dy + (xy − cos x)dx = 0 is a first order differential equation with degree 1 
since the equation can be rewritten as 
\(\frac{dy}{dx}\) + xy - cos x = 0

The given differential equation is \(3\left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) ={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ \frac { 3 }{ 2 } }\)Squaring both sides, we get 
\(9{ \left( \frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \right) }^{ 2 }={ \left[ 4+{ \left( \frac { dy }{ dx } \right) }^{ 2 } \right] }^{ 3 }\)
In this equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 2.
Therefore, the given differential equation is of order 2 and degree 2.

In the given differential equation, the highest order derivative is \(\frac { { d }^{ 2 }y }{ { dx }^{ 2 } } \) whose power is 1.
Therefore, the given differential equation is of order 2.
The given differential equation is not a polynomial equation in its derivatives and so its degree is not defined.

Separating the variables we get,
\(\frac { dy }{ \sqrt { 1-{ y }^{ 2 } } } \frac { dx }{ \sqrt { 1-{ x }^{ 2 } } } \)
Taking Integration on both sides, we get
\(\int \frac{d y}{\sqrt{1-y^{2}}}=\int \frac{d x}{\sqrt{1-x^{2}}}\)
sin^-1y = sin^-1 x + c

To solve the given differential equation, we make the substitution 3x + y + 4 = z.
Differentiating with respect to x, we get \(\frac { dy }{ dx } =\frac { dz }{ dx } \)-3. 
So the given differential equation becomes \(\frac { dz }{ dx } \) = z²+ 3.
In this equation variables are separable. So, separating the variables and integrating, we get the general solution of the given differential equation as \(\frac { 1 }{ \sqrt { 3 } } { tan }^{ -1 }\left( \frac { 3x+y+4 }{ \sqrt { 3 } } \right) =x+C\)