\(\frac{d y}{d x}=\frac{24 x^{2}-y^{2}}{x y}\) is a homogeneous equation x and y
y = vx
\( \frac{d y}{d x} =v+x \frac{d v}{d x} \)
\(v+x \frac{d v}{d x} =\frac{24 x^{2}-v^{2} x^{2}}{x(v x)} \\ v+x \frac{d v}{d x} =\frac{24-v^{2}}{v} \)
\( x \frac{d v}{d x}=\frac{24-v^{2}}{v}-v \)
\(x \frac{d v}{d x}=\frac{24-v^{2}-v^{2}}{v} \)
\(\int \frac{v}{24-2 v^{2}} d v=\int \frac{d x}{x} \)
\(\int \frac{-4 v}{24-2 v^{2}} d v=-4 \int \frac{d x}{x} \)
\(\log \left(24-2 \mathrm{v}^{2}\right)=-4 \log x+\log \mathrm{c} \)
\(\log \left(24-\frac{2 y^{2}}{x^{2}}\right)+4 \log x=\log \mathrm{c} \)
\(\log \left(\frac{24 x^{2}-2 y^{2}}{x^{2}}\right)+\log x^{4}=\log \mathrm{c} \)
\(\log \left(\frac{24 x^{2}-2 y^{2}}{x^{2}}\right) x^{4}=\log \mathrm{c} \)
\(x^{2}\left(24 x^{2}-2 \mathrm{y}^{2}\right)=\mathrm{c} \)
\(\text {If } x=2, \mathrm{y}=4 \)
\(4(96-32)=\mathrm{c} \Rightarrow \mathrm{c}=256\)
Required equation \(x^{2}\left(24 x^{2}-2 y^{2}\right)=256\)
Dividing by 2 
\(x^{2}\left(12 x^{2}-y^{2}\right)=128\)

\(\frac{d c}{d q}=\frac{c^{2}+q^{2}}{2 c q}\) is a homogeneous equation
c = vq
\(\frac{d c}{d q}=v+q \frac{d v}{d q}\)
\( v+q \frac{d v}{d q} =\frac{v^{2} q^{2}+q^{2}}{2(1 v) q}=\frac{v^{2}+1}{2 v} \)
\(q \frac{d v}{d q} =\frac{v^{2}+1}{2 v}-v \)
\( =\frac{v^{2}+1-2 v^{2}}{2 v}=\frac{1-v^{2}}{2 v} \)
\(-\int \frac{2 v}{1-v^{2}} =-\int \frac{d q}{q} \)
\(\log \left(1-v^{2}\right) =-\log q+\log \mathrm{k} \)
\(\left(1-\frac{v^{2}}{q^{2}})+\log q\right. =\log \mathrm{k} \)
\(\mathrm{ks}\left(\frac{q^{2}-c^{2}}{q^{2}}\right) q =\log \mathrm{k} \)
\(\frac{q^{2}-c^{2}}{q} =\mathrm{k} \)
\(q^{2}-c^{2} =k q \)
\(\text {If } \mathrm{c}=4, \mathrm{q}=2 \)
\(4-16 =2 \mathrm{k} \Rightarrow \mathrm{k}=-6 \)
\(\mathrm{q}^{2}-\mathrm{c}^{2} =-6 \mathrm{q} \)
\(\mathrm{c}^{2}-\mathrm{q}^{2} =6 \mathrm{q} \)

\( \frac{d y}{d x} =\frac{y}{x}-\frac{y^{2}}{x^{2}} \)
\(\mathrm{y} =\mathrm{v} x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x} \)
\(v+x \frac{d v}{d x} =\mathrm{v}-\mathrm{v}^{2} \Rightarrow x \frac{d v}{d x}=-\mathrm{v}^{2} \)
\(-\int \frac{d v}{v^{+2}} =\int \frac{d x}{x} \)
\(-\int v^{-2} d v =\log x+\log \mathrm{c} \)
\(\frac{-v^{-1}}{-1} =\log c x \)
\(\frac{1}{v} =\log c x \Rightarrow \frac{x}{y}=\log c x \)
\(c x =\mathrm{e}^{x / y} \Rightarrow \mathrm{c}=x \mathrm{e}^{-x / y} \)

