y'(x)) \, dx, J [ y ] = ∫ x 1 x 2 L ( x , y ( x ) , y ′ ( x ) ) d x ,
where:
L
(
x
,
y
,
y
′
)
L
(
x
,
y
,
y
′
)
L(x,y,y^(‘)) L(x, y, y’) L ( x , y , y ′ ) is the Lagrangian , a function of
x
x
x x x ,
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) , and
y
′
(
x
)
y
′
(
x
)
y^(‘)(x) y'(x) y ′ ( x ) ,
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) is the function to be determined,
y
′
(
x
)
=
d
y
d
x
y
′
(
x
)
=
d
y
d
x
y^(‘)(x)=(dy)/(dx) y'(x) = \frac{dy}{dx} y ′ ( x ) = d y d x ,
the Euler-Lagrange equation is:
∂
L
∂
y
−
d
d
x
(
∂
L
∂
y
′
)
=
0.
∂
L
∂
y
−
d
d
x
∂
L
∂
y
′
=
0.
(del L)/(del y)-(d)/(dx)((del L)/(dely^(‘)))=0. \frac{\partial L}{\partial y} – \frac{d}{dx} \left( \frac{\partial L}{\partial y’} \right) = 0. ∂ L ∂ y − d d x ( ∂ L ∂ y ′ ) = 0.
Derivation Idea:
The Euler-Lagrange equation arises from requiring that the first variation of the functional
J
[
y
]
J
[
y
]
J[y] J[y] J [ y ] vanishes, i.e.,
δ
J
[
y
]
=
0
δ
J
[
y
]
=
0
delta J[y]=0 \delta J[y] = 0 δ J [ y ] = 0 .
Example Application:
If
L
(
x
,
y
,
y
′
)
=
1
2
(
y
′
)
2
−
V
(
y
)
L
(
x
,
y
,
y
′
)
=
1
2
(
y
′
)
2
−
V
(
y
)
L(x,y,y^(‘))=(1)/(2)(y^(‘))^(2)-V(y) L(x, y, y’) = \frac{1}{2}(y’)^2 – V(y) L ( x , y , y ′ ) = 1 2 ( y ′ ) 2 − V ( y ) , the Euler-Lagrange equation gives:
d
2
y
d
x
2
+
d
V
d
y
=
0.
d
2
y
d
x
2
+
d
V
d
y
=
0.
(d^(2)y)/(dx^(2))+(dV)/(dy)=0. \frac{d^2y}{dx^2} + \frac{dV}{dy} = 0. d 2 y d x 2 + d V d y = 0.
Question:-01(c)
If
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z| < 1 | z | < 1 and
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left|\frac{z}{1-z}\right| < 1 | z 1 − z | < 1 , then write the value of
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
a
,
b
;
c
;
1
2
2F_(1)(a,b;c;(1)/(2)) 2 F_1\left(a, b ; c ; \frac{1}{2}\right) 2 F 1 ( a , b ; c ; 1 2 ) .
Answer:
To evaluate
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
(
a
,
b
;
c
;
1
2
)
2F_(1)(a,b;c;(1)/(2)) 2 F_1(a, b; c; \frac{1}{2}) 2 F 1 ( a , b ; c ; 1 2 ) , let’s analyze the given conditions and the structure of the hypergeometric function
2
F
1
2
F
1
_(2)F_(1) {}_2F_1 2 F 1 .
Problem Setup
The hypergeometric function
2
F
1
(
a
,
b
;
c
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
_(2)F_(1)(a,b;c;z) {}_2F_1(a, b; c; z) 2 F 1 ( a , b ; c ; z ) is defined as:
2
F
1
(
a
,
b
;
c
;
z
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
z
n
n
!
,
2
F
1
(
a
,
b
;
c
;
z
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
z
n
n
!
,
_(2)F_(1)(a,b;c;z)=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))(z^(n))/(n!), {}_2F_1(a, b; c; z) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{z^n}{n!}, 2 F 1 ( a , b ; c ; z ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n z n n ! ,
where:
(
a
)
n
(
a
)
n
(a)_(n) (a)_n ( a ) n is the Pochhammer symbol:
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(
a
)
n
=
a
(
a
+
1
)
(
a
+
2
)
⋯
(
a
+
n
−
1
)
(a)_(n)=a(a+1)(a+2)cdots(a+n-1) (a)_n = a (a+1)(a+2)\cdots(a+n-1) ( a ) n = a ( a + 1 ) ( a + 2 ) ⋯ ( a + n − 1 ) ,
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z| < 1 | z | < 1 ensures convergence of the series.
