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]

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

Given Gradient = \(\frac { x-a }{ y-b } \)
⇒ \(\frac { dy }{ dx } =\frac { x-a }{ y-b } \)
Separating the variables we get,
(y - b) dy = (x - a) dx
Integrating both sides we get,
\(\int { (y-b) } dy=\int { (x-a) } \)dx
\(\frac { { y }^{ 2 } }{ 2 } -by=\frac { { x }^{ 2 } }{ 2 } -ax\)+c     ......(1)
Since the curve passes through the origin (0, 0), we get
0-0 = 0-0+c ⇒ c=0
∴ (1) be comes,
\(\frac { { y }^{ 2 } }{ 2 } -by=\frac { { x }^{ 2 } }{ 2 } -ax\)

y²-2by = x²- 2ax
Adding and subtracting b² in the L.H.S and a² in the R.H.S we get

(y-b)²-b² = (x-a)²-a²
⇒ (y-b)² = (x-a)²+b²-a²

Given \(\frac { dy }{ dx } \) = ae⁷
Separating the variables we get,
\(\frac { dy }{ { e }^{ y } } \) =adx ⇒ e-y dy = adx
Integrating both sides we get,
\(\int { e^{ -y }dy } =a\int { dx } \)
-e^-y= ax+c
ax + e^-y+ c = 0

Separating the variables, we get
\(\frac { { \sec }^{ x }x }{ \tan x } dx+\frac { { \sec }^{ 2 } }{ { \tan y } } dy=0\)
Integrating, we get
ഽ\(\frac { { \sec }^{ x }x }{ \tan x } \)dx + ഽ\(\frac { { \sec }^{ 2 } }{ { \tan y } } dy\) = c
log tan x + log tan y = log c
log(tan x tan y) = log c
tan x tan y = c

Given  \(y=\frac { a }{ x } +b\)
Differentiating w.r.t x , we get
\(\frac { dy }{ dx } =\frac { -a }{ { x }^{ 2 } } \)
\({ x }^{ 2 }\frac { dy }{ dx } =-a\)
Again differentiating w.r.t x we get
\({ x }^{ 2 }\frac { dy }{ dx } +2x\frac { dy }{ dx } =0\)
⇒ \({ x }^{ 2 }\frac { dy }{ dx } +2x\frac { dy }{ dx } =0\) which is the required differential equation

Given equation is (x - α)² + (y - β)² = r²
Differentiating w.r.t. 'x' we get,
2 (x - α)² + (y - β)\(\frac { dy }{ dx } \) = r²
⇒ (x - α) + (y - β)\(\frac { dy }{ dx } \) = 0....(2)
Differentiating again w.r.t. 'x' we get
1+(y-β) \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\frac { dy }{ dx } .\frac { dy }{ dx } \) = 0
⇒ 1+(y-β) \(\frac { d^{ 2 }y }{ dx^{ 2 } } +\left( \frac { dy }{ dx } \right) ^{ 2 }\)= 0
⇒ y-β = \(\frac { -1\left( 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right) }{ \frac { d^{ 2 }y }{ dx^{ 2 } } } \) .....(3)
Substituting this value in (2) we get
(x-α) = \(\left( \frac { 1+\left( \frac { dy }{ dx } \right) ^{ 2 }\frac { dy }{ dx } }{ \frac { d^{ 2 }y }{ dx^{ 2 } } } \right) \) ......(4)
Substituting (3) and (4) in (1) we get
\(\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 }\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } +\frac { \left\{ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right\} ^{ 2 } }{ \left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 } } \)
=\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] \left[ \left( \frac { dy }{ dx } \right) ^{ 2 }+1 \right] ={ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 3 }\)
= \({ r }^{ 2 }\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) ^{ 2 }\)

