Since only two values of yare given, the polynomial which fits the data is of degree 1.
Hence 2nd differences are zeros
∴ Δ²(y₀) = 0
⇒ (E-1)²y_o=0
⇒(E² - 2E + 1) y_o = 0
⇒y₂ - 2y₁ +y_o = 0
⇒34 - 2y₁ + 51 = 0
​​​​​​​⇒85 - 2y₁ =0
​​​​​​​⇒2y₁ + 85 ​​​​​​​⇒ y₁ = \(\frac{85}{2}\)
​​​​​​​⇒y₁ = 42.5 

LHS = Δ∇ = (E-1) \(\left( \frac { E-1 }{ E } \right) \)

= Δ - ∇ [∵ Δ = E -1 & ∇ = \(\frac { E-1 }{ E } \) ]
= RHS
Hence Proved.

⇒ \(\frac { dy }{ dx } \) = -e^x(1+y)
Separating the variables we get,
⇒ \(\frac { dy }{ 1+y } \) = -e^x dx
⇒ log(1+y) = -e^x+c

(i) Δe^ax = e^a(x+h)−e^x
= e^aX.e^h-e^ax [∵ a^m+n = a^m.aⁿ]
= e^ax[e^h-1]
 (ii) Δ²e^x = Δ.[Δe^x]
= Δ[e^x+h - e^x]
= Δ[e^xe^h - e^x]
= Δe^x [e^h - 1]
= (e^h - 1)Δe^x 
= (e^h−1).(e^h−1).e^x
= (e^h−1)².e^x
(iii) Δ log x = log(x+h) − log x
= log \(\frac{x + h}{x}\)
= log \(\left( \frac { x }{ x } +\frac { h }{ x } \right) \)
= log \(\left( 1+\frac { h }{ x } \right) \)

The auxiliary equation is m² + 16 = 0
m₂ = -16
⇒ m₂ = ±\(\sqrt { -16 } \) = ±4i
Hence α = 0 and β = 4
∴ Complementary function CF is
e^ax = [A cos βx + B sin βx]
CF = e^0x[A cos 4x + B sin 4x]
= A cos 4x + B sin 4x
[∵ e^o= 1]
∴ The general solution is y = A cos 4x + B sin 4x

The highest derivative is first order and its power is one
∴ order : 1 
degree : 1

Since four values of f(x) are given
\( \Delta^{4} \mathrm{y}_{0}=0 \\ (\mathrm{E}-1)^{4} \mathrm{y}_{0}=0 \)
\( \left(E^{4}-4 E^{3}+6 E^{2}-4 E+1\right) y_{0}=0 \)
\(y_{4}-4 y_{3}+6 y_{2}-4 y_{1}+y_{0}=0 \)
\(32-4(14)+6 y_{2}-4(5)+2=0 \)
\(6 y_{2}=56+20-32-2=42 \)
\(y_{2}=7 \)

Given f(x) = x²+ 3h
LHS = Δ f(x)
= f(x + h) - f(x)
= [(x + h)² + 3 (x + h)] - [x² + 3x]

= h² + 2xh + 3h
when h = 1,
LHS = 1² + 2x(1) + 3(1)
= 1+2x+3
= 2x+4= RHS
Hence proved.

y = f(x) = x³+2x+1 for x = 1,2,3,4,5

The auxiliary equation is 4m² + 4m - 3 = 0
∴ (2m +3)(2m - 1) = 0
⇒ m = \(\frac { -3 }{ 2 } \) and m =\(\frac { 1 }{ 2 } \)

∴ Complementary function CF is \(Ae^{ \frac { -3x }{ 2 } }+Be^{ \frac { 1x }{ 2 } }\)
Particular Integral PI =\(\frac { 1 }{ \phi (D) } \)
PI=\(\frac { 1 }{ 4{ D }^{ 2 }+4D-3 } \).e^2x
=\(\frac { 1 }{ (2D+3)(2D-1) } \).e^2x
=\(\frac { 1 }{ 4\left( D+\frac { 3 }{ 2 } \right) \left( D-\frac { 1 }{ 2 } \right) } \).e^2x
=\(\frac { e^{ 2x } }{ 4\left( 2+\frac { 3 }{ 2 } \right) \left( 2-\frac { 1 }{ 2 } \right) } \)
=\(\frac { e^{ 2x } }{ 4\left( \frac { 7 }{ 2 } \right) \left( \frac { 5 }{ 2 } \right) } =\frac { { e }^{ 2x } }{ 21 } \)
General solution is y = CF + PI
⇒ y=\({ Ae }^{ \frac { -3x }{ 2 } }+Be^{ \frac { x }{ 2 } }=\frac { { e }^{ 2x } }{ 21 } \).

