NCERT Solutions for Class 12 Maths Chapter 9 Exercise 9.1 - Differential Equations

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

Answer:

Given function is
$y' + 5y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'}$
Therefore, the order of the given differential equation $y' + 5y = 0$ is 1
Now, the given differential equation is a polynomial equation in its derivatives and its highest power raised to y ' is 1
Therefore, it's a degree is 1.

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

Answer:

Given function is
$\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$
We can rewrite it as
$(s^{'})^4+3s.s^{''} =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $s^{''}$

Therefore, the order of the given differential equation $\left(\frac{\mathrm{d} s}{\mathrm{d} t} \right )^4 + 3s \frac{\mathrm{d}^2 s}{\mathrm{d} t^2} = 0$ is 2
Now, the given differential equation is a polynomial equation in its derivatives and power raised to s '' is 1
Therefore, it's a degree is 1

Question:4 Determine order and degree (if defined) of differential equation.

$\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$

Answer:

Given function is
$\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$
We can rewrite it as
$(y^{''})^2+\cos y^{''} =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, the order of the given differential equation $\left(\frac{d^2y}{dx^2} \right )^2 + \cos\left(\frac{dy}{dx} \right )= 0$ is 2
Now, the given differential equation is not a polynomial equation in its derivatives
Therefore, it's a degree is not defined

Question:5 Determine order and degree (if defined) of differential equation.

$\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$

Answer:

Given function is
$\frac{d^2y}{dx^2} = \cos 3x + \sin 3x$
$\Rightarrow \frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}\left ( \frac{d^2y}{dx^2} \right )$

Therefore, order of given differential equation $\frac{d^2y}{dx^2}- \cos 3x - \sin 3x = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $\frac{d^2y}{dx^2}$ and power raised to $\frac{d^2y}{dx^2}$ is 1
Therefore, it's degree is 1

Question:6 Determine order and degree (if defined) of differential equation $(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$

Answer:

Given function is
$(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'''}$

Therefore, order of given differential equation $(y''')^2 + (y'')^3 + (y')^4 + y^5= 0$ is 3
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'''} , y^{''} \ and \ y^{'}$ and power raised to $y^{'''}$ is 2
Therefore, it's degree is 2

Question:7 Determine order and degree (if defined) of differential equation

$y''' + 2y'' + y' =0$

Answer:

Given function is
$y''' + 2y'' + y' =0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'''}$

Therefore, order of given differential equation $y''' + 2y'' + y' =0$ is 3
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'''} , y^{''} \ and \ y^{'}$ and power raised to $y^{'''}$ is 1
Therefore, it's degree is 1

Question:8 Determine order and degree (if defined) of differential equation

$y' + y = e^x$

Answer:

Given function is
$y' + y = e^x$
$\Rightarrow$ $y^{'}+y-e^x=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{'}$

Therefore, order of given differential equation $y^{'}+y-e^x=0$ is 1
Now, the given differential equation is a polynomial equation in it's dervatives $y^{'}$ and power raised to $y^{'}$ is 1
Therefore, it's degree is 1

Question:9 Determine order and degree (if defined) of differential equation

$y'' + (y')^2 + 2y = 0$

Answer:

Given function is
$y'' + (y')^2 + 2y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + (y')^2 + 2y = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $y^{''} \ and \ y^{'}$ and power raised to $y^{''}$ is 1
Therefore, it's degree is 1

Question:10 Determine order and degree (if defined) of differential equation

$y'' + 2y' + \sin y = 0$

Answer:

Given function is
$y'' + 2y' + \sin y = 0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $y'' + 2y' + \sin y = 0$ is 2
Now, the given differential equation is a polynomial equation in it's dervatives $y^{''} \ and \ y^{'}$ and power raised to $y^{''}$ is 1
Therefore, it's degree is 1

Question:11 The degree of the differential equation $\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is

(A) 3

(B) 2

(C) 1

(D) not defined

Answer:

Given function is
$\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$
We can rewrite it as
$(y^{''})^3+(y^{'})^2+\sin y^{'}+1=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $\left(\frac{d^2y}{dx^2} \right )^3 + \left(\frac{dy}{dx} \right )^2 + \sin\left(\frac{dy}{dx}\right ) + 1= 0$ is 2
Now, the given differential equation is a not polynomial equation in it's dervatives
Therefore, it's degree is not defined

Therefore, answer is (D)

Question:12 The order of the differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is

(A) 2

(B) 1

(C) 0

(D) Not Defined

Answer:

Given function is
$2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$
We can rewrite it as
$2x.y^{''}-3y^{'}+y=0$
Now, it is clear from the above that, the highest order derivative present in differential equation is $y^{''}$

Therefore, order of given differential equation $2x^2\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + y = 0$ is 2

Therefore, answer is (A)