(3D² + D - 14)y = 4 - 13\({ e }^{\frac{-7}{3}x}\)
The auxiliary equation is
3m² + m − 14 = 0
(3m+7)(m−2) = 0
m = \(\frac{-7}{3}\),2
C.F = \({ Ae }^{ \frac { -7 }{ 3 } x }+{ Be }^{ 2x }\)
\(=\frac { 1 }{ 3{ D }^{ 2 }+D-14 } (4)+\frac { 1 }{ { 3D }^{ 2 }+ } \left( { 4-13e }^{ \frac { -7 }{ 3 } x } \right) \)
= PI₁+ PI₂
P.I₁ = \(\frac { 1 }{ 3{ D }^{ 2 }+D-14 } 4{ e }^{ 0x }\) 
=\(\frac { 1 }{ 0+0-14 } { 4 }e^{ 0x }\)(Replace D by 0)
P.I₁ = \(\frac{-4}{14} =\frac{-2}{7} \)
P.I₂ = \(\frac { 1 }{ 3{ D }^{ 2 }+D-14 } \times { (-13)e }^{ \frac { -7 }{ 3 } x }\)
Replace D by \(\frac{-7}{3}\) Here 3D² + D − 14 = 0 when D = −\(\frac{7}{3}\)
∴ P.I₂ = x.\(\frac { 1 }{ 6D+1 } \left( -{ 13e }^{ \frac { -7 }{ 3 } x } \right) \)
Replace D by \(\frac{-7}{3}\) 
∴ P.I₂ = \(x\frac { 1 }{ 6\left( \frac { -7 }{ 3 } \right) +1 } \left( -{ 13 }e^{ \frac { -7 }{ 3 } x } \right) \)
= x\(e^{ \frac { -7 }{ 3 } x }\)
The general solution is y = C.F. + P.I₁ + P.I₂
y = \({ Ae }^{ \frac { -7 }{ 3 } x }+B{ e }^{ 2x }-\frac { 2 }{ 7 } +{ xe }^{ \frac { -7 }{ 3 } x }\)

The given differential equation is of the form
\(\frac { dy }{ dx } \)+ Py = Q where
P = \(\frac { 1 }{ x } \); Q = xe^x
∴ \(\int { p } dx=\int { \frac { 1 }{ x } } \) = log x
∴ Integrating factor (I. F) = \(e^{ \int { p } dx }=e^{ logx }=x\)
Hence the solution is
\(ye^{ \int { p } dx }=\int { Q } .e^{ \int { p } dx }dx+c\)
⇒ yx = \(\int { x{ e }^{ x }dx+c } \)
⇒ xy =\(\int { x{ e }^{ x }dx+c } \)....(1)
Let u = x²; dv = e^x
u' = 2x; v = e^x
u" = 2; v₁ = e^x
v₂ = e^x
Using Bernoulli's formula,
\(\int { u } dv\) = uv-u'v₁+u"v₂
∴ (1) becomes
xy = x²e^x - 2xe^x + 2e^x+c
⇒ xy = e^x(x²-2x+2) + c

The given equation can be reduced to
\(\frac { dy }{ dx } +\frac { 2x }{ { x }^{ 2 }+1 } y=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
It is of the form \(\frac { dy }{ dx } \) + Py = Q
Here P \(=\frac { 2x }{ { x }^{ 2 }+1 } ,Q=\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)
ഽPdx = ഽ\(\frac { 2x }{ { x }^{ 2 }+1 } \)dx = log(x² +1)
I.F = e^ഽpdx = e^log(x2+1) = x² + 1
The required solution is y(IF) = ഽQ(I.F)dx + c
y(x² +1) = ഽ\(\frac { 4{ x }^{ 2 } }{ { x }^{ 2 }+1 } \)(x² + 1)dx + c
y(x² +1) = \(\frac { 4{ x }^{ 3 } }{ 3 } \) + c

Given (y³−2yx²)dx + (2xy²−x³)dy = 0
(y³−2yx²)dx = - (2xy²−x³)dy
(y³−2yx²)dx = - (x³-2xy²)dy
⇒ \(\frac { dy }{ dx } =\frac { { y }^{ 3 }-2yx^{ 2 } }{ { x }^{ 3 }-2xy^{ 2 } } \)
Since the numerator and denominator are homogeneous functions of degree 3,
put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

