VMOU MT-03 SOLVED ASSIGNMENT | MA/M.SC. MT- 03 (Differential Equations, Calculus of Variations and Special Functions) | July-2024 & January-2025

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.
  1. (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|z1z|<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)2F1(a,b;c;12).
    (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.
  1. 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^2dydx=22y+y2
Whose one particular solution is ( 1 + tan x ) ( 1 + tan x ) (1+tan x)(1+\tan x)(1+tanx).
  1. 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)ddx[2F1(a,b;c;x)]=abc2F1(a+1,b+1;c+1;x)
  1. 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)=0y^{\prime \prime}+\lambda y=0, \quad y^{\prime}(-\pi)=0, \quad y^{\prime}(\pi)=0y+λy=0,y(π)=0,y(π)=0
  1. 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)Ln+1(x)=(2n+1x)Ln(x)nLn1(x) Section – C (Long Answer Questions)
Note :- Answer any one question. Each answer should be given in 800 words. Each question carries 08 marks.
  1. (a) Find the curve with fixed boundary revolves such that its rotation about x x xxx-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=0x(1-x) \frac{d^2 y}{d x^2}+(1+5 x) \frac{d y}{d x}-4 y=0x(1x)d2ydx2+(1+5x)dydx4y=0
  1. (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)dtB(\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 tB(λ,cλ)2F1(a,b;c;z)=01tλ1(1t)cλ12f1(a,b;λ;zt)dt
(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)(2n+1)(x21)Pn=n(n+1)(Pn+1Pn1)
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)}11(x21)Pn+1(x)Pn(x)dx=2n(n+1)(2n+1)(2n+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),dydx=a(x)y2+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 xxx.

Special Cases:

  1. If a ( x ) = 0 a ( x ) = 0 a(x)=0a(x) = 0a(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).dydx=b(x)y+c(x).
  2. If b ( x ) = 0 b ( x ) = 0 b(x)=0b(x) = 0b(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),dydx=a(x)y2+c(x),
    which is a Bernoulli equation when c ( x ) = 0 c ( x ) = 0 c(x)=0c(x) = 0c(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]=x1x2L(x,y(x),y(x))dx,
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 xxx, 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)=dydx,
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.Lyddx(Ly)=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] = 0δJ[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)=12(y)2V(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.d2ydx2+dVdy=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|z1z|<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)2F1(a,b;c;12).

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})2F1(a,b;c;12), let’s analyze the given conditions and the structure of the hypergeometric function 2 F 1 2 F 1 _(2)F_(1){}_2F_12F1.

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)2F1(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!},2F1(a,b;c;z)=n=0(a)n(b)n(c)nznn!,
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+n1),
  • | z | < 1 | z | < 1 |z| < 1|z| < 1|z|<1 ensures convergence of the series.
Given:
  1. | z | < 1 | z | < 1 |z| < 1|z| < 1|z|<1,
  2. | z 1 z | < 1 z 1 z < 1 |(z)/(1-z)| < 1\left| \frac{z}{1-z} \right| < 1|z1z|<1.

Interpretation and Simplification

Step 1: Analyze z 1 z z 1 z (z)/(1-z)\frac{z}{1-z}z1z

The condition | z 1 z | < 1 z 1 z < 1 |(z)/(1-z)| < 1\left| \frac{z}{1-z} \right| < 1|z1z|<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||1z|<1|z|<|1z|.
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)2F1(a,b;c;z) converges, as z z zzz remains well within the unit disk.

Step 2: Evaluate 2 F 1 2 F 1 _(2)F_(1){}_2F_12F1 at z = 1 2 z = 1 2 z=(1)/(2)z = \frac{1}{2}z=12

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})2F1(a,b;c;12) 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!}.2F1(a,b;c;12)=n=0(a)n(b)n(c)n(1/2)nn!.
For a general evaluation, the exact closed form of this series depends on the parameters a , b , c a , b , c a,b,ca, b, ca,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})2F1(a,b;c;12), so the final value will be twice the sum.

Final Answer

Without specific values for a , b , c a , b , c a,b,ca, b, ca,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})22F1(a,b;c;12) 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!}.2n=0(a)n(b)n(c)n(1/2)nn!.

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,(1x2)d2ydx22xdydx+n(n+1)y=0,
where:
  • n n nnn 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)Pn(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).Pn(x)=12nn!dndxn((x21)n).

Applications:

  1. Legendre’s equation arises in solving Laplace’s equation in spherical coordinates.
  2. 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^2dydx=22y+y2
Whose one particular solution is ( 1 + tan x ) ( 1 + tan x ) (1+tan x)(1 + \tan x)(1+tanx).

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,dydx=22y+y2,
with a known particular solution y p ( x ) = 1 + tan x y p ( x ) = 1 + tan x y_(p)(x)=1+tan xy_p(x) = 1 + \tan xyp(x)=1+tanx, 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)=yp(x)+1z(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)=yp(x)+1z(x),
where y p ( x ) = 1 + tan x y p ( x ) = 1 + tan x y_(p)(x)=1+tan xy_p(x) = 1 + \tan xyp(x)=1+tanx 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)=yp(x)+1z(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}.dydx=dypdx1z2dzdx.
Substitute d y d x d y d x (dy)/(dx)\frac{dy}{dx}dydx 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 xy_p(x) = 1 + \tan xyp(x)=1+tanx 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.dypdx=22yp+yp2.
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)=yp(x)+1z 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}dydx=dypdx1z2dzdx 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.dydx=22y+y2.
Simplify step by step:
  1. Substitute y = y p + 1 z y = y p + 1 z y=y_(p)+(1)/(z)y = y_p + \frac{1}{z}y=yp+1z:
    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)^2.22y+y2=22(yp+1z)+(yp+1z)2.
  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(yp+1z)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}.(yp+1z)2=yp2+2ypz+1z2.
  3. 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}.22(yp+1z)+yp2+2ypz+1z2.
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^2dypdx=22yp+yp2, the terms involving y p y p y_(p)y_pyp 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}.