Given (x+y)²\(\frac { dy }{ dx } \) = 1
put x+y = z
⇒ 1+\(\frac { dy }{ dx } =\frac { dz }{ dx } \)
\(\frac { dy }{ dx } =\frac { dz }{ dx } \)-1
∴ (1) becomes,
z²\(\left( \frac { dz }{ dx } -1 \right) \)= 1
⇒ z²\(\frac { dz }{ dx } \)-z² = 1
⇒ z²\(\frac { dz }{ dx } \) = 1 + z²
Separating the variables we get,
\(\left( \frac { { z }^{ 2 } }{ 1+{ z }^{ 2 } } \right) \)dz = dx
Adding and Subtracting 1 in the numerator, we get
\(\left( \frac { 1+{ z }^{ 2 }-1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
⇒ \(\left( \frac { 1+{ z }^{ 2 } }{ 1+{ z }^{ 2 } } -\frac { 1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
⇒ \(\left( z-\frac { 1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
Integrating, \(\int { dz } -\int { \frac { dz }{ 1+{ z }^{ 2 } } } \)
[∵ \(\int { \frac { dz }{ 1+{ z }^{ 2 } } } \) = tan^-1x+y]
⇒ (z-tan^-1(2) = x+C
⇒ (x+y)-tan^-1(x+y) = x + C
⇒ y-tan^-1(x+y) = C

From the graph, it is clear that when x = 27, the value of y is 24.8

Given non-homogeneous equations are
x- 3y + 4z = 3, 2x - 5y + 7z = 6, 3x - 8y + 11z = 1

Augmented matrix  [A, B]	Elementary Transformation
\(\left( \begin{matrix} 1 & -3 & 4 \\ 2 & -5 & 7 \\ 3 & -8 & 11 \end{matrix}\begin{matrix} 3 \\ 6 \\ 1 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & -3 & 4 \\ 0 & 1 & -1 \\ 0 & 1 & -1 \end{matrix}\begin{matrix} 3 \\ 0 \\ -8 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ 3R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & -3 & 4 \\ 0 & 1 & -1 \\ 0 & 0 & -0 \end{matrix}\begin{matrix} 3 \\ 0 \\ -8 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-R_{ 2 }\)

Clearly \(\rho (A)=2\) and \(\rho (A,B)=3\)
\(\rho (A,B)\neq \rho (A)\)
Hence, the given system is inconsistent and has no solution.

Since only four values of f(x) are given, the polynomial which fits the data is of degree three. Hence fourth differences are zeros.
(ie) Δ⁴y₀ = 0, ∴ (E−1)⁴y = 0
(E⁴ - 4E³ + 6E² - 4E +1)y₀ = 0
E⁴y₀ - 4E³y₀+ 6E²y₀ - 4Ey₀ +y₀  = 0
y₄ − 4y₃ + 6y₂ − 4y₁ + y₀ = 0
67.4 − 4y₃+ 6(54.1) − 4(4.2) + 45 = 0
240.2 = 4y₃ ∴ y₃ = 60.05

The given differential equation is of the follows
\(\frac { dy }{ dx } \)+ Py =Q where
P = cos x, Q = sinx cosx
∴ \(\int { P } dx=\int { cosx } dx\)
∴ Integrating factor (I.F) =\(e^{ \int { p.dx } }=e^{ sinx }\)
Hence, the solution is
\(ye^{ \int { p.dx } }=\int { Qe^{ \int { p.dx } } } dx+c\)
⇒ \(y.e^{ sinx }=\int { sinx } cosx.e^{ sinx }dx+c\)
put t = sin x ⇒ dt =cos x dx
⇒ \(ye^{ sinx }=\int { t{ e }^{ t } } dt\)  ....(1)

put u = t; dv = et dt
du = dt; v = et
Using integration by parts
\(\int { u } dv=uv-\int { v } du\)
⇒ \(\int { t } e^{ t }dt=te^{ t }-\int { e^{ t }dt } \) = te^t-e^t....(2)
Substituting (2) in (1) we get,
ye^sinx = t e^t-e^t + c
⇒ y e^sinx = e^t(t-1)+c
⇒ y e^sinx = e^sinx (sin x-1)+c   [∵ t = sinx]

