Let p(t) denotes the amount of money in the account at time t. Then the differential equation governing the growth of money is
\(\frac { dp }{ dt } =\frac { 12 }{ 100 } p\Rightarrow \frac { dp }{ dt } =0.12p\)
Separating the variables,
\(\frac { dp }{ dt } =0.12\)
Integrating, \(\int { \frac { dp }{ P } } =\int { 0.12 } \)dt+c
⇒ log p = 0.12t + C
⇒ P = e^0.12t+c ⇒ P = e^0.12t . e^c
⇒ P = e^0.12t.C1  .....(1)
When t = 0, p = 0 ⇒ p = e⁰ (c₁) ⇒ c₁ = p
When t = 5, and p = 25000
25000 = e^12(5).p       [∵ c₁ = p]
⇒ 25000 = e^0.6p
⇒ \(\frac { 2500 }{ { e }^{ 0.6 } } \) = p
⇒ 25000 (e^-0.6) = p
⇒ 25000 (0.5488) = p
⇒ Rs. 13720
Hence, to get an amount of Rs. 25000 at the end of 5 years, Rs. 13720 must be invested.

The auxiliary equation is m² + 1 = 0
⇒  m² = -1
⇒ m = ±\(\sqrt { -1 } \) = ±i
Here α = 0, β = 1
∴ CF is e^0x [A cosx + B sinx]
∴ The general solution is
y = A cos x + B sin x ...(1)
Given when x = 0, y = 2
∴ 2 = A cos 0 + B sin 0
⇒ 2 = A+0 ⇒ A = 2
[∵ cos0 = 1 and sin0 = 0]
Also, when x = \(\frac { \pi }{ 2 } \), y = -2
∴ -2 = A\(cos\frac { \pi }{ 2 } +Bsin\frac { \pi }{ 2 } \)
⇒ -2 = A(0)(+B(1) ⇒ B = -2
[∵ \(cos\frac { \pi }{ 2 } \) = 0 and \(sin\frac { \pi }{ 2 } \)= 1]
Substituting the values of A & B in (1) we get,
y = 2 cos x - 2 sin x
⇒ y = 2 (cosx - sin x)

Given slope = y + 2x
⇒ \(\frac { dy }{ dx } \) = y+2x
⇒ \(\frac { dy }{ dx } \)- y = 2x
This is of the form \(\frac { dy }{ dx } \)+ Py = Q where P = -1, Q = 2x
\(\\ \int { P } dx=\int { -1 } dx\) = -x
∴ I.F. = \(e^{ \int { P } dx }\) = e^-x
∴ The solution is y.\(e^{ \int { P } dx }=\int { Q } .e^{ \int { P } dx }\)dx+C
⇒ y.e^-x = \(\int { 2x } \).e^-xdx+C
Let u = x; dv = e^-x
u² = 1; v = -e^-x
v₁ = e^-x
⇒ ye^-x = 2[-xe^-x-1(e^-x)]+C
[Bernoulli's formula]
⇒ ye^-x = -2x e^-x -2e^-x+ C...(1)
Since the Curve passes through (0, 0), we get
⇒ 0 = 0-2e⁰ + C ⇒ C = 2
(1) becomes,
∴ ye^-x = -2xe^-x - 2e^-x + 2
ye^-x = -2xe^-x - 2e^-x + 2e^x.e^-x
= e^-x(2 e^x-2x-2)
ye^-x = 2e^-x(e^x-x-1)

Given slope is 1+\(\frac { y }{ x } \)
⇒ \(\frac { dy }{ dx } =1+\frac { y }{ x } \Rightarrow \frac { dy }{ dx } =\frac { x+y }{ x } \)
The numerator and denominator homogeneous functions of degree 1
So put y = vx and \(\frac { dy }{ dx } =v+x\frac { dv }{ dx } \)

⇒ x\(\frac { dv }{ dx } \) = 1
Separating the variables we get,
dv = \(\frac { dx }{ x } \)
Integrating, \(\int { dv } =\int { \frac { dv }{ x } } \)
⇒ v = log x + log c
⇒ v = log x c
Replacing V by \(\frac { y }{ x } \) we get,
\(\frac { y }{ x } \) log x c ⇒ y - x log c x...(1)
Since the Curve passes through (1, 0),
0 = 1 logc ⇒ log c =0 ⇒ c = e⁰ =1
∴ c = 1
∴ (1) becomes, y = x logx.

Given (x+y)²\(\frac { dy }{ dx } \) = 1
put x+y = z
⇒ 1+\(\frac { dy }{ dx } =\frac { dz }{ dx } \)
\(\frac { dy }{ dx } =\frac { dz }{ dx } \)-1
∴ (1) becomes,
z²\(\left( \frac { dz }{ dx } -1 \right) \)= 1
⇒ z²\(\frac { dz }{ dx } \)-z² = 1
⇒ z²\(\frac { dz }{ dx } \) = 1 + z²
Separating the variables we get,
\(\left( \frac { { z }^{ 2 } }{ 1+{ z }^{ 2 } } \right) \)dz = dx
Adding and Subtracting 1 in the numerator, we get
\(\left( \frac { 1+{ z }^{ 2 }-1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
⇒ \(\left( \frac { 1+{ z }^{ 2 } }{ 1+{ z }^{ 2 } } -\frac { 1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
⇒ \(\left( z-\frac { 1 }{ 1+{ z }^{ 2 } } \right) \)dz = dx
Integrating, \(\int { dz } -\int { \frac { dz }{ 1+{ z }^{ 2 } } } \)
[∵ \(\int { \frac { dz }{ 1+{ z }^{ 2 } } } \) = tan^-1x+y]
⇒ (z-tan^-1(2) = x+C
⇒ (x+y)-tan^-1(x+y) = x + C
⇒ y-tan^-1(x+y) = C