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

Given (x²-yx²)dy + (y²+xy²)dx = 0
⇒ x²(1-y)dy+y²(1+x)dx = 0
⇒ x²(1-y)dy = -y²(1+x)dx
Separating the variables we get,
\(\frac { (1-y) }{ y^{ 2 } } dy=-\frac { (1+x) }{ x^{ 2 } } \)dx
⇒ \(\frac { 1 }{ { y }^{ 2 } } dy-\frac { 1 }{ y } dy=-\frac { 1 }{ x^{ 2 } } dx-\frac { 1 }{ x } dx\)
Integrating, \(\int { { y }^{ -2 } } dy-\int { \frac { 1 }{ y } } dy=-\int { \frac { 1 }{ x^{ 2 } } } dx-\int { \frac { 1 }{ x } } \)
\(-\frac { 1 }{ y } -logy=\frac { 1 }{ x } \) -log x + C
⇒ log x - log y = \(\frac { 1 }{ x } +\frac { 1 }{ y } \)+C
⇒ log \(log\left( \frac { x }{ y } \right) =\frac { x+y }{ xy } \)+C
⇒ \(\frac { x }{ y } =e^{ \frac { x+y }{ xy } +C }\)
⇒ \(\frac { x }{ y } =K.e^{ \frac { x+y }{ xy } }\)  [where e^C = K]

Given cos²x dy + y.e^tanx dx=0
⇒ cos²x dy = -y e^tanx dx   [∵ t = tanx, dt = sec²x dx, ∴ \(\int { { e }^{ t } } dt\) = e^tanx]
⇒ \(\frac { dy }{ y } =-\frac { { e }^{ tanx } }{ cos^{ 2 }x } \)dx
⇒ \(\frac { dy }{ y } \) = -sec²x.e^tanx dx
Integrating, \(\int { \frac { dy }{ y } } =-\int { sec^{ 2 }x } .e^{ tanx }dx\)
log y = -e^tanx + C
⇒ log y + e^tanx = C

Given Sec 2x dy = sin 5x sec² y dx
Separating the variables we get,
\(\frac { dy }{ sec^{ 2 }y } =\frac { sin5x }{ sec2x } \)dx
⇒ cos²ydy = sin5x.cos2xdx
⇒ Integrating, \(\int { cos^{ 2 }y } dy=\int { sin5x } cos2x\)dx
⇒ \(\int { \frac { 1+cos2y }{ 2 } } dy=\int { \left( \frac { sin7x+sin3x }{ 2 } \right) dx } \)+C
[∵ cos 2y = 2 cos²y-1 and sin C sin D = \(\frac{1}{2}\)[sin(C+D)+sin(C-D)]
⇒ y+\(\frac { sin2y }{ 2 } =\frac { cos7x }{ 7 } +\frac { cos3x }{ 3 } \)+C

Given y = (A + Bx)e^3x ....(1)
Differentiating w.r.t 'x' we get,
\(\frac { dy }{ dx } \)= (A+Bx)e^3x(3)+e^3x(B)
⇒ \(\frac { dy }{ dx } \) = 3y + Be^3x  [Using (1)]
⇒ Be^3x = \(\frac { dy }{ dx } \)-3y
Differentiating again w.r.t 'x' we get,
\(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) \)+ Be3x(3)
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left[ \frac { dy }{ dx } -3y \right] \) [Using (2)]
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =3\left( \frac { dy }{ dx } \right) +3\left( \frac { dy }{ dx } \right) \)-9y
⇒ \(\frac { d^{ 2 }y }{ { dx }^{ 2 } } =6\left( \frac { dy }{ dx } \right) \)-9y which is the required differential equation.

Given \(\frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } \)= 1
⇒ \(\frac { { b }^{ 2 }x^{ 2 }+{ a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ y }^{ 2 } } \)= 1
⇒ b²x²+a²y² = a²b²
Differentiating again w.r.t 'x' we get,
2b²x + 2a²y\(\frac { dy }{ dx } \)=0 ⇒ b²x+a²yy₁ = 0 ....(2)
Differentiating w.r.t 'x' we get,
b² + a²[yy₂ + y₁y₁] = 0
⇒ b² + a²[yy₂ + y₁²] = 0 ...(3)
Eliminating a² and b² from (1) and (3) we get
\(\left| \begin{matrix} x & y{ y }_{ 1 } \\ 1 & { y }_{ 1 }^{ 2 }+y{ y }_{ 2 } \end{matrix} \right| \) = 0
⇒ x(y₁²+yy₂) - yy₁ = 0
⇒ x\(\left( \left( \frac { dy }{ dx } \right) ^{ 2 }+y.\frac { d^{ 2 }y }{ { dx }^{ 2 } } \right) -y\left( \frac { dy }{ dx } \right) \)= 0 which is the required differential equation.

Given x²+y²+2gx = 0 ...(1)
where g is the arbitrary constant.
Differentiating w.r.t 'x' we get,
2x+2y\(\frac { dy }{ dx } \)+2g = 0
⇒ 2g = -2x-2y\(\frac { dy }{ dx } \) ...(2)
Substituting (2) in (1) we get,
x²+y²+x\(\left( -2x-2y\frac { dy }{ dx } \right) \)= 0
⇒ x²+y²-2x²+2xy\(\left( \frac { dy }{ dx } \right) \)= 0
⇒ y²-x²+2xy\(\left( \frac { dy }{ dx } \right) \) = 0 which is the required differential equation.