Let \(I=\int _{ -1 }^{ 1 }{ { x }^{ 2 } } e^{ -2x }dx\)
u = x²    dv = e-^2x
u¹ = 2x    \(v=\cfrac { { e }^{ -2x } }{ -2 } =\cfrac { -e^{ -2x } }{ 2 } \)
u⁴ = 2  \({ v }_{ 1 }=+\cfrac { { e }^{ -2x } }{ 4 } \)
\({ v }_{ 2 }=-\cfrac { { e }^{ -2x } }{ 8 } \)
Using Bernoulli's formula
I = uv = u¹v₁ + u⁴v₂
= \(\left[ { x }^{ 2 }\left( \cfrac { -e^{ -2x } }{ 2 } \right) -2x\left( \cfrac { { e }^{ -2x } }{ 4 } \right) +2\left( \cfrac { -e^{ -2x } }{ 8 } \right) \right] _{ -1 }^{ 1 }\)
= \(\left( { e }^{ -2x }\left[ \cfrac { -{ x }^{ 2 } }{ 2 } -\cfrac { x }{ 2 } -\cfrac { 1 }{ 4 } \right] \right) \)

\(\left( { e }^{ -2x }\left[ \cfrac { -{ x }^{ 2 } }{ 2 } -\cfrac { x }{ 2 } -\cfrac { 1 }{ 4 } \right] \right) \)
= \({ e }^{ -2 }\left( -\cfrac { 5 }{ 4 } \right) -{ e }^{ 2 }\left( -\cfrac { 1 }{ 4 } \right) \)
= \(\cfrac { -5 }{ 4 } { e }^{ -2 }+\cfrac { 1 }{ 4 } { e }^{ 2 }\)
= \(\cfrac { 1 }{ 4 } \left( { e }^{ 2 }-\cfrac { 5 }{ { e }^{ 2 } } \right) =\cfrac { 1 }{ 4 } \left( \cfrac { { e }^{ 4 }-5 }{ { e }^{ 2 } } \right) \)

Given that \(\int _{ a }^{ b }{ dx } =1\)
\({ \left[ x \right] }_{ a }^{ b }\) = 1
b − a = 1 … (1)
Now, \(\int _{ a }^{ b }{ xdx } =1\)
\({ \left[ \frac { { x }^{ 2 } }{ a } \right] }_{ a }^{ b }\) = 1
b² − a² = 2
(b + a)(b − a) = 2
b + a = 2 … (2) [∵ b -a =1]
(1) + (2) ⇒ 2b = 3
ஃ \(b=\frac { 3 }{ 2 } \)
Now, \(\frac { 3 }{ 2 } \) - a = 1 [∵ from (1)]
 ஃ a = \(\frac { 1 }{ 2 } \)

Saving Cost S(t) = \(\int _{ 0 }^{ t }{ 20000t } \ dt\)
= 10000 t²
To recoup the total price,
10000 t² = 640000
t² = 64
t = 8
When t = 8 years, one can recoup the price.

∫ sin³x dx
We know that sin 3x = 3sin x - 4 sin³x
⇒ 4 sin³x = 3sinx - sin 3x
⇒ sin³x = \(\frac { 1 }{ 4 } \left( 3\sin x-\sin3x \right) \)
sin³x = \(\frac { 1 }{ 4 } \left( 3 \sin x-\not 3\sin 3x \right) \)dx
\(=\frac { 3 }{ 4 } \int { \sin x\ dx-\frac { 1 }{ 4 } } \int { \sin 3x dx } \)
\(=\frac { 3 }{ 4 } \left( -\cos x \right) -\frac { 1 }{ 4 } \left( -\frac { \cos 3x }{ 3 } \right) +c \quad\left[ \because \int { \sin ax \ dx=\frac { -1 }{ a } \cos ax+c } \right] \)
\(=-\frac { 3 }{ 4 } \cos x+\left( \frac { \cos 3x }{ 12 } \right) +c\)

Let ‘x’ be the cost of a pencil
Let ‘y’ be the cost of an eraser
\(\therefore \) By given data, we get the following equations
11x + 3y = 64
8x + 3y = 49
\(\triangle =\left| \begin{matrix} 11 & 3 \\ 8 & 3 \end{matrix} \right| =9\neq 0,\) It has unique solution.
\({ \triangle }_{ x }\left| \begin{matrix} 64 & 3 \\ 49 & 3 \end{matrix} \right| =45\)
\({ \triangle }_{ y }\left| \begin{matrix} 11 & 64 \\ 8 & 49 \end{matrix} \right| =27\)
\(\therefore \) By Cramer’s rule
\(x={ \frac { \triangle x }{ \triangle } =\frac { 45 }{ 9 } =5 }\)
\(y={ \frac { \triangle y }{ \triangle } =\frac { 27 }{ 9 } =3 }\)
\(\therefore \) The cost of a pencil is Rs. 5 and the cost of an eraser is Rs. 3.