Given \(\left(6 x^{2}+2 y^{2}\right) \mathrm{d} x-\left(x^{2}+4 x y\right) \mathrm{dy}=0\)
\(\frac{d y}{d x}=\frac{6 x^{2}+2 y^{2}}{x^{2}+4 x y}\) is a homogeneous equation in x and y
y = vx
\( \frac{d y}{d x} =v+x \frac{d v}{d x} v+x \frac{d v}{d x}=\frac{6 x^{2}+2 v^{2} x^{2}}{x^{2}+4 x(v x)} =\frac{6+2 v^{2}}{1+4 v} \)
\(x \frac{d v}{d x} =\frac{6+2 v^{2}}{1+4 v}-v \ =\frac{6+2 v^{2}-v-4 v^{2}}{1+4 v} \)
\(x \frac{d v}{d x} =\frac{6-v-2 v^{2}}{1+4 v} \)
\(-\int \frac{(1+4 v)}{6-v-2 v^{2}} d v =-\int \frac{d x}{x} \)
\(\log \left(6-v-2 v^{2}\right) =-\log x+\log \mathrm{k} \)
\( \log \left(6-\frac{y}{x}-\frac{2 y^{2}}{x^{2}}\right)+\log x =\log \mathrm{k} \)
\(\log \left(\frac{6 x^{2}-x y-2 y^{2}}{x^{2}}\right) x =\log \mathrm{k} \)

\(\frac{d c}{d q}=\frac{c^{2}+2 c q}{q^{2}}\)
This is a homogeneous equation in c and q
\( \mathrm{c} =\mathrm{vq} \)
\(\frac{d c}{d q} =v+q \frac{d v}{d q} \\ v+q \frac{d v}{d q} =\frac{v^{2} \cdot q^{2}+2(v q) q}{q^{2}} \)
\(v+q \frac{d v}{d q} =\mathrm{v}^{2}+2 \mathrm{v} \)
\(q \frac{d v}{d q} =\mathrm{v}^{2}+\mathrm{v} \)
\(\int \frac{d v}{v^{2}+v} =\int \frac{d q}{q} \)............(1)
Applying partial fraction to \(\frac{1}{v^{2}+v}\)
\( \frac{1}{v^{2}+v}=\frac{1}{v(v+1)}=\frac{A}{v}+\frac{B}{v+1} \)
\(1=\mathrm{A}(\mathrm{v}+1)+\mathrm{B} v\)
\( \\ \text {If } \mathrm{v}=0, \mathrm{~A}=1 \)
\(\text {If } \mathrm{v}=-1, \mathrm{B}=-1 \)
\(\frac{1}{v(v+1)}=\frac{1}{v}-\frac{1}{v+1} \)
\(\frac{1}{v(v+1)}=\frac{1}{v}-\frac{1}{v+1} \)
\( \frac{\frac{c}{q}}{\frac{c}{q}+1} =\mathrm{qk}\)
\( \mathrm{c} =(\mathrm{c}+\mathrm{q}) \mathrm{qk} \)
\(\text { If } \mathrm{c} =1, \mathrm{q}=1 \)
\(1 =2 \mathrm{k} \Rightarrow \mathrm{k}=1 / 2 \)
\(c=\frac{q(q+c)}{2}\) is the relationship between c and q

\(\frac{d p}{d x}=\frac{2 p^{3}-x^{3}}{3 x p^{2}}\)
It is a differential equation in x and p of homogeneous type.
\( \therefore \mathrm{p} =\mathrm{v} x \\ \frac{d p}{d x} =v+x \frac{d v}{d x} \)
\(v+x \frac{d v}{d x} =\frac{2 v^{3} x^{3}-x^{3}}{3 x\left(v^{2} x^{2}\right)}=\frac{2 v^{3}-1}{3 v^{2}} \)
\(x \frac{d v}{d x} =\frac{2 v^{3}-1}{3 v^{2}}-v \)
\( =\frac{2 v^{3}-1-3 v^{3}}{3 v^{2}}=-\left(\frac{1+v^{3}}{3 v^{2}}\right) \)
\(\int \frac{3 v^{2}}{1+v^{3}} d v =-\int \frac{d x}{x} \)
\(\log \left(1+v^{3}\right) =-\log x+\log \mathrm{c} \)
\(\log \left(1+\frac{p^{3}}{x^{3}}\right)+\log x =\log \mathrm{c} \)
\(\log \left(\frac{x^{3}+p^{3}}{x^{3}}\right) x =\log \mathrm{c} \)
\(x^{3}+\mathrm{p}^{3} =c^{2} \)
\(\text {If } \mathrm{p} =20, x=10 \)
\( 1000+8000 =c(100) \)
\(\Rightarrow =\frac{9000}{100}=90 \)
Required equation is \(x^{3}+\mathrm{p}^{3}=90 x^{2}\)

