\(\frac { { d }^{ 2 }x }{ d{ t }^{ 2 } } -3\frac { dx }{ dt } +2x=0\)
Given (D²−3D+2) x = 0 where D = \(\frac{d}{dt}\)
A.E is m²− 3m + 2 = 0
(m−1) (m−2) = 0
m = 1, 2
C.F = Ae^t+ Be^2t
The general solution is x = Ae^t+ Be^2t   (1)
Now when t=0, x=0 (given)
(1) ⇒ 0 = A + B   (2)
Differentiating (1) w.r.t ‘t’
\(\frac { dx }{ dt } \)= Ae^t+ 2Be^2t
When t = 0, \(\frac { dx }{ dt } \) = 1
A + 2B = 1
Thus we have A + B = 0 and A + 2B = 1
Solving, we get A = –1, B = 1
∴ (1) ⇒ x = –e^t + e^2t
(i.e.) x = e^2t–e^t
Type II: \(f(x)=e^{a x}(\text { i.e }) \phi(D) y=e^{a x}\)
\(\text { P.I }=\frac{1}{\phi(D)} e^{a x}\)
Replace D by a, provided \(\phi(D) \neq 0\) when D = a
If \(\phi(D)=0\) when D = a, then
\(\text { P.I }=x \frac{1}{\phi^{\prime}(D)} e^{a x}\)
Replace D by a, provided \(\phi(D) \neq 0\) when D = a
If \(\phi^{\prime}(D)=0\) when D = a, then
\(\text { P.I }=x^{2} \frac{1}{\phi^{\prime \prime}(D)} e^{a x}\) and so on 

Given \(\frac { dy }{ dx } +\frac { 1 }{ x } y={ x }^{ 3 }\)
It is of the form \(\frac { dy }{ dx } +Py=Q\)
Here \(P=\frac { 1 }{ x } ,Q={ x }^{ 3 }\)
ഽPdx = ഽ\(\frac { 1 }{ x } \) dx = log x
I. F = eഽ^pdx = e^{log x} = x
The required solution is y(I.F) = ഽQ(I.F)dx + c
yx = ഽx³.x dx + c
= ഽx⁴dx + c
= \(\frac { { x }^{ 5 } }{ 5 } \) + c
yx = \(\frac { { x }^{ 5 } }{ 5 } \) + c

(1 − x)dy − (1 + y)dx
Separating the variables we get,
\(\frac { dy }{ 1+y } =\frac { dx }{ 1-x } \)
Integrating both sides we get,
\(\int { \frac { dy }{ 1+y } } =\int { \frac { dx }{ 1-x } } \)
log(1+y) =\(\frac { log(1-x) }{ -1 } \) + log c
log(1+y) = -log(1-x) + log c
⇒ log(1+y) + log(1-x) = log c
⇒ log(1+y)(1-x) = log c
⇒ (1+y)(1-x) = c
Multiplying by a negative sign we get,
(x-1)(y+1) = -c = C where C = -c

cos x(1 + cos y)dx − sin y(1 + sin x)dy
Separating the variables we get
\(\frac { \cos x }{ 1+\sin x } dx=\frac { \sin y }{ 16 \cos y } dy\)
Integrating both sides we get
\(\int { \frac { \cos x }{ 1+\sin x } } dx=\int { \frac { \sin y }{ 16 \cos y } } dy\)
put 1 + sin x = t ⇒ cos x dx = dt
Also 1 + cosy = s ⇒ -siny dy = ds
⇒ siny dy = -ds
⇒ \(\int { \frac { dt }{ t } } =-\int { \frac { ds }{ s } } \)
⇒ log t = log s + log c
⇒ log t = log\(\left( \frac { c }{ s } \right) \)
[∵ log m- logn=log\(\frac{m}{n}\)]
⇒ t=\(\frac { c }{ s } \)
⇒ 1+sinx =\(\frac { c }{ 1+cosy } \)
[∵ t = 1 + sin x & s = 1 + cos y]
⇒ (1 + sin x)(1 + cos y) = c

Equation of family of parabolas with foci at the origin and axis along the x-axis is
y² = 4a(x+a) ...(1)
[ ∵ the focus is at the origin its vertex will be (-a, 0) and latus rectum is 4a]
Differentiating w.r.t. 'x' we get,
2y\(\left( \frac { dy }{ dx } \right) \) = 4a(1) ....(1)
⇒ 2y\(\left( \frac { dy }{ dx } \right) \) = 4a ...(2)
Also \(\frac { 2y }{ 4 } \left( \frac { dy }{ dx } \right) \) = a
⇒ \(\frac { y }{ 2 } \left( \frac { dy }{ dx } \right) \) = a...(3)
Substituting (2) and (3) in (1) we get,
y² = 2y\(\left( \frac { dy }{ dx } \right) \left[ x+\frac { y }{ 2 } \left( \frac { dy }{ dx } \right) \right] \)
⇒ y² = 2xy\(\left( \frac { dy }{ dx } \right) +{ y }^{ 2 }\left( \frac { dy }{ dx } \right) ^{ 2 }\)
Dividing by Y we get,
\(y=2x\frac { dy }{ dx } { +y\left( \frac { dy }{ dx } \right) }^{ 2 }\).