Let A= \(\left( \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right) \)
Order of A is 3 \(\times\) 3.
∴\(\rho \)(A)\(\le \)3
Consider the third order minor \(\left| \begin{matrix} 5 & 3 & 0 \\ 1 & 2 & -4 \\ -2 & -4 & 8 \end{matrix} \right| =0\)
Since the third order minor vanishes, therefore \(\rho (A)\neq 3\)
Consider a second order minor \(\left| \begin{matrix} 5 & 3 \\ 1 & 2 \end{matrix} \right| =7\neq 0\)
There is a minor of order 2, which is not zero.
\(\therefore \rho (A)=2\)

Let I =  ഽ x³ sin (x⁴) dx
Put t = x⁴
⇒ dt = 4x³ dfx
\(\Rightarrow \frac { dt }{ 4 } ={ x }^{ 3 }dx\)
\(\therefore I=\int { sin } t.\frac { dt }{ 4 } =\frac { 1 }{ 4 } sint\quad dt\)
= \(-\frac { 1 }{ 4 } cos\quad t+c\)
= \(-\frac { 1 }{ 4 } cos\left( { x }^{ 4 } \right) +c\)      \(\left[ \because t={ x }^{ 4 } \right] \)

\({ x }^{ 2 }\frac { dc }{ dx } -10xc=-10\)
\(\div { x }^{ 2 },\frac { dc }{ dx } -\frac { 10c }{ x } =\frac { 10 }{ { x }^{ 2 } } \frac { dc }{ dx } +Pc=Q\)
This is a first order linear differential equation of the form \(\frac{dc}{dx}\) + Pc = Q where
\(P=\frac { 10 }{ x }\) and \(Q=-\frac { 10 }{ { x }^{ 2 } } \)
\(\int { pdx } =-\int { \frac { 10 }{ x } =10logx=log\left( \frac { 1 }{ { x }^{ 10 } } \right) } \)
∴ I.F. = \({ e }^{ \int { pdf } }{ = }^{ { e }^{ log{ 1/x }^{ 10 } } }=\frac { 1 }{ { x }^{ 10 } } \)
∴ General solution is
\({ Ce }^{ \int { px } }=\int { Q.{ e }^{ \int { pdf } }dx+k } \)
\(\Rightarrow c.\left( \frac { 1 }{ { x }^{ 10 } } \right) =\int { -\frac { 10 }{ { x }^{ 2 } } . } \frac { 1 }{ { x }^{ 10 } } dx+k\)
\(=-10\int { \frac { 1 }{ { x }^{ 12 } } } dx+k\)
\(\Rightarrow \frac { c }{ { x }^{ 10 } } =-10\int { { x }^{ -12 } } dx+k\)
\(\Rightarrow \frac { c }{ { x }^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } \right) +k\)
\(\Rightarrow \frac { { C }_{ 0 } }{ { x }_{ 0 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }_{ 0 }^{ 11 } } \right) +k\)
When c = c₀, x = x₀
\(\Rightarrow k=\frac { c }{ { x }_{ 0 }^{ 10 } } -\frac { 10 }{ 11.{ x }_{ 0 }^{ 11 } } \)
∴ The solution is
\(\Rightarrow \frac { c }{ x^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } \right) +k\left( \frac { c }{ { x }_{ 0 }^{ 10 } } -\frac { 10 }{ 11{ x }_{ 0 }^{ 11 } } \right) \)
\(\Rightarrow \frac { c }{ x^{ 10 } } -\frac { c }{ { x }_{ 0 }^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } -\frac { 10 }{ { x }_{ 0 }^{ 11 } } \right) \).