Given:
|
z
|
<
1
|
z
|
<
1
|z| < 1 |z| < 1 | z | < 1 ,
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left| \frac{z}{1-z} \right| < 1 | z 1 − z | < 1 .
Interpretation and Simplification
Step 1: Analyze
z
1
−
z
z
1
−
z
(z)/(1-z) \frac{z}{1-z} z 1 − z
The condition
|
z
1
−
z
|
<
1
z
1
−
z
<
1
|(z)/(1-z)| < 1 \left| \frac{z}{1-z} \right| < 1 | z 1 − z | < 1 implies:
|
z
|
|
1
−
z
|
<
1
⟹
|
z
|
<
|
1
−
z
|
.
|
z
|
|
1
−
z
|
<
1
⟹
|
z
|
<
|
1
−
z
|
.
(|z|)/(|1-z|) < 1quadLongrightarrowquad|z| < |1-z|. \frac{|z|}{|1-z|} < 1 \quad \implies \quad |z| < |1-z|. | z | | 1 − z | < 1 ⟹ | z | < | 1 − z | .
This suggests the series for
2
F
1
(
a
,
b
;
c
;
z
)
2
F
1
(
a
,
b
;
c
;
z
)
_(2)F_(1)(a,b;c;z) {}_2F_1(a, b; c; z) 2 F 1 ( a , b ; c ; z ) converges, as
z
z
z z z remains well within the unit disk.
Step 2: Evaluate
2
F
1
2
F
1
_(2)F_(1) {}_2F_1 2 F 1 at
z
=
1
2
z
=
1
2
z=(1)/(2) z = \frac{1}{2} z = 1 2
The value of
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
(
a
,
b
;
c
;
1
2
)
_(2)F_(1)(a,b;c;(1)/(2)) {}_2F_1(a, b; c; \frac{1}{2}) 2 F 1 ( a , b ; c ; 1 2 ) is derived from the series expansion:
2
F
1
(
a
,
b
;
c
;
1
2
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
2
F
1
(
a
,
b
;
c
;
1
2
)
=
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
_(2)F_(1)(a,b;c;(1)/(2))=sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))((1//2)^(n))/(n!). {}_2F_1(a, b; c; \frac{1}{2}) = \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{(1/2)^n}{n!}. 2 F 1 ( a , b ; c ; 1 2 ) = ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n ( 1 / 2 ) n n ! .
For a general evaluation, the exact closed form of this series depends on the parameters
a
,
b
,
c
a
,
b
,
c
a,b,c a, b, c a , b , c . However, in some cases, known results for specific parameter values can be used.
Step 3: Factor of 2
The problem asks for
2
F
1
(
a
,
b
;
c
;
1
2
)
2
F
1
(
a
,
b
;
c
;
1
2
)
2F_(1)(a,b;c;(1)/(2)) 2 F_1(a, b; c; \frac{1}{2}) 2 F 1 ( a , b ; c ; 1 2 ) , so the final value will be twice the sum.
Final Answer
Without specific values for
a
,
b
,
c
a
,
b
,
c
a,b,c a, b, c a , b , c , the value of
2
2
F
1
(
a
,
b
;
c
;
1
2
)
2
2
F
1
(
a
,
b
;
c
;
1
2
)
2_(2)F_(1)(a,b;c;(1)/(2)) 2 {}_2F_1(a, b; c; \frac{1}{2}) 2 2 F 1 ( a , b ; c ; 1 2 ) is written as:
2
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
2
∑
n
=
0
∞
(
a
)
n
(
b
)
n
(
c
)
n
(
1
/
2
)
n
n
!
.
2sum_(n=0)^(oo)((a)_(n)(b)_(n))/((c)_(n))((1//2)^(n))/(n!). 2 \sum_{n=0}^\infty \frac{(a)_n (b)_n}{(c)_n} \frac{(1/2)^n}{n!}. 2 ∑ n = 0 ∞ ( a ) n ( b ) n ( c ) n ( 1 / 2 ) n n ! .
Question:-01(d)
Write down the Legendre equation.
