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

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

Ans: The given differential equation is $\dfrac{{{d^4}y}}{{d{x^4}}} + \sin \left( {y'''} \right) = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$ \Rightarrow y'''' + \sin \left( {y'''} \right) = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y''''$, that is, $\dfrac{{{d^4}y}}{{d{x^4}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is found to be four.

Now, it is seen that, in its derivatives, the above differential equation is not a polynomial equation because of the trigonometric term involved. As a result, its degree is unknown.

Therefore, the required order of the given differential equation $\dfrac{{{d^4}y}}{{d{x^4}}} + \sin \left( {y'''} \right) = 0$ is found to be four and the degree is unknown. 

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

Ans: The given differential equation is $y' + 5y = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$\Rightarrow y' + 5y = 0$

$\Rightarrow \dfrac{{dy}}{{dx}} + 5y = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y'$, that is, $\dfrac{{dy}}{{dx}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is found to be one.

Now, it is seen that the given differential equation is a polynomial equation in $y'$, thus, the highest power that is raised to $y'$ is 1. 

Hence, the degree of the given differential equation is found to be one.

Therefore, the required order of the given differential equation $y' + 5y = 0$ is obtained as one and the degree is also found to be one. 

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

Ans: The given differential equation is ${\left( {\dfrac{{ds}}{{dt}}} \right)^4} + 3s\dfrac{{{d^2}s}}{{d{t^2}}} = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$ \Rightarrow {\left( {\dfrac{{ds}}{{dt}}} \right)^4} + 3s\dfrac{{{d^2}s}}{{d{t^2}}} = 0$

$\Rightarrow {\left( {s'} \right)^4} + 3s\left( {s''} \right) = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $s''$, that is, $\dfrac{{{d^2}s}}{{d{t^2}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is obtained as two.

Now, it is seen that, the given differential equation is a polynomial equation in $\dfrac{{{d^2}s}}{{d{t^2}}}$ and $\dfrac{{ds}}{{dt}}$, thus, the highest power that is raised to $\dfrac{{{d^2}s}}{{d{t^2}}}$ is 1. 

Hence, the degree of the given differential equation is found to be one.

Therefore, the required order of the given differential equation ${\left( {\dfrac{{ds}}{{dt}}} \right)^4} + 3s\dfrac{{{d^2}s}}{{d{t^2}}} = 0$ is obtained as two and the degree is also found to be one. 

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

Ans: The given differential equation is ${\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^2} + \cos \left( {\dfrac{{dy}}{{dx}}} \right) = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$ \Rightarrow {\left( {y''} \right)^2} + \cos \left( {y'} \right) = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y''$, that is, $\dfrac{{{d^2}y}}{{d{x^2}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is obtained as two.

Now, it is seen that, in its derivatives, the above differential equation is not a polynomial equation because of the trigonometric term involved. As a result, its degree is unknown.

Therefore, the required order of the given differential equation ${\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^2} + \cos \left( {\dfrac{{dy}}{{dx}}} \right) = 0$ is found as two and the degree is unknown. 

5. Determine order and degree (if defined) of differential equation $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\cos 3x+\sin 3x$.

Ans: The given differential equation is $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\cos 3x+\sin 3x$.

It is known that the differential expression $\dfrac{dy}{dx}$ can be expressed as $y'$, the differential expression $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\cos 3x+\sin 3x$

$\Rightarrow y''=\cos 3x+\sin 3x$

$\Rightarrow y''-\left( \cos 3x+\sin 3x \right)=0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y''$, that is, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is two.

Now, it is seen that the given differential equation is a polynomial equation in $y''$, thus, the highest power that is raised to $y''$ is 1.

Hence, the degree of the given differential equation is obtained one.

Therefore, the order of the differential equation $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}=\cos 3x+\sin 3x$ is found as two and the degree is found to be one. 

6. Determine order and degree (if defined) of differential equation ${\left( {y'''} \right)^2} + {\left( {y''} \right)^3} + {\left( {y'} \right)^4} + {y^5} = 0$.

Ans: The given differential equation is ${\left( {y'''} \right)^2} + {\left( {y''} \right)^3} + {\left( {y'} \right)^4} + {y^5} = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$\Rightarrow {\left( {y'''} \right)^2} + {\left( {y''} \right)^3} + {\left( {y'} \right)^4} + {y^5} = 0$

$\Rightarrow {\left( {\dfrac{{{d^3}y}}{{d{x^3}}}} \right)^2} + {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\dfrac{{dy}}{{dx}}} \right)^4} + {y^5} = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y'''$, that is, $\dfrac{{{d^3}y}}{{d{x^3}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is obtained as three.

Now, it is seen that the given differential equation is a polynomial equation in $y'''$,$y''$ and $y'$, thus, the highest power that is raised to $y'''$ is 2. 

Hence, the degree of the given differential equation is obtained as two.

Therefore, the required order of the given differential equation ${\left( {y'''} \right)^2} + {\left( {y''} \right)^3} + {\left( {y'} \right)^4} + {y^5} = 0$ is found as three and the degree is also found to be two.

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

Ans: The given differential equation is $y''' + 2y'' + y' = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

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

$\Rightarrow \dfrac{{{d^3}y}}{{d{x^3}}} + 2\dfrac{{{d^2}y}}{{d{x^2}}} + \dfrac{{dy}}{{dx}} = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y'''$, that is, $\dfrac{{{d^3}y}}{{d{x^3}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is obtained as three.