For market clearance, Q_d = Q_s
\(\Rightarrow 42-4p-4\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ { dt }^{ 2 } } =-6+8p\)
\(\Rightarrow 48-12p-4\frac { dp }{ dt } +\frac { { d }^{ 2 }p }{ { dt }^{ 2 } } =0\)
\(\Rightarrow \frac { { d }^{ 2 }p }{ { dt }^{ 2 } } =4\frac { dp }{ dt } -12p=-48\)
The auxiliary equation is m² - 4m - 12 = 0
⇒ (m - 6) (m + 2) = 0
⇒ m = -2, 6
The roots are real and different
∴ C.F. is Ae^-2t + Be^6t
P.I. = \(\frac { 48 }{ (D-6)(D+2) } { e }^{ 0t }=\frac { -48 }{ (0-6)(0+2) } \)
= \(\frac { -48 }{ -12 } \)
∴ The general solution is
P = C.F. + P.I.
⇒ P = Ae^-2t + Be^6t + 4

\({ x }^{ 2 }\frac { dc }{ dx } -10xc=-10\)
\(\div { x }^{ 2 },\frac { dc }{ dx } -\frac { 10c }{ x } =\frac { 10 }{ { x }^{ 2 } } \frac { dc }{ dx } +Pc=Q\)
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 (6x²+2y²)dx-(x²+4xy)dy = 0
⇒ (6x²+2y²)dx = (x²+4xy)dy
⇒ \(\frac { dy }{ dx } =\frac { 6{ x }^{ 2 }+2{ y }^{ 2 } }{ { x }^{ 2 }+4xy } \)
This is a homogeneous function of degree 2.
∴ Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

⇒ v+x\(\frac { dv }{ dx } =\frac { 6+2{ v }^{ 2 } }{ 1+4v } \)
⇒ x\(\frac { dv }{ dx } =\frac { 6+2v^{ 2 } }{ 1+4v } -v=\frac { 6+2v^{ 2 }-v(1+4v) }{ 1+4v } \)
=\(\frac { 6+2{ v }^{ 2 }-v-4{ v }^{ 2 } }{ 1+4v } \) ....(1)
⇒ x\(\frac { dv }{ dx } =\frac { 6-v-2v^{ 2 } }{ 1+4v } \)
Separating the variables we get,
\(\frac { (1+4v)dv }{ 6-v-2{ v }^{ 2 } } =\frac { dx }{ x } \)
[∵ t = 6-v-2v² ⇒ dt = (-1-4v)dv, \(\int { \frac { dt }{ t } } \) = log t]
∴ -\(\int { \frac { (-1-4v)dv }{ 6-v-2v^{ 2 } } } =\int { \frac { dx }{ x } } \)
⇒ -log(6-v-2v²) = log x + log k
⇒ \(\frac { 1 }{ 6-v-2{ v }^{ 2 } } \) = kx
Replacing v by \(\frac{y}{x}\) we get, \(\frac { 1 }{ 6-\frac { y }{ x } -\frac { 2y^{ 2 } }{ { x }^{ 2 } } } \) = kx
⇒ \(\frac { { x }^{ 2 } }{ 6x^{ 2 }-xy-2y^{ 2 } } \) = kx
⇒ x = k(6x²-xy-2y²)
When x = 1, y = 2 ⇒
1 = k(6-2-8) ⇒ 1 = k(-4) ⇒ k = -\(\frac{1}{4}\)
⇒ x = \(\frac{1}{4}\)(6x²-xy-2y²)
⇒ 4x = 2y²+xy-6x²

Given \(\frac { dc }{ dq } =\frac { { c }^{ 2 }+2cq }{ { q }^{ 2 } } \)
This is a homogeneous equation in e and q of order 2
∴ Put c = vq and \(\frac { dc }{ dq } =v+q\frac { dv }{ dq } \)
∴ v + q\(\frac { dv }{ dq } \) = v²a² + 2vq₀q =\(\frac { { q }^{ 2 }({ v }^{ 2 }+2v) }{ { q }^{ 2 } } \)v²+2vv
q\(\frac { dv }{ dq } \) = v²+2v-v = v²+v
Separating the variables we get,
\(\frac { dv }{ v+v } =\frac { dq }{ q } \)
Integrating \(\int { \frac { dv }{ v(v+1) } } =\int { \frac { dq }{ q } } \)
[ \(\frac { 1 }{ v(v+1) } =\frac { A }{ v } +\frac { B }{ v+1 } \)
⇒ 1 = A(v+1)+B
put v = -1
1 = -B ⇒ B = -1
put v = 0
⇒ 1 = A ]
\(\int { \left( \frac { 1 }{ v } -\frac { 1 }{ v+1 } \right) dv } =\int { \frac { dq }{ q } } \)
⇒ log v -log (v + 1) = log q + log k
⇒ log\(\frac { v }{ v+1 } \) = logq.k
⇒ \(\frac { v }{ v+1 } \) = q.k
Replacing v by \(\frac { c }{ q } \), we get
\(\frac { c/q }{ c/q+1 } \) = q.k
⇒ \(\frac { c }{ c+q } \) = kq
⇒ c = kq(c+q)  .....(1)
Given when c = 1 and q = 1
⇒ 1 = k(1) (1+1) ⇒ 1 = 2 k ⇒ k = \(\frac { 1 }{ 2 } \)
∴ (1) ⇒ c = \(\frac{q}{2}\)(c+q)
∴ ⇒ 2c = q(c+q)