Given x_o = 0, x₁ = 3, x₂ = 5, x₃ = 6, x₄ = 8
y_o = 276, y₁ = 460, y₂ = 414, y₃ = 343, y₄ = 110
Lagrange's formula is
y = \(\frac { (x-{ x }_{ 1 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 0 }-{ x }_{ 1 })({ x }_{ 0 }-{ x }_{ 2 })({ x }_{ 0 }-{ x }_{ 3 })({ x }_{ 0 }-{ x }_{ 4 }) } { y }_{ 0 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 1 }-{ x }_{ 0 })({ x }_{ 1 }-{ x }_{ 2 })({ x }_{ 1 }-{ x }_{ 3 })({ x }_{ 1 }-{ x }_{ 4 }) } { y }_{ 1 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 2 }-{ x }_{ 0 })({ x }_{ 2 }-{ x }_{ 1 })({ x }_{ 2 }-{ x }_{ 3 })({ x }_{ 2 }-{ x }_{ 4 }) } { y }_{ 2 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 1 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 3 }-{ x }_{ 0 })({ x }_{ 3 }-{ x }_{ 1 })({ x }_{ 3 }-{ x }_{ 2 })({ x }_{ 3 }-{ x }_{ 4 }) } { y }_{ 3 }+\frac { (x-{ x }_{ 0 })(x-{ x }_{ 2 })(x-{ x }_{ 3 })(x-{ x }_{ 4 }) }{ ({ x }_{ 4 }-{ x }_{ 0 })({ x }_{ 4 }-{ x }_{ 2 })({ x }_{ 4 }-{ x }_{ 3 })({ x }_{ 4 }-{ x }_{ 4 }) } { y }_{ 4 }\)
⇒ 276 \(\frac { (1)(-1)(-2)(-4) }{ (-3)(-5)(-6)(-8) } +460\frac { (4)(-1)(-2)(-4) }{ (3)(-2)(-3)(-5) } +414\frac { (4)(1)(-1)(-4) }{ (5)(2)(-1)(-3) } +343\frac { (4)(1)(-1)(-2) }{ (6)(3)(1)(-2) } +110\frac { (4)(1)(-1)(-2) }{ (8)(5)(3)(2) } \)
⇒ y = -3.066 + 163.555 + 441.6 - 152.44 + 3.666
⇒ y = 453.311.

Let P be the principal at time ‘t’
\(\frac { dP }{ dt } =\frac { 5 }{ 100 } P=0.05P\)
⇒ ഽ\(\frac { dP }{ P } \) = ഽ0.05 dt + c
log_e P = 0.05t + c
P = e^0.05te^c
P = c₁e^0.05t   (1)
Given P = 2000 when t = 0
⇒  c₁ = 2000
∴ (1) ⇒ P = 2000e^0.05t
To find t , when P = 4000
(2) ⇒ 4000 = 2,000e^0.05t
2 = e^0.05t
0.05t = log2
t = \(\frac { 0.0931 }{ 0.05 } \) = 14 years (approximately)

Given non-homogeneous equations are
\(3x-y+\lambda z=1\)
\(2x+y+z=2\)
\(x+2y-\lambda z=-1\)
The matrix equation corresponding to the given system is

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 3 & -1 & \lambda \\ 2 & 1 & 1 \\ 1 & 2 & - \end{matrix}\begin{matrix} 1 \\ 2 \\ -1 \end{matrix} \right) \)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 2 & 1 & 1 \\ 3 & -1 & \lambda \end{matrix}\begin{matrix} -1 \\ 2 \\ 1 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -3 & 1+2\lambda \\ 3 & -1 & \lambda \end{matrix}\begin{matrix} -1 \\ 4 \\ 1 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-2{ R }_{ 1 }\)
\(\left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -3 & 1+2\lambda \\ 0 & -7 & 4\lambda \end{matrix}\begin{matrix} -1 \\ 4 \\ 4 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-3{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -1 & \frac { 1+2\lambda }{ 3 } \\ 0 & - & \frac { 4\lambda }{ 7 } \end{matrix}\begin{matrix} -1 \\ \frac { 4 }{ 3 } \\ \frac { 4 }{ 7 } \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }\div 3\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }\div 7\)
\(\left( \begin{matrix} 1 & 2 & -\lambda \\ 0 & -1 & \frac { 1+2 }{ 3 } \\ 0 & 0 & \frac { -7-2\lambda }{ 21 } \end{matrix}\begin{matrix} -1 \\ \frac { 4 }{ 3 } \\ \frac { -16 }{ 21 } \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-{ R }_{ 2 }\)