Now, it is seen that the given differential equation is a polynomial equation in $y'''$,$y''$ and $y'$, thus, the highest power that is raised to $y'''$ is 1. 

Hence, the degree of the given differential equation is found to be one.

Therefore, the required order of the given differential equation $y''' + 2y'' + y' = 0$ is three and the degree is also found to be one. 

8. Determine order and degree (if defined) of differential equation $y' + y = e'$.

Ans: The given differential equation is $y' + y = e'$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$\Rightarrow y' + y = e'$

$\Rightarrow y' + y - e' = 0$

$\Rightarrow \left( {\dfrac{{dy}}{{dx}}} \right) + y - e' = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y'$, that is, $\dfrac{{dy}}{{dx}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is obtained asone.

Now, it is seen that the given differential equation is a polynomial equation in $y'$, thus, the highest power that is raised to $y'$ is 1. 

Hence, the degree of the given differential equation is found to be one.

Therefore, the required order of the given differential equation $y' + y = e'$ is obtained as one and the degree is also found to be one. 

9. Determine order and degree (if defined) of differential equation $y''+{{\left( y' \right)}^{2}}+2y=0$.

Ans: The given differential equation is $y''+{{\left( y' \right)}^{2}}+2y=0$.

It is known that the differential expression $\dfrac{dy}{dx}$ can be expressed as $y'$, the differential expression $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$\Rightarrow y''+{{\left( y' \right)}^{2}}+2y=0$

$\Rightarrow \dfrac{{{d}^{2}}y}{d{{x}^{2}}}+{{\left( \dfrac{dy}{dx} \right)}^{2}}+2y=0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y''$, that is, $\dfrac{{{d}^{2}}y}{d{{x}^{2}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is found to be two.

Now, it is seen that the given differential equation is a polynomial equation in $y''$ and $y'$,thus, the highest power that is raised to $y''$ is 1.

Hence, the degree of the given differential equation is found to be one.

Therefore, the order of the differential equation $y''+{{\left( y' \right)}^{2}}+2y=0$ is obtained as two and the degree is also found to be one. 

10. Determine order and degree (if defined) of differential equation $y'' + 2y' + \sin y = 0$.

Ans: The given differential equation is $y'' + 2y' + \sin y = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

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

$\Rightarrow \dfrac{{{d^2}y}}{{d{x^2}}} + 2\left( {\dfrac{{dy}}{{dx}}} \right) + \sin y = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y''$, that is, $\dfrac{{{d^2}y}}{{d{x^2}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is found to be two.

Now, it is seen that the given differential equation is a polynomial equation in $y''$ and $y'$, thus, the highest power that is raised to $y''$ is 1. 

Hence, the degree of the given differential equation is found to be one.

Therefore, the required order of the given differential equation $y'' + 2y' + \sin y = 0$ is obtained as two and the degree is found to be one. 

11. Determine degree (if defined) of differential equation ${\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) + 1 = 0$ is 

(A) 3

(B) 2

(C) 1

(D) Not defined

Ans: The given differential equation is ${\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) + 1 = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$\Rightarrow {\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) + 1 = 0$

$\Rightarrow {\left( {y''} \right)^3} + {\left( {y'} \right)^2} + \sin \left( {y'} \right) + 1 = 0$

It is seen that, in its derivatives, the above differential equation is not a polynomial equation because of the trigonometric term involved. As a result, its degree is not defined or unknown.

Therefore, the degree of the differential equation ${\left( {\dfrac{{{d^2}y}}{{d{x^2}}}} \right)^3} + {\left( {\dfrac{{dy}}{{dx}}} \right)^2} + \sin \left( {\dfrac{{dy}}{{dx}}} \right) + 1 = 0$ is not defined or unknown. Hence, the correct answer is found to be (D).

12. Determine order of the differential equation $2{x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + 3\dfrac{{dy}}{{dx}} + y = 0$

(A) 2

(B) 1

(C) 0

(D) Not defined

Ans: The given differential equation is $2{x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + 3\dfrac{{dy}}{{dx}} + y = 0$.

It is known that the differential expression $\dfrac{{dy}}{{dx}}$ can be expressed as $y'$, the differential expression $\dfrac{{{d^2}y}}{{d{x^2}}}$  is $y''$, and so on.

Thus, above equation can also be rearranged and written as follows,

$\Rightarrow 2{x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + 3\dfrac{{dy}}{{dx}} + y = 0$

$\Rightarrow 2{x^2}\left( {y''} \right) + 3\left( {y'} \right) + y = 0$

It can be observed from the above equation that the highest order derivative that is present in the differential equation is $y''$, that is, $\dfrac{{{d^2}y}}{{d{x^2}}}$.

The order of the differential equation is determined as the highest order of the derivative present in that differential equation.

Hence, the order of the given differential equation is found to be two.

Therefore, the required order of the given differential equation $2{x^2}\dfrac{{{d^2}y}}{{d{x^2}}} + 3\dfrac{{dy}}{{dx}} + y = 0$ is found to be 2. Hence, option (A) is found to be the correct answer.

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

Opting for the NCERT solutions for Ex 9.1 Class 12 Maths is considered as the best option for the CBSE students when it comes to exam preparation. This chapter consists of many exercises. Out of which we have provided the Exercise 9.1 Class 12 Maths NCERT solutions on this page in PDF format. You can download this solution as per your convenience or you can study it directly from our website/ app online.

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