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

(x² + y²)dy = xydx 
⇒ \(\frac { dy }{ dx } +\frac { xy }{ x^{ 2 }+{ y }^{ 2 } } \)....(1)
Since the numerator and denominators are homogeneous functions of degree 2,
put y = vx, and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
(1) becomes

= \(\frac { v }{ 1+v^{ 2 } } \)
⇒ v+\(x\frac { dv }{ dx } =\frac { v }{ 1+v^{ 2 } } -v=\frac { v-v(1+v^{ 2 }) }{ 1+{ v }^{ 2 } } \)
= \(\frac { v-v-{ v }^{ 3 } }{ 1+v } \)
= \(\frac { -v^{ 3 } }{ 1+v^{ 2 } } \)
Separating the variables we get,
\(x\frac { dv }{ dx } =\frac { -v^{ 3 } }{ 1+v^{ 2 } } \)
⇒ \(\frac { (1+v^{ 2 })dv }{ { v }^{ 3 } } =\frac { -dx }{ x } \)
⇒ \(\frac { 1 }{ { v }^{ 3 } } dv+\frac { { v }^{ 2 } }{ { v }^{ 3 } } dv=-\int { \frac { dx }{ x } } \)
⇒ \(\frac { 1 }{ { v }^{ 3 } } dv+\frac { dv }{ v } =-\frac { dx }{ x } \)
Integrating both sides we get,
\(\int { v^{ -3 }dv } +\int { \frac { dv }{ v } } =-\int { \frac { dx }{ x } } \)
\(\frac { -1 }{ 2v^{ 2 } } \) + log v = -log x + log c
⇒ \(\frac { 1 }{ 2v^{ 2 } } \) = log v + log x - log c
⇒ \(\frac { 1 }{ 2v^{ 2 } } =log\frac { vx }{ c } \)
Replace v by \(\frac { y }{ x } \) we get,

⇒ \(\frac { { x }^{ 2 } }{ 2{ y }^{ 2 } } =log\frac { y }{ c } \)
⇒ \(e^{ { x }^{ 2 }/2{ y }^{ 2 } }=\frac { y }{ c } \)
⇒ y = \(ce^{ { x }^{ 2 }/2{ y }^{ 2 } }\).

Given marginal cost function is (x² + xy)dy + (3xy + y²)dx = 0 
\(\frac { dy }{ dx } =\frac { -(3xy+{ y }^{ 2 }) }{ { x }^{ 2 }+xy } \)    (1)
Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \) in (1)
\(v+x\frac { dv }{ dx } =\frac { -(3xvx+{ v }^{ 2 }{ x }^{ 2 }) }{ { x }^{ 2 }+xvx } \)
\(=\frac { -(3v+{ v }^{ 2 }) }{ 1+v } \)
Now, \(x\frac { dv }{ dx } =\frac { -3v-{ v }^{ 2 } }{ 1+v } -v\)
\(=\frac { -3v-{ v }^{ 2 }-v-{ v }^{ 2 } }{ 1+v } \)
\(x\frac { dv }{ dx } =\frac { -4v-{ 2v }^{ 2 } }{ 1+v } \)
\(\frac { 1+v }{ 4v+2{ v }^{ 2 } } dv=\frac { -dx }{ x } \)
On Integration
\(\int { \frac { 1+v }{ 4v+2{ v }^{ 2 } } } =-\int { \frac { dx }{ x } } \)
Now, multiply 4 on both sides
\(\int { \frac { 1+v }{ 4v+2{ v }^{ 2 } } } =-4\int { \frac { dx }{ x } } \)
log (4v+2v²) = −4 logx+logc
4v + 2v² = \(\frac { c }{ { x }^{ 4 } } \)
x⁴(4v + 2v²) = c
Replace \(v=\frac { y }{ x } \)
\({ x }^{ 4 }\left( 4\frac { y }{ x } +2\frac { { y }^{ 2 } }{ { x }^{ 2 } } \right) =c\)
\({ x }^{ 4 }\left[ \frac { 4xy+2{ y }^{ 2 } }{ { x }^{ 2 } } \right] \) = c
c = 2x²(2xy + y²)   (2)
Cost of producing a pair of shoes = Rs. 12
(i.e) y = 12 when x = 2
c = 8[48 + 144]= 1536
∴ The cost function is x²(2xy + y²) = 768

y²dx + (xy + x²)dy = 0
(xy +x²)dy = −y²dx
\(\frac { dy }{ dx } =\frac { -{ y }^{ 2 } }{ xy+{ x }^{ 2 } } \)  (1)
It is a homogeneous differential equation, same degree in x and y
Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) becomes
\(v+x\frac { dv }{ dx } =\frac { { v }^{ 2 }{ x }^{ 2 } }{ xvx+{ x }^{ 2 } } \)
= \(\frac { -{ v }^{ 2 } }{ v+1 } \)
\(x\frac { dv }{ dx } =\frac { { -v }^{ 2 } }{ v+1 } -v\)
\(=\frac { { -v }^{ 2 }-{ v }^{ 2 } }{ v+1 } -v\)
\(x\frac { dv }{ dx } =\frac { -\left( v+{ 2v }^{ 2 } \right) }{ 1+v }\)
Now, separating the variables
\(\frac { 1+v }{ v(1+2v) } dv=\frac { -dx }{ x } \)
\(\frac { (1+2v)-v }{ v(1+2v) } dv=\frac { -dx }{ x } \)  (∵ 1 + v = 1 + 2v − v)
\(\frac { 1 }{ v } -\frac { 1 }{ 1+2v } dv=\frac { -dx }{ x } \)
On Integration we have
ഽ\(\left( \frac { 1+v }{ v(1+2v) } \right) dv\) = -ഽ\(\frac { dx }{ x } \)
log v \(\frac12\) log (1 + 2v) = -log x + log c
log \(\left( \frac { v }{ \sqrt { 1+2v } } \right) \) = log \(\left( \frac { c }{ x } \right) \)
\(\frac { v }{ \sqrt { 1+2v } } =\frac { c }{ x } \)
Replace \(v=\frac { y }{ x } \) we get
\(\frac { \frac { y }{ x } }{ \sqrt { 1+\frac { 2y }{ x } } } =\frac { c }{ x } \)
\(\frac { y\sqrt { x } }{ \sqrt { x+2y } } =c\)
\(\frac { { y }^{ 2 } }{ x+2y } =k\)
where k = c²

Given 3e^x tan y dx + (1 + e^x)sec²y dy = 0
3e^x tan y dx = −(1 + e^x)sec² y dy
\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx=-\frac { { sec }^{ 2 }y }{ tany } dy\)
Integrating, we get 3ഽ\(\frac { { 3e }^{ x } }{ 1+{ e }^{ x } } dx\) = -ഽ\(\frac { { sec }^{ 2 }y }{ tany } dy\) + c
3log(1+ e^x ) = −log tan y + log c   \(\left[ \therefore \int { \frac { f'(x) }{ f(x) } dx=logf(x) } \right] \)
log(1+e^x)³ + log tan y = log c
log [(1 + e^x)³ tan y ] = log c
(1+e^x)³ tan y = c (1)
Given y(0) = \(\frac { \pi }{ 4 } \) (i.e) y = \(\frac { \pi }{ 4 } \) at x =0
(1) ⇒ (1 +e⁰ )³ tan \(\frac { \pi }{ 4 } \) = c
2³ (1) = c
⇒ c = 8
Hence the required solution is (1  +e^x)³ tan y = 8