Since
\(\cfrac { 4\lambda }{ 7 } -\cfrac { 1+2\lambda }{ 3 } \)
= \(\cfrac { 12\lambda -7-14\lambda }{ 21 } =\cfrac { -7-2\lambda }{ 21 } \)
and \(\cfrac { 4 }{ 7 } -\cfrac { 4 }{ 3 } =\cfrac { 12-28 }{ 21 } \)
= \(\cfrac { -16 }{ 21 } \)
\(\therefore\) Since the system is fail to have unique solution either it can have infinitely many solution or it may be inconsistent.
This can happen only when \(\cfrac { -7-2\lambda }{ 21 } =0\)
\(\Rightarrow -7-2\lambda =0\)
\(\Rightarrow -7=2\lambda \)
\(\Rightarrow \lambda =\cfrac { -7 }{ 2 } \)

The non-homogeneous equation are
2x - y + z = 7, 3x + y - 5z = 13, x + y + z = 5

Augmented matrix [A,B]	Elementary Transformation
\(\left( \begin{matrix} 2 & -1 & 1 \\ 3 & 1 & -5 \\ 1 & 1 & 1 \end{matrix}\begin{matrix} 7 \\ 13 \\ 5 \end{matrix} \right) \)
\(-\left( \begin{matrix} 1 & 1 & 1 \\ 3 & 1 & -5 \\ 2 & -1 & 1 \end{matrix}\begin{matrix} 5 \\ 13 \\ 7 \end{matrix} \right) \)	\({ R }_{ 1 }\leftrightarrow { R }_{ 3 }\)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -2 & -8 \\ 0 & -3 & -1 \end{matrix}\begin{matrix} 5 \\ -2 \\ -3 \end{matrix} \right) \)	\({ R }_{ 2 }\rightarrow { R }_{ 2 }-3{ R }_{ 1 }\) \({ R }_{ 3 }\rightarrow { R }_{ 3 }-2{ R }_{ 1 }\)
\(\sim \left( \begin{matrix} 1 & 1 & 1 \\ 0 & -2 & -8 \\ 0 & 0 & 11 \end{matrix}\begin{matrix} 5 \\ -2 \\ 0 \end{matrix} \right) \)	\({ R }_{ 3 }\rightarrow { R }_{ 3 }-\cfrac { 3 }{ 2 } { R }_{ 2 }\)

Clearly \(\rho (A)=3\) and \(\rho (A,B)\) = 3 = Number of unknowns 
\(\therefore\) The given system is consistent and has unique solution.

Given \(\int _{ 0 }^{ a }{ { 3x }^{ 2 } } dx=8\)

⇒ \({ \left[ { x }^{ 3 } \right] }_{ 0 }^{ a }=8\)
⇒ a³ - 0 = 8
⇒ a³ = 8
⇒ a³ = 2³
⇒ a = 2
∴ a = 2

Squaring both sides we get,
\(\left[ \sqrt { 1+\left( \frac { dy }{ dx } \right) ^{ 2 } } \right] ^{ 2 }\)= (4x)² ⇒ 1+\(\left( \frac { dy }{ dx } \right) ^{ 2 }\)= 16x²
The highest derivative is of order 1and its power is 2
∴ order is 1 and degree is 2.

We know Δ (f(x)) = f(x + h) -f(x)
Since h = 1, ∆ (f)) = f(x + 1) - f(x)
[∵ (x + 1)³ = x³ + 3x² + 3x + 1]
⇒ Δ (x³) = (x + 1)³ - x³

⇒ Δ (x³) = 3x² + 3x + 1

Given (9D²−12D + 4)y = 0
The auxiliary equation is (3m - 2)² = 0
(3m−2) (3m−2) = 0 ⇒ m = \(\frac 23,\frac 23\)
Roots are real and equal
The C.F. is (Ax + B)\({ e }^{ \frac { 2 }{ 3 } x }\)
The general solution is y = (Ax + B)\({ e }^{ \frac { 2 }{ 3 } x }\)

ഽ\(\sqrt { { x }^{ 2 }-16 } \) dx = ഽ\(\sqrt { { x }^{ 2 }-{ 4 }^{ 2 } } \) dx
= \(\frac { x }{ 2 } \sqrt { { x }^{ 2 }-{ 4 }^{ 2 } } -\frac { { 4 }^{ 2 } }{ 2 } \log\left| x+\sqrt { { x }^{ 2 }-{ 4 }^{ 2 } } \right| +c\)
= \(\frac { x }{ 2 } \sqrt { { x }^{ 2 }-16 } -8 \log\left| x+\sqrt { { x }^{ 2 }-16 } \right| +c\)