Differential Equations

Class 12 NCERT

NCERT

1   Determine order and degree (if defined) of differential equation$\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4}+\sin(y''')=0$

Solution :

$\dfrac{\mathrm{d}^4y}{\mathrm{d}x^4}+\sin(y''')=0$$\\$ $\implies y''''+\sin(y''')=0$$\\$ The highest order derivative present in the differential equation is $y''''.$ Therefore, its order is four.$\\$ The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.

2   Determine order and degree (if defined) of differential equation $y '+ 5y = 0$

Solution :

The given differential equation is: $y '+5y = 0$$\\$ The highest order derivative present in the differential equation is $y’.$ Therefore, its order is one. It is a polynomial equation in $y'.$ The highest power raised to $y'$ is $1.$ Hence, its degree is one.

3   Determine order and degree (if defined) of differential equation$(\dfrac{\mathrm{d}s}{\mathrm{d}t})^4+3s\dfrac{\mathrm{d^2}s}{\mathrm{d}t^2}=0$

Solution :

$(\dfrac{\mathrm{d}s}{\mathrm{d}t})^4+3s\dfrac{\mathrm{d^2}s}{\mathrm{d}t^2}=0$$\\$ The highest order derivative present in the given differential equation is$\dfrac{\mathrm{d^2}s}{\mathrm{d}t^2}.$Therefore, its order is two.$\\$ It is a polynomial equation in $\dfrac{\mathrm{d^2}s}{\mathrm{d}t^2} and \dfrac{\mathrm{d}s}{\mathrm{d}t}$The power raised to $ \dfrac{\mathrm{d^2}s}{\mathrm{d}t^2}$ is $1$. Hence , its degree is one.

4   Determine order and degree (if defined) of differential equation $(\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2})^2+\cos(\dfrac{\mathrm{d}y}{\mathrm{d}x})=0$

Solution :

$(\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2})^2+\cos(\dfrac{\mathrm{d}y}{\mathrm{d}x})=0$$\\$ The highest order derivative present in the given differential equation is$\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2}.$Therefore, its order is 2.$\\$ The given differential equation is not a polynomial equation in its derivatives.$\\$ Hence, its degree is not defined.

5   Determine order and degree (if defined) of differential equation $(\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2})^2=\cos3x+\sin3x$

Solution :

$(\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2})^2=\cos3x+\sin3x$$\\$ $\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2}=\cos3x+\sin3x$$\\$ The highest order derivative present in the differential equation is $\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2}$ Therefore, its order is two. It is a polynomial equation $\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2}$ in and the power raised to $\dfrac{\mathrm{d^2}y}{\mathrm{d}x^2}$ is $1$.$\\$ Hence, its degree is one.