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.