Answer:
Legendre’s Differential Equation:
The Legendre equation is a second-order linear differential equation of the form:
(
1
−
x
2
)
d
2
y
d
x
2
−
2
x
d
y
d
x
+
n
(
n
+
1
)
y
=
0
,
(
1
−
x
2
)
d
2
y
d
x
2
−
2
x
d
y
d
x
+
n
(
n
+
1
)
y
=
0
,
(1-x^(2))(d^(2)y)/(dx^(2))-2x(dy)/(dx)+n(n+1)y=0, (1 – x^2) \frac{d^2y}{dx^2} – 2x \frac{dy}{dx} + n(n+1)y = 0, ( 1 − x 2 ) d 2 y d x 2 − 2 x d y d x + n ( n + 1 ) y = 0 ,
where:
n
n
n n n is a non-negative integer (degree of the associated Legendre polynomial),
y
=
y
(
x
)
y
=
y
(
x
)
y=y(x) y = y(x) y = y ( x ) is the unknown function,
x
∈
(
−
1
,
1
)
x
∈
(
−
1
,
1
)
x in(-1,1) x \in (-1, 1) x ∈ ( − 1 , 1 ) .
General Solution:
The solutions to Legendre’s equation are the
Legendre polynomials
P
n
(
x
)
P
n
(
x
)
P_(n)(x) P_n(x) P n ( x ) , which form a complete orthogonal set over
[
−
1
,
1
]
[
−
1
,
1
]
[-1,1] [-1, 1] [ − 1 , 1 ] and are defined recursively or by the Rodrigues formula:
P
n
(
x
)
=
1
2
n
n
!
d
n
d
x
n
(
(
x
2
−
1
)
n
)
.
P
n
(
x
)
=
1
2
n
n
!
d
n
d
x
n
(
x
2
−
1
)
n
.
P_(n)(x)=(1)/(2^(n)n!)(d^(n))/(dx^(n))((x^(2)-1)^(n)). P_n(x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} \left( (x^2 – 1)^n \right). P n ( x ) = 1 2 n n ! d n d x n ( ( x 2 − 1 ) n ) .
Applications:
Legendre’s equation arises in solving Laplace’s equation in spherical coordinates.
It is used extensively in physics, particularly in potential theory and quantum mechanics.
Question:-02
Find the general solution of the Riccati’s equation:
d
y
d
x
=
2
−
2
y
+
y
2
d
y
d
x
=
2
−
2
y
+
y
2
(dy)/(dx)=2-2y+y^(2) \frac{d y}{d x} = 2 – 2 y + y^2 d y d x = 2 − 2 y + y 2
Whose one particular solution is
(
1
+
tan
x
)
(
1
+
tan
x
)
(1+tan x) (1 + \tan x) ( 1 + tan x ) .
Answer:
To solve the Riccati equation:
d
y
d
x
=
2
−
2
y
+
y
2
,
d
y
d
x
=
2
−
2
y
+
y
2
,
(dy)/(dx)=2-2y+y^(2), \frac{dy}{dx} = 2 – 2y + y^2, d y d x = 2 − 2 y + y 2 ,
with a known particular solution
y
p
(
x
)
=
1
+
tan
x
y
p
(
x
)
=
1
+
tan
x
y_(p)(x)=1+tan x y_p(x) = 1 + \tan x y p ( x ) = 1 + tan x , we will transform the equation into a linear equation by substituting
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y(x)=y_(p)(x)+(1)/(z(x)) y(x) = y_p(x) + \frac{1}{z(x)} y ( x ) = y p ( x ) + 1 z ( x ) .
Step 1: General Substitution
Let:
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
,
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
,
y(x)=y_(p)(x)+(1)/(z(x)), y(x) = y_p(x) + \frac{1}{z(x)}, y ( x ) = y p ( x ) + 1 z ( x ) ,
where
y
p
(
x
)
=
1
+
tan
x
y
p
(
x
)
=
1
+
tan
x
y_(p)(x)=1+tan x y_p(x) = 1 + \tan x y p ( x ) = 1 + tan x is the given particular solution. Substituting this into the Riccati equation, we aim to derive a linear equation for
z
(
x
)
z
(
x
)
z(x) z(x) z ( x ) .
Step 2: Differentiate
y
(
x
)
y
(
x
)
y(x) y(x) y ( x ) :
Differentiating
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y
(
x
)
=
y
p
(
x
)
+
1
z
(
x
)
y(x)=y_(p)(x)+(1)/(z(x)) y(x) = y_p(x) + \frac{1}{z(x)} y ( x ) = y p ( x ) + 1 z ( x ) , we get:
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
.
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
.
(dy)/(dx)=(dy_(p))/(dx)-(1)/(z^(2))(dz)/(dx). \frac{dy}{dx} = \frac{dy_p}{dx} – \frac{1}{z^2} \frac{dz}{dx}. d y d x = d y p d x − 1 z 2 d z d x .
Substitute
d
y
d
x
d
y
d
x
(dy)/(dx) \frac{dy}{dx} d y d x into the Riccati equation.
