a) \( y=m x \)
\( \frac{d y}{d x} =\mathrm{m} \)
\(\mathrm{y} =x \frac{d y}{d x}\)
\(b) \ \mathbf{y}=m x+\frac{a}{m} , \) (m is arbitrary constant)
\( \frac{d y}{d x} =\mathrm{m} \)
\(\mathrm{y} =x \frac{d y}{d x}+\frac{a}{\frac{d y}{d x}} \)
\(\Rightarrow \ x\left(\frac{d y}{d x}\right)^{2}-y \frac{d y}{d x}+a =0\)
\(c) \ y=a \cos 3 x+b \sin 3 x \)
\( \frac{d y}{d x}=-3 \mathrm{a} \sin 3 x+3 \mathrm{~b} \cos 3 x \)
\( \frac{d^{2} y}{d x^{2}}=-9 \mathrm{a} \cos 3 x-9 \mathrm{~b} \sin 3 x \)
\( \frac{d^{2} y}{d x^{2}}=-9 y \)
\(d) \ x^{2}+y^{2}=a^{2} \)
\( 2 x+2 y \frac{d y}{d x}=0 \)
\( divided\ by\ 2\\ x+y \frac{d y}{d x}=0 \)

\( y =a \cos (m x+b) \)
\(\frac{d y}{d x} =-a m \sin (m x+b) \)
\(\frac{d^{2} y}{d x^{2}} =-a m^{2} \cos (m x+b) \)
\(\frac{d^{2} y}{d x^{2}}+m^{2} y =0\)

\( \mathrm{y} =\mathrm{A} \cos 5 x+\mathrm{B} \sin 5 x \)
\(\frac{d y}{d x} =-5 \mathrm{~A} \sin 5 x+5 \mathrm{~B} \cos 5 x \)
\(\frac{d^{2} y}{d x^{2}} =-25 \mathrm{~A} \cos 5 x-25 \mathrm{~B} \sin 5 x \)
\(\frac{d^{2} y}{d x^{2}} =-25(\mathrm{~A} \cos 5 x+\mathrm{B} \sin 5 x) \)
\(\frac{d^{2} y}{d x^{2}} =-25 \mathrm{y} \Rightarrow \frac{d^{2} y}{d x^{2}}+25 y=0\)

The auxiliary equation is m² - 6m + 25 = 0
Here a = 1, b = -6, c = 25
∴ m = \(\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } =\frac { 6\pm \sqrt { 36-4(1)(25) } }{ 2 } \)
= \(\frac { 6\pm \sqrt { 36-100 } }{ 2 } =\frac { 6\pm \sqrt { -64 } }{ 2 } =\frac { 6\pm 8i }{ 2 }\)

= 3 ± 4i
∴ α = 3, β = 4
Complementary function CF is e^ax
[A cosβx + B sinβx]
⇒ CF = e^3x[A cos4x + B sn4x]
∴ The general solution is e^3x
[A cos4x + B sin4x]

The auxiliary equation is 3m² - 5m + 2 = 0
⇒ (m - 1)(3m - 2) = 0
⇒ m = 1, \(\frac { 2 }{ 3 } \)
The roots are real and different
∴ Complementary function CF is Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\)
∴ The general solution is y = Ae^x + \({ Be }^{ \frac { 2 }{ 3 } x }\).

Given \(\frac { dC }{ dq } =a-\frac { c }{ q } \)
⇒ \(\frac { dC }{ dq } +\frac { C }{ q } \) = a
The given differential equation is of the form
\(\frac { dC }{ dq } \)+PC = Q where
P = \(\frac { 1 }{ q } \) and Q = a
\(\int { p } dq=\int { \frac { 1 }{ q } } \)
Integrating factor I.F = e^{log q} =q
∴ The solution is C y\(e^{ \int { P } dq }=\int { Q } e^{ \int { P } dq }\)+C
⇒ C(q) = \(\int { a } .qdq+C\)
⇒ C.q = a\(\left( \frac { { q }^{ 2 } }{ 2 } \right) \)+C
⇒ 2Cq = aq² + K  where K = 2C.

The given equation is of the form \(\frac { dy }{ dx } \)+ py = Q
where P = a and Q = e^x
\(\int { P } dx=\int { a } dx\) = ax
Integrating factor (L.F.) = \(e^{ \int { P } dx }\) = eax
∴ The solution is y \(e^{ \int { P } dx }=\int { Q } e^{ \int { P } dx }dx\)+C
⇒ y.e^ax =\(\int { { e }^{ x }.{ e }^{ ax } } dx+C\)
⇒ y.e^ax = \(\int { e^{ (a+1)x }dx+C } \)
⇒ y.e^ax = \(\frac { { e }^{ (a+1)x } }{ a+1 } \)+C

x²dx - aydx = axdy - y²dy
⇒ \(\int { { x }^{ 2 } } dx+\int { { y }^{ 2 } } dy=a\left[ \int { xdy } +\int { y } dx \right] \)
=\(\frac { { x }^{ 3 } }{ 3 } +\frac { { y }^{ 3 } }{ 3 } \) = a(xy)+C [∵ d(xy) = x.dy + y.dy]

x dy = -y dx
Separating the variables we get
\(\frac { dy }{ y } =-\frac { dx }{ x } \)
Integrating, \(\int { \frac { dy }{ y } } =-\int { \frac { dx }{ x } } \)
⇒ log y = -log x + log C
⇒ log y = log\(\left( \frac { C }{ x } \right) \Rightarrow y=\frac { C }{ x } \) ⇒ xy = C.

Equation of family of rectangular hyperbolas whose asymptotes are the Co-ordinate axis is
xy - c²
Differentiating w.r.t. 'x' we get,
x.\(\frac { dy }{ dx } \)+y(1) = 0
⇒ x\(\left( \frac { dy }{ dx } \right) \)+y(1) = 0 which is the required differential equation.

Given y = mx + \(\frac { a }{ m } \) ...(1)
Differentiating w.r.t. 'x' we get,
\(\frac { dy }{ dx } \) = m(1)+0 ⇒ m = \(\frac { dy }{ dx } \) ...(2)
Substituting (2) in (1) we get,
y = \(\left( \frac { dy }{ dx } \right) x+\frac { a }{ \frac { dy }{ dx } } \Rightarrow y=\frac { \left( \frac { dy }{ dx } \right) ^{ 2 }x+a }{ \left( \frac { dy }{ dx } \right) } \)
⇒ y\(\left( \frac { dy }{ dx } \right) =x\left( \frac { dy }{ dx } \right) ^{ 2 }\)+ a which is the required differential equation.

Taking power 3 both sides, we get
\(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ \frac { 2 }{ 3 } \times 3 }=\left( \frac { d^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }\)
⇒ \(\left[ 1+\left( \frac { dy }{ dx } \right) ^{ 2 } \right] ^{ 2 }=\left( \frac { d^{ 2 }y }{ { dx }^{ 2 } } \right) ^{ 3 }\)
The highest derivative is of order 2 and its power is 3
∴ order is 2 and degree is 3.

Squaring both sides we get,
\(\left[ \sqrt { 1+\left( \frac { dy }{ dx } \right) ^{ 2 } } \right] ^{ 2 }\)= (4x)² ⇒ 1+\(\left( \frac { dy }{ dx } \right) ^{ 2 }\)= 16x²
The highest derivative is of order 1and its power is 2
∴ order is 1 and degree is 2.

The highest derivative if of order 3 and its power is 1
∴ order is 3 and degree is 1.

The highest derivative is of order 1 and its power is 3
∴ order is 1 and degree is 3.