(D²−4D−1)y = e^−3x
The auxiliary equation is
m²−4m−1 = 0
(m−2)²−4−1 = 0
(m− 2)² = 5
\(m-2=±\sqrt { 5 } \)
\(m=2±\sqrt5\)
C.F = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}
\(PI=\frac { 1 }{ \phi (D) } f(x)\)
= \(\frac { 1 }{ { D }^{ 2 }-4D-1 } { e }^{ -3x }\)
\(=\frac { 1 }{ { (-3) }^{ 2 }-4(-3)-1 } { e }^{ -3x }\) (Replace D by −3)
\(=\frac { 1 }{ 9+12-1 } ​​{ e }^{ -3x }\)
\(=\frac { e^{ -3x } }{ 20 } \)
Hence the general solution is y = C.F+P.I
⇒ y = Ae^{\((2+\sqrt5)x\)} + Be^{\((2-\sqrt5)x\)}+\(\frac { e^{ -3x } }{ 20 } \)

(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

Given y = ae^4x + be^−x.    (1)
Here a and b are arbitrary constants
From (1), \(\frac { dy }{ dx } \) = 4ae^4x− be^−x     (2)
and \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 16ae^4x+ be^−x          (3)
(1) + (2) ⇒ \(y+\frac { dy }{ dx } \) = 5ae^4x      (4)
= (2) + (3) ⇒ \(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) = 20ae^4x
= 4(5ae^4x)
= \(4\left( y+\frac { dy }{ dx } \right) \)
\(\frac { dy }{ dx } +\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } =4y+4\frac { dy }{ dx } \)
⇒ \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } -3\frac { dy }{ dx } -4y=0\) which is the required differential equation

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 the family of paraboles with latus rectum 4a and whose axes are parallel to the x-axis is (y- k)² = 4a(x- h)
[Where (h, k) is the centre of the parabola]
Differentiating w.r.t. 'x' we get,
2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a(1)
⇒ 2(y-k)\(\left( \frac { dy }{ dx } \right) \) = 4a           (1)
Differentiating again w.r.t x we get,
2(y-k)\(\left( \frac { { d }^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) (2)\frac { dy }{ dx } \) = 0  (Product rule)
(y-k)\(\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 2 }\) = 0 [Divided by 2]
y-k= \(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \)    (2)
Substituting (2) in (1) we get,
2\(\frac { -\left( \frac { dy }{ dx } \right) ^{ 2 } }{ \frac { { d }^{ 2 }y }{ dx^{ 2 } } } \left( \frac { dy }{ dx } \right) \)= 4a
⇒ \(-2\left( \frac { dy }{ dx } \right) ^{ 3 }=4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) \)
⇒ \(4a\left( \frac { d^{ 2 }y }{ dx^{ 2 } } \right) +\left( \frac { dy }{ dx } \right) ^{ 3 }\)= 0
\(2a\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } +{ \left( \frac { dy }{ dx } \right) }^{ 3 }=0\).

Given \(\frac { dy }{ dx } \) − (3 cot x) y = sin 2x
It is of the form \(\frac { dy }{ dx } \) + Py = Q
Here P= − 3 cot x,Q = sin 2x
ഽPdx = ഽ-3 cot xdx = -3 log sin x = - log sin³x = log \(\frac { 1 }{ { sin }^{ 3 }x } \)
I.F. = \({ e }^{ log\frac { 1 }{ { sin }^{ 3 }x } }=\frac { 1 }{ { sin }^{ 3 }x } \)
The required solution is y (I.F) = ഽQ(I.F)dx+c
\(y\frac { 1 }{ { sin }^{ 3 }x } =\int { sin2x } \frac { 1 }{ { sin }^{ 3 }x } dx+c\)
\(\int { \frac { 1 }{ { sin }^{ 3 }x } } =\int { 2sinxcosx\times \frac { 1 }{ { sin }^{ 3 }x } } dx+c\)
= \(2\int { \frac { 1 }{ sinx } \times \frac { cosx }{ sinx } } dx+c\)
= ഽcos ecx cot xdx + c
\(y\frac { 1 }{ { sin }^{ 3 }x } =-2cosecx+c\)
Now y = 2 when x = \(\frac { \pi }{ 2 } \)
(1) ⇒ 2\(\frac {1 }{ 1 } \) = −2×1+c ⇒ c= 4
∴ (1) ⇒ \(y\frac { 1 }{ { sin }^{ 3 }x } \) = −2cosecx + 4

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 (D²−3D−4)y = 0
The auxiliary equations is
m²− 3m − 4 = 0
⇒ (m − 4)(m + 1) = 0
m = −1, 4
Roots are real and different
∴ The complementary function is Ae^−x+ Be^4x
The general solution is y = Ae^−x + Be^4x