Given \(\frac { dp }{ dx } =\frac { 2{ p }^{ 3 }-{ x }^{ 3 } }{ 3x{ p }^{ 2 } } \)
The numerator and denominator are homogeneous functions of 3,
∴ Put p=vx and \(\frac { dp }{ dx } =v+x\frac { dv }{ dx } \)

= \(\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } \)
⇒ \(\frac { dv }{ dx } =\frac { 2{ v }^{ 3 }-1 }{ 3{ v }^{ 2 } } -v=\frac { 2{ v }^{ 3 }-1-3{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
= \(\frac { -1-{ v }^{ 3 } }{ 3{ v }^{ 2 } } \)
⇒ \(\left( \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } \right) dv=-\frac { dx }{ x } \)
Integrating, \(\int { \frac { 3{ v }^{ 2 } }{ 1+{ v }^{ 3 } } } dv=-\int { \frac { dx }{ x } } \)
⇒ log(1+v³) = -log x + log c
⇒ 1+v³ = \(\frac { c }{ x } \)
Replacing v by \(\frac { p }{ x } \) we get
\(1+\frac { { p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \Rightarrow \frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 3 } } =\frac { c }{ x } \)
⇒ \(\frac { { x }^{ 3 }+{ p }^{ 3 } }{ { x }^{ 2 } } \)= c ⇒ x³+p³ = cx²...(1)
When x = 10, p = 20
⇒ 10³ + 20³ = c(10)² ⇒ 1000 + 8000 = 100 c
⇒ 9000 = 100 c
⇒ c = 90
∴ (1) becomes,
x³+p³ = 90x²
⇒ p³ = 90x²-x³
⇒ p³ = x²(90-x) which is the required relationship.

The auxilary equation is m² + 14m + 49 = 0
⇒ (m + 7)² = 0
⇒ m = -7, -7
The roots are real and equal
∴ CF is (Ax+ B) e^-7x
[∴ (D-7)² = 0, when D = 7]
PI₁ = \(\frac { e^{ -7x } }{ (D-7)^{ 2 } } =\frac { { x }^{ 2 } }{ 2 } \).e^-7x
PI₂ = \(\frac { 4.{ e }^{ 0x } }{ (D-7)(D-7) } =\frac { 4.e^{ 0x } }{ (0-7)(0-7) } =\frac { 4 }{ 49 } \)
∴ The general solution is y = CF + PI₁ + PI₂
⇒ y = (Ax+B)e-7x + \(\frac { { x }^{ 2 } }{ 2 } e^{ -7x }+\frac { 4 }{ 49 } \).