Given y = ae^4x + be^−x.    (1)
Here a and b are arbitrary constants
From (1), \(\frac { dy }{ dx } \) = 4ae^4x− be^−x     (2)
and \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 16ae^4x+ be^−x          (3)
(1) + (2) ⇒ \(y+\frac { dy }{ dx } \) = 5ae^4x      (4)
= (2) + (3) ⇒ \(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 20ae^4x
= 4(5ae^4x)
= \(4\left( y+\frac { dy }{ dx } \right) \)
\(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =4y+4\frac { dy }{ dx } \)
⇒ \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } -4y=0\) which is the required differential equation

Let the equation of straight lines passing through the origin be
y = mx ..(1)
where m is the arbitrary constant
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \)= m(1) ⇒ \(\frac { dy }{ dx } \) = m  .....(2)
Substituting (2) in (1) we get,
\(y=x\frac { dy }{ dx } \).

Given

 Find f(2.8)
Since the required value 2.8 is at the end of the table, apply Newton's backward interpolation formula.
x_n + nh = 2.8 ⇒ 3 + n (1) = 2.8
⇒  n = 2.8 - 3 = -0.2
The difference table is

Newton's backward interpolation formula is
y(x = x_n + nh) = \(\frac { n }{ 1! } { \triangledown y }_{ n }+\frac { n(n+1) }{ 2! } { \triangledown }^{ 2 }{ y }_{ n }+\frac { n(n+1)(n+2) }{ 3! } { \triangledown }^{ 3 }{ y }_{ n }\)
⇒ y(2.8) = 34 + (-0.2) (23) + \(\frac { (-0.2)(-0.2+1) }{ 2 } (14)+\frac { (0.2)(-0.2+1)(-0.2+2) }{ 6 } \)(16)
⇒ y(2.8) = 34 - 4.6 + (-0.2) (0.8) (7) + (-0.2) (0.8) (1.8)
⇒ y(2.8) = 34 - 4.6 - 1.12- 0.288
⇒ y(2.8) = 27.992

The forward interpolation formula is
\({ y }_{ ({ x=x }_{ 0 }+nh) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
Here x₀ + nh = x ⇒ x₀ = 0, h = 1
∴ 0 + n = x ⇒ n = x.
The difference table is

\({ y }_{ (n=x) }={ y }_{ 0 }+\frac { n }{ 1! } =\triangle { y }_{ 0 }+\frac { n(n-1) }{ 2! } { \triangle }^{ 2 }{ y }_{ 0 }+\frac { n(n-1)(n-2) }{ 3! } { \triangle }^{ 3 }{ y }_{ 0 }+...\)
\(y=1+\frac { n }{ 1! } (1)+\frac { n(n-1) }{ 2! } (-2)+\frac { n(n-1)(n-2) }{ 6 } (12)\)
⇒ y = 1 + x + (x² - x)(-1) + x(x² - 3x + 2) (2)
⇒ y = 1 + x - x² + x + 2x³ - 6x² + 4x
⇒ y = 1+ 6x - 7x² + 2x³
Hence, the cubic polynomial is 2x³ - 7x²+ 6x + 1.

Let the missing entries by y₁ and y₄
Since only four values of f(x) are given, the polynomial which fits the data is of degree 3.
Hence fourth differences are zero.
∴ Δ⁴y_k = 0 ⇒ (E - 1)⁴y_k = 0
(E⁴ - 4E³ + 6E²- 4E + 1) y_k = 0
Put k = 0 in (1) we get,
(E₄ - 4E₃ + 6E₂ - 4E + 1)y₀ = 0
⇒ y₄ - 4y₃ + 5y₂ - 4y₁ + y₀ = 0
⇒ y₄ - 4(15) + 6(8) - 4(y₁) + 0 = 0
⇒ y₄ - 60 + 48 - 4y₁ = 0
⇒ y₄ - 4y₁ - 12 = 0
⇒ y₄ - 4y₁ = 12  (2)
put k = 1 in (1) we get
(E⁴- 4E³ + 6E² - 4E + 1) y₁ = 0
⇒ y₅ - 4y₄ + 6y₃ - 4y₂ + y₁ = 0
⇒ 35 - 4y₄ + 6 (15) - 4 (8) +y₁ = 0
⇒ 35 - 4y₄ + 90 - 32 +y₁ = 0
⇒ 4y₄ +y₁ = -93  (3)

⇒ y₁ = \(\frac{-45}{-15}\) = 3
⇒ y₁ = 3.
Substituting y₁ = 3 in (2) we get,
y₄ - 4(3) = 12
y₄ - 12 = 12
⇒ y₄ = 12 + 12
⇒ y₄ = 24
Hence the missing entries are 3 and 24.

\(\Delta \)\(\left[ \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } \right] \)
By Partial fraction method
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\frac { A }{ x+3 } +\frac { B }{ x+2 } \)
\(A=\frac { 5x+12 }{ x+12 } [x=-3]=\frac { -15+12 }{ -1 } =\frac { -3 }{ -1 } =-3\)
\(B=\frac { 5x+12 }{ x+3 } \)[x = -2] \(=\frac { 2 }{ 1 } =2\)
\(\frac { 5x+12 }{ { x }^{ 2 }+5x+6 } = \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(\Delta \frac { 5x+12 }{ { x }^{ 2 }+5x+6 } =\Delta \left[ \frac { 3 }{ x+3 } +\frac { 2 }{ x+2 } \right] \)
\(=\left[ \frac { 3 }{ x+1+3 } -\frac { 3 }{ x+3 } \right] +\left\{ \frac { 2 }{ x+1+2 } -\frac { 2 }{ x+2 } \right\} \)
\(=3\left[ \frac { 1 }{ x+4 } -\frac { 1 }{ x+3 } \right] +2\left[ \frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \right] \)
\(=\left[ \frac { -3 }{ (x+4)(x+3) } -\frac { 2 }{ (x+3)(2+3) } \right] \)
\(=\frac { -5x-14 }{ (x+2)(x+3)(x+4) } \)