Equation of all circles passing through the origin and having their centres on the y-axis.
x² + (y - k)² = k² ....(1)
[where (0, k) is the centre of the cirde which lies on the y-axis and radius is k].
Differentiating w.r.t. 'x
⇒ 2x + 2(y-k)\(\frac { dy }{ dx } \) = 0
⇒ (y-k) =\(\frac { -x }{ \frac { dy }{ dx } } \) ....(2)
Also, y+\(\frac { x }{ \frac { dy }{ dx } } \) = k ....(3)
Substituting (2) & (3) in (1) we get
\({ x }^{ 2 }+\left( \frac { -x }{ \frac { dy }{ dx } } \right) ^{ 2 }=\left( y+\frac { x }{ \frac { dy }{ dx } } \right) ^{ 2 }\)

⇒ x² = y²+\(\\ \frac { 2xy }{ \frac { dy }{ dx } } \)
⇒ \(y^{2}-x^{2}-2 x y \frac{d y}{d x}=0\)

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
\(\Rightarrow \frac { c }{ x^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } \right) +k\left( \frac { c }{ { x }_{ 0 }^{ 10 } } -\frac { 10 }{ 11{ x }_{ 0 }^{ 11 } } \right) \)
\(\Rightarrow \frac { c }{ x^{ 10 } } -\frac { c }{ { x }_{ 0 }^{ 10 } } =\frac { 10 }{ 11 } \left( \frac { 1 }{ { x }^{ 11 } } -\frac { 10 }{ { x }_{ 0 }^{ 11 } } \right) \).

The given equation can be written as \(\frac { dy }{ dx } +\frac { 1 }{ { cos }^{ 2 }x } y=\frac { tanx }{ { cos }^{ 2 }x } \)
\(\frac { dy }{ dx } \) + y sec²x = tan x sec²x
It is of the form \(\frac{dy}{dx}\) + Py + Q
Here P = sec²x,Q = tanx sec²x
ഽPdx = ഽsec² x dx = tanx
I.F = eഽ^pdx = e^{tan x}
The required solution is y(I.F) = ഽQ(I.F)dx + c
ye^{tan x} = ഽtan x sec²xe^tan xdx + c
Put tan x = t
Then sec² xdx = dt
∴ ye^{tan x} = ഽte^t dt + c
= ഽtd(e^t) + c
= te^t − e^t + c
= tanx e^{tan x}−e^{tan x}+ c
ye^{tan x} = e^{tan x}(tan x − 1) + c

\(\frac { dy }{ dx } =\frac { x+y }{ x } \)
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.

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

x²dy + y(x + y) dx = 0
x²dy = −y( x+y) dx
\(\frac { dy }{ dx } =\frac { -(xy+{ y }^{ 2 }) }{ { x }^{ 2 } } \)
Put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \) in (1)
\(v+x\frac { dv }{ dx } =\frac { (xvx+{ v }^{ 2 }{ x }^{ 2 }) }{ { x }^{ 2 } } \)
= −(v + v²)
\(x\frac { dv }{ dx } \) = −v²−v− v
= −(v² + 2v)
On separating the variables
\(\frac { dv }{ { v }^{ 2 }+2v } =-\frac { dx }{ x } \)
\(\frac { dv }{ v(v+2) } =\frac { -dx }{ x } \)
\(\frac { 1 }{ 2 } \left[ \frac { (v+2)-v }{ v(v+2) } \right] dv=\frac { -dx }{ x } \)
\(\frac { 1 }{ 2 } ഽ\left[ \frac { 1 }{ v } -\frac { 1 }{ v+2 } \right] dv=-ഽ\frac { dx }{ x } \)
\(\frac12\) [log v - log (v + 2)] = - log x + log c
\(\frac { 1 }{ 2 } log\frac { v }{ v+2 } =log\frac { c }{ x } \)
We have
\(\frac { v }{ v+2 } =\frac { { c }^{ 2 } }{ { x }^{ 2 } } \)
Replace v = \(\frac{y}{x}\), we get
\(\frac { y }{ x\left( \frac { y }{ x } +2 \right) } =\frac { k }{ { x }^{ 2 } } \)  where c² =k
\(\frac { y{ x }^{ 2 } }{ y+2x } =k\)
When x = 1, y = 1
∴ (2) ⇒ k = \(\frac { 1 }{ 1+2 } \)
k = \(\frac { 1 }{ 3 } \)
∴ The solution is 3x² y = 2x + y

(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.

Given  \(\frac { { d }^{ 2 }y }{ d{ x }^{ 2 } } -4\frac { dy }{ dx } +5y\) = 0
(D²−4D+5)y = 0
The auxiliary equation is m²−4m + 5 = 0
⇒ (m−2)²−4 + 5 = 0
(m− 2)² = –1
m - 2 = 土\(\sqrt{-1}\)
m = 2 土 i , it is if the form α 土 iβ
∴ C.F = e^2x[A cos x + B sin x]
The general solution is y = e^2x[A cos x + B sin x]