⇒ \(v+x\frac { dv }{ dx } =\frac { { v }^{ 3 }-2v }{ 1-2v^{ 2 } } \)
⇒ \(x\frac { dv }{ dx } =\frac { { v }^{ 3 }-2v }{ 1-2v^{ 2 } } -v=\frac { { v }^{ 3 }-2v-v(1-2v^{ 2 }) }{ 1-2v^{ 2 } } \)
⇒ \(x\frac { dv }{ dx } =\frac { { v }^{ 3 }-2v-v+2v^{ 3 } }{ 1-2v^{ 2 } } =\frac { { v }^{ 3 }+2v^{ 3 }-3v }{ 1-2v^{ 2 } } \)
= \(\frac { 3v^{ 3 }-3v }{ 1-2v^{ 2 } } \)
Separating the variables we get,
\(\frac { 1-2v^{ 2 } }{ 3v^{ 3- }3v } dv=\frac { dx }{ x } \)
⇒ \(\frac { 1-2v^{ 2 } }{ { v }^{ 3 }-v } dv=3\frac { dx }{ x } \)
\(\frac { 1-2v^{ 2 } }{ { v }^{ 3 }-v } =\frac { 1-2v^{ 2 } }{ v({ v }^{ 3 }-1) } \)
= \(\frac { 1-2v^{ 2 } }{ v(v+1)(v-1) } \)
= \(\frac { A }{ v } +\frac { B }{ v+1 } +\frac { C }{ v-1 } \)
1-2v² = A(v+1)(v-1)+Bv(v-1)+Cv(v+1)
put v = 1 ⇒ -1 = 2c ⇒ c = -\(\frac { 1 }{ 2 } \)
put v = -1 ⇒ -1 = -B = -\(\frac { 1 }{ 2 } \)
put v = -1 ⇒ -1 = -B(-2)
⇒ -1 = 2B ⇒ B = -\(\frac { 1 }{ 2 } \)
put v = 0 ⇒ 1 = -A + 0 + 0
⇒ A = -1
⇒ \(\int { \left( \frac { -1 }{ v } \frac { -\frac { 1 }{ 2 } }{ v+1 } \frac { -\frac { 1 }{ 2 } }{ v-1 } \right) } dv=3\int { \frac { dx }{ x } } \)
⇒ -log v -\(\frac { 1 }{ 2 } \)log(v+1) -\(\frac { 1 }{ 2 } \)log(v-1)
= 3log x + log c
⇒ -\(\frac { 1 }{ 2 } \)log(v-1) = 3log x + log c
⇒ log v+\(\frac { 1 }{ 2 } \)log(v+1)+\(\frac { 1 }{ 2 } \)log(v-1)
= -3log x + log c
⇒ log v.\(\sqrt { v+1 } \sqrt { v-1 } =log\left( \frac { 1 }{ x^{ 3 } } \right) \).c
⇒ v\(\sqrt { v^{ 2 }-1 } =\frac { c }{ { x }^{ 3 } } \)
Replace v by \(\frac { y }{ x } \) we get,
\(\frac { y }{ x } \sqrt { \frac { { y }^{ 2 } }{ x^{ 2 } } -1 } =\frac { c }{ { x }^{ 3 } } \)
⇒ \(\frac { y }{ x } \sqrt { \frac { { y }^{ 2 }-{ x }^{ 2 } }{ x } } =\frac { c }{ { x }^{ 3 } } \)
⇒ \(\sqrt { { y }^{ 2 }-x^{ 2 } } =\frac { c }{ x } \Rightarrow \sqrt { { y }^{ 2 }-x^{ 2 } } \)= c        (1)
Since the curve passes through (1, 2) we get
(1) (2) \(\sqrt { 4-1 } =c\Rightarrow c=2\sqrt { 3 } \)
Substituting c = \(2\sqrt { 3 } \) in (1) we get
\(xy\sqrt { { y }^{ 2 }-x^{ 2 } } =2\sqrt { 3 } \).

\(\frac { dy }{ dx } =\frac { x-y }{ x+y } \)  ..... (1)
This is a homogeneous differential equation.
Now put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)
∴ (1) ⇒ \(v+x\frac { dv }{ dx } =\frac { x-vx }{ x+vx } \)
\(=\frac { 1-v }{ 1+v } \)
\(x\frac { dv }{ dx } =\frac { 1-v }{ 1+v } -v\)
\(=\frac { 1-2v-{ v }^{ 2 } }{ 1+v } \)
\(\frac { 1+v }{ { v }^{ 2 }+2v-1 } dv=\frac { -dx }{ x } \)
Multiply 2 on both sides
\(\frac { 2+v }{ { v }^{ 2 }+2v-1 } dv=-2\frac { -dx }{ x } \)
On Integration
ഽ\(\frac { 2+v }{ { v }^{ 2 }+2v-1 } \)dv = -2ഽ\(\frac { -dx }{ x } \)
log(v²+2v − 1) = −2log x + log c
v²+2v − 1 = \(\frac { c }{ { x }^{ 2 } } \)
x²(v²+2v−1) = c
Now, Replace \(v=\frac { y }{ x } \)
\({ x }^{ 2 }\left[ \frac { { { y }^{ 2 } } }{ { x }^{ 2 } } +\frac { 2y }{ x } -1 \right] =c\)
y² + 2xy − x² = c is the solution

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)