Maths 1 Marks

4. 1

Solution:

Y = Ax + A⁴

This equation is a linear Differential equation

$=\frac{\text{dy}}{\text{dx}}=\text{A}$

Here the highest order Derivative is y

The Degree of this Derivative is 1

The equation of the famliy of circles touching x-axis at the origin is 

$(\text{x}-0)^{2}+(\text{y}-\text{a})^{2}=\text{a}^{2}$

$\text{x}^{2}+\text{y}^{2}-2\text{ay}=0\ ...(\text{i})$

Here, a is the parameter.

Since, this equation contain only one conatant, we differentiate it only once.

 $2\text{x}+2\text{y}\frac{\text{dy}}{\text{dx}}-2\text{a}\frac{\text{dy}}{\text{dx}}=0$

$\text{a}=\frac{\text{x}+\text{y}(\frac{\text{dy}}{\text{dx}})}{\frac{\text{dy}}{\text{dx}}}\ ...(\text{ii})$

Putting the value of a from (i) in (i), we get

$\text{x}^{2}+\text{y}^{2}=2\text{y}\left\{\frac{\text{x}+\text{y}(\frac{\text{dy}}{\text{dx}})}{\frac{\text{dy}}{\text{dx}}}\right\}$

$(\text{x}^{2}+\text{y}^{2})\frac{\text{dy}}{\text{dx}}=2\text{xy}$

So, this is the differential equation.

Here, order of the diffrential equation is 1.

$\frac{\text{d}^{4}{\text{y}}}{\text{d}\text{x}^{4}}+\text{sin}(\text{y"'})=0$

The highest order derivative present in the differential equation is $\frac{\text{d}^{4}\text{y}}{​​\text{dx}^{4}}$

$\therefore$ its order is 4

The given differential equation is not a polynomial equation in its derivative and so its degree is not defined.

In this differential equation, the order of the highest order derivative is 2.

Clearly, the R.H.S. of the differential equation cannot be expressed as a polynomial in $\frac{\text{d}^2\text{y}}{\text{dx}^2}$ So, its degree is not defined.

The order of the differential equation is 2 and its degree is not defined.

It is a non-linear differential equation, as one of its differential co-efficients, that is, $\Big(\frac{\text{dy}}{\text{dx}}\Big)$ has exponent 2, which is more than 1.