Step 3: Simplify Using the Particular Solution:
Using the fact that
y
p
(
x
)
=
1
+
tan
x
y
p
(
x
)
=
1
+
tan
x
y_(p)(x)=1+tan x y_p(x) = 1 + \tan x y p ( x ) = 1 + tan x satisfies the Riccati equation, we know:
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
.
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
.
(dy_(p))/(dx)=2-2y_(p)+y_(p)^(2). \frac{dy_p}{dx} = 2 – 2y_p + y_p^2. d y p d x = 2 − 2 y p + y p 2 .
Now substitute
y
(
x
)
=
y
p
(
x
)
+
1
z
y
(
x
)
=
y
p
(
x
)
+
1
z
y(x)=y_(p)(x)+(1)/(z) y(x) = y_p(x) + \frac{1}{z} y ( x ) = y p ( x ) + 1 z and
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
d
y
d
x
=
d
y
p
d
x
−
1
z
2
d
z
d
x
(dy)/(dx)=(dy_(p))/(dx)-(1)/(z^(2))(dz)/(dx) \frac{dy}{dx} = \frac{dy_p}{dx} – \frac{1}{z^2} \frac{dz}{dx} d y d x = d y p d x − 1 z 2 d z d x into the equation:
d
y
d
x
=
2
−
2
y
+
y
2
.
d
y
d
x
=
2
−
2
y
+
y
2
.
(dy)/(dx)=2-2y+y^(2). \frac{dy}{dx} = 2 – 2y + y^2. d y d x = 2 − 2 y + y 2 .
Simplify step by step:
Substitute
y
=
y
p
+
1
z
y
=
y
p
+
1
z
y=y_(p)+(1)/(z) y = y_p + \frac{1}{z} y = y p + 1 z :
2
−
2
y
+
y
2
=
2
−
2
(
y
p
+
1
z
)
+
(
y
p
+
1
z
)
2
.
2
−
2
y
+
y
2
=
2
−
2
y
p
+
1
z
+
y
p
+
1
z
2
.
2-2y+y^(2)=2-2(y_(p)+(1)/(z))+(y_(p)+(1)/(z))^(2). 2 – 2y + y^2 = 2 – 2\left(y_p + \frac{1}{z}\right) + \left(y_p + \frac{1}{z}\right)^2. 2 − 2 y + y 2 = 2 − 2 ( y p + 1 z ) + ( y p + 1 z ) 2 .
Expand
(
y
p
+
1
z
)
2
y
p
+
1
z
2
(y_(p)+(1)/(z))^(2) \left(y_p + \frac{1}{z}\right)^2 ( y p + 1 z ) 2 :
(
y
p
+
1
z
)
2
=
y
p
2
+
2
y
p
z
+
1
z
2
.
y
p
+
1
z
2
=
y
p
2
+
2
y
p
z
+
1
z
2
.
(y_(p)+(1)/(z))^(2)=y_(p)^(2)+(2y_(p))/(z)+(1)/(z^(2)). \left(y_p + \frac{1}{z}\right)^2 = y_p^2 + \frac{2y_p}{z} + \frac{1}{z^2}. ( y p + 1 z ) 2 = y p 2 + 2 y p z + 1 z 2 .
Collect terms:
2
−
2
(
y
p
+
1
z
)
+
y
p
2
+
2
y
p
z
+
1
z
2
.
2
−
2
y
p
+
1
z
+
y
p
2
+
2
y
p
z
+
1
z
2
.
2-2(y_(p)+(1)/(z))+y_(p)^(2)+(2y_(p))/(z)+(1)/(z^(2)). 2 – 2\left(y_p + \frac{1}{z}\right) + y_p^2 + \frac{2y_p}{z} + \frac{1}{z^2}. 2 − 2 ( y p + 1 z ) + y p 2 + 2 y p z + 1 z 2 .
Substituting
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
d
y
p
d
x
=
2
−
2
y
p
+
y
p
2
(dy_(p))/(dx)=2-2y_(p)+y_(p)^(2) \frac{dy_p}{dx} = 2 – 2y_p + y_p^2 d y p d x = 2 − 2 y p + y p 2 , the terms involving
y
p
y
p
y_(p) y_p y p cancel out, leaving:
−
1
z
2
d
z
d
x
=
−
2
z
.
−
1
z
2
d
z
d
x
=
−
2
z
.
-(1)/(z^(2))(dz)/(dx)=-(2)/(z). -\frac{1}{z^2} \frac{dz}{dx} = -\frac{2}{z}.