Given Q_d = 13 - 6p + 2\(\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ dt^{ 1 } } \)  Q_s = - 3 + 2p
At equilibrium  Q_d = Q_s
⇒ 13 - 6p + 2\(\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ dt^{ 1 } } \) = - 3 + 2p
⇒ \(\frac { { d }^{ 2 }p }{ dt^{ 2 } } +2\frac { dp }{ dt } \) - 6p + 13 + 3 - 2p = 0 
⇒ \(\frac { { d }^{ 2 }p }{ dt^{ 2 } } +2\frac { dp }{ dt } \) - 8p + 16 = 0
⇒  \(\frac { { d }^{ 2 }p }{ dt^{ 2 } } +2\frac { dp }{ dt } \) - 8p = - 16
The auxiliary equation is m² + 2m - 8 = 0
⇒ (m + 4) (m - 2) = 0
⇒ m = - 4, 2
The roots are real and different.
∴ Complementary function CF is Ae^-4t + Be^2t
Particular Integral PI = \(\\ \frac { 1 }{ \phi (D) } \).f(t)
=\(\frac { 1 }{ { D }^{ 2 }+2D-8 } .(-16)=\frac { -16.e^{ 0x } }{ (D+4)(D-2) } \)
PI = \(\frac { -16 }{ (0+4)(0-2) } =\frac { -16 }{ -8 } \) = 2
∴ P = CF + PI
∴ The general solution is
P = Ae^-4t + Be^2t + 2

The auxiliary equation is m² + m - 6 = 0
⇒ (m + 3) (m - 2) = 0
⇒ m = -3, 2
The roots are real and different.
∴ The complementary function CF is Ae^-3x + B^ex
Particular Integral
PI₁=\(\frac { 1 }{ \phi (D) } \)f₁(x)
=\(\frac { 1 }{ ({ D }^{ 2 }+D-6) } \)e^3x
PI₁=\(\frac { { e }^{ 3x } }{ (D+3)(D-2) } =\frac { { e }^{ 3x } }{ (3+3)(3-2) } \)
= \(\frac { { e }^{ 3x } }{ 6(1) } =\frac { e^{ 3x } }{ 6 } \)
PI₂ = \(\frac { 1 }{ \phi (D0 } { f }_{ 2 }(x)=\frac { e^{ -3x } }{ (D+3)(D-2) } \)
= x\(\frac { { e }^{ -3x } }{ (-3-2) } \) [∵ when D = - 3, D + 3 = 0]
PI₂ = \(-\frac { x }{ 5 } \)e^-3x
∴ y = CF+PI₁+PI₂
∴ The general solution is
y = Ae^-3x+Be^2x+\(\frac { { e }^{ 3x } }{ 6 } -\frac { x }{ 5 } \)e^-3x

\(\frac { dy }{ dx } =\frac { x+y }{ x } \)
Since the numerator and denominator are homogeneous functions of degree 1,
put y = vx ⇒ \(\frac { dy }{ dx } =v(1)+x.\frac { dv }{ dx } \)
∴ v + x\(\frac { dv }{ dx } \) = \(\frac { x+vx }{ x } \)

⇒ x\(\frac { dv }{ dx } \) = 1 + v - v = 1
⇒ x\(\frac { dv }{ dx } \) = 1
Separating the variables we get
\(\frac { dv }{ 1 } =\frac { dx }{ x } \)
Integrating both sides we get,
\(\int { dv } =\int { \frac { dx }{ x } } \)
\(\int { \frac { dx }{ x } } =\int { dv } \) ⇒ log x = v + log c
⇒ log\(\left( \frac { x }{ c } \right) =v\Rightarrow \frac { x }{ c } \) = e^v
x = c.ev
Replace v by y/x we get,
x = ce^y/x.

(i) m is an arbitrary constant
y = mx + c ...(1)
Differentiating w.r. to x ,
we get \(\frac{dy}{dx}\) = m  ...(2)
Now we eliminate m from (1) and (2)
For this substitute (2) in (1)
y = x \(\frac{dy}{dx}\) + c
x \(\frac{dy}{dx}\) - y + c = 0 which is the required differential equation of first order

(ii) c is an arbitrary constant
Differentiating (1), we get \(\frac{dy}{dx}\) = m
Here c is eliminated from the given equation
∴ \(\frac{dy}{dx}\) = m is the required differential equation.

(iii) both m and c are arbitrary constants
Since m and c are two arbitrary constants differentiating (1) twice we get
\(\frac{dy}{dx}\) = m
\(​​\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 0
Here m and c are eliminated from the given equation.
\(​​\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 0 which is the required differential equation.