(y-x)\(\frac { dy }{ dx } \) = a²
⇒ \(\frac { dy }{ dx } =\frac { { a }^{ 2 } }{ y-x } \Rightarrow \frac { dx }{ dy } =\frac { y-x }{ { a }^{ 2 } } \)
⇒ \(\frac { dx }{ dy } =\frac { y }{ { a }^{ 2 } } -\frac { x }{ a^{ 2 } } \)
⇒ \(\frac { dx }{ dy } +\frac { x }{ { a }^{ 2 } } =\frac { 1 }{ { a }^{ 2 } } \)
This is of the form \(\frac { dx }{ dy } \)+Px = Q
where P = \(\frac { 1 }{ { a }^{ 2 } } \) and Q = \(\frac { 1 }{ { a }^{ 2 } } \)y
∴ \(\int { P } dy=\int { \frac { 1 }{ { a }^{ 2 } } dy } =\frac { 1 }{ { a }^{ 2 } } \)
I.F = \(e^{ \int { pdy } }=e^{ y/{ a }^{ 2 } }\)
∴ The solution is \(\int { x. } e^{ \int { pdy } }=\int { Q.e^{ \int { pdy } } } dy\)       
⇒ x.e^y/a2 =\(\int { \frac { 1 }{ { a }^{ 2 } } y } \) e^y/a2dy+C....(1)
Put \(\frac { 1 }{ { a }^{ 2 } } \)y = t ⇒ dy = a²dt
[∵ u = t; d = e^t]
u¹= 1; v = e^t
v₁ = e^t
\(\int { u } dv\) = uv-u¹v₁]
∴ (1) ⇒ x.e^y/a2 = a²\(\int { te^{ t } } \)dt
= a²[te^t-e^t]+C
\(xe^{ \frac { y }{ { a }^{ 2 } } }\) = a².e^t(t-1)+C
\(xe^{ \frac { y }{ { a }^{ 2 } } }=a^{ 2 }.e^{ \frac { y }{ { a }^{ 2 } } }\left( \frac { y }{ { a }^{ 3 } } -1 \right) \)   [∵ t = \(\frac { y }{ { a }^{ 2 } } \)]

Given x²\(\frac { dy }{ dx } \)= y²+2xy
⇒ \(\frac { dy }{ dx } =\frac { { y }^{ 2 }+2xy }{ x^{ 2 } } \)
The numerator and denominator are homogeneous function of degree
∴ put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

= v²+2v
⇒ v + x\(\frac { dv }{ dx } \) = v²+ 2v
⇒ x\(\frac { dv }{ dx } \) = v²+ 2v-v = v²+ v
Separating the variables,
\(\frac { dv }{ { v }^{ 2 }+v } =\frac { dx }{ x } \Rightarrow \frac { dv }{ v(v+1) } =\frac { dx }{ x } \)
[\(\frac { 1 }{ v(v+1) } =\frac { A }{ v } +\frac { B }{ v+1 } \)
1 = A(v+1)+Bv
put v = -1
1 = -B
put v = 0
1 = A
∴ \(\left( \frac { 1 }{ v } +\frac { 1 }{ v+v } \right) dv=\frac { dx }{ x } \)
Integrating
\(\int { \frac { dv }{ v } } -\int { \frac { dv }{ v+1 } } =\int { \frac { dx }{ x } } \)
⇒ log v - log (v + 1) - log x + log c
⇒ log\(\left( \frac { v }{ v+1 } \right) \)= log(xc)
⇒ \(\frac { v }{ v+1 } \)
Replacing v by \(\frac { y }{ x } \) we get,
\(\frac { \frac { y }{ x } }{ \frac { y }{ x } +1 } =xc\Rightarrow \frac { \frac { y }{ x } }{ \frac { x+y }{ x } } \) = xc
⇒ \(\frac { y }{ x+y } \) = xc
⇒ y = cx(x+y)  ....(1)
Given, when x = -1, y = 1
∴ 1 = c(1) (1+1) ⇒ = 2c ⇒ c = \(\frac { 1 }{ 2 } \)
∴ (1) becomes, y = \(\frac { x }{ 2 } \)(x+y)
⇒ 2y = x(x+y)

Given \(\frac { dy }{ dx } \) = sin(x+y)....(1)
put x + y = z
⇒ 1 + \(\frac { dy }{ dx } =\frac { dz }{ dx } \)
⇒ \(\frac { dy }{ dx } =\frac { dz }{ dx } \)-1
∴ (1) becomes, \(\frac { dz }{ dx } \)-1 = sin z
⇒ \(\frac { dz }{ dx } \)= 1 + sin z
Separating the variables we get
\(\\ \frac { dz }{ 1+sinz } \) = dx
Multiplying & dividing by (1 + sin z) we get,
\(\\ \frac { (1+sinx)dz }{ (1+sinz)(1-sinz) } \) = dx
⇒ \(\frac { (1+sinz)dz }{ 1-sin^{ 2 }z } \) = dx
⇒ \(\frac { (1+sinz) }{ cos^{ 2 }z } \)dz = dx  [∵ sin²x + cos²x = 1]
⇒ \(\left( \frac { 1 }{ cos^{ 2 }z } +\frac { sinz }{ cos^{ 2 }z } \right) \)dz = dx
⇒ (sec²z + tanz secz)dz = dx
Integrating \(\int { sec^{ 2 } } zdz+\int { tanz } \)secz dz =\(\int { dx } \)
⇒ tanz - sec z = x + C
⇒ tan(x + y) - sec(x + y) = x + C [∵ z = x + y]