# NCERT Solutions

**Differential Equation Exercise 9.1**

**Determine order and degree (if defined) of differential equations given in Questions 1 to 10:**

**Question 1.** $\frac{{{d}^{4}}y}{d{{x}^{4}}}+\sin \left( y”’ \right)=0$

### Answer:

Given: $\frac{{{d}^{4}}y}{d{{x}^{4}}}+\sin \left( y”’ \right)=0$

The highest order derivative present in the differential equation is $\frac{{{d}^{4}}y}{d{{x}^{4}}}$ and its order is 4.

The given differential equation is not a polynomial equation in derivatives as the term $\sin \left( y”’ \right)$ is a T-function of derivative $y”’.$ Therefore the degree is not defined.

Hence, order is 4 and degree is not defined.

**Question 2.** $y’+5y=0$

### Answer:

Given: $y’+5y=0$

The highest order derivative present in the differential equation is $y’=\frac{dy}{dx}$ and its order is 1.

The given differential equation is a polynomial equation in derivative$y’$ and the highest power raised to highest order derivative $y’$ is one, so its degree is 1.

Hence, order is 1 and degree is 1.

**Question 3. **${{\left( \frac{ds}{dt} \right)}^{4}}+3s\frac{{{d}^{2}}s}{d{{t}^{2}}}=0$

### Answer:

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

The highest order derivative present in the differential equation is $\frac{{{d}^{2}}s}{d{{t}^{2}}}$ and its order is 2. The given differential equation is a polynomial equation in derivatives and the highest power raised to highest order derivative $\frac{{{d}^{2}}s}{d{{t}^{2}}}$ is one, so its degree is 1.

Hence, order is 2 and degree is 1.

**Question 4.** ${{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{2}}+\cos \frac{dy}{dx}=0$

### Answer:

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

The highest order derivative present in the differential equation is $\frac{{{d}^{2}}y}{d{{x}^{2}}}$ and its order is 2.

The given differential equation is not a polynomial equation in derivatives as the term $\cos \frac{dy}{dx}$ is a T-function of derivative $\frac{dy}{dx}.$ Therefore the degree is not defined.

Hence, order is 2 and degree is not defined.

**Question 5.** $\frac{{{d}^{2}}y}{d{{x}^{2}}}=\cos 3x+\sin 3x$

### Answer:

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

The highest order derivative present in the differential equation is $\frac{{{d}^{2}}y}{d{{x}^{2}}}$ and its order is 2.

The given differential equation is a polynomial equation in derivatives and the highest power raised to highest order $\frac{{{d}^{2}}y}{d{{x}^{2}}}={{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{1}}$ is one, so its degree is 1.

Hence, order is 2 and degree is 1.

**Question 6.** ${{\left( y”’ \right)}^{2}}+{{\left( y” \right)}^{3}}+{{\left( y’ \right)}^{4}}+{{y}^{5}}=0$

### Answer:

Given: ${{\left( y”’ \right)}^{2}}+{{\left( y” \right)}^{3}}+{{\left( y’ \right)}^{4}}+{{y}^{5}}=0$

The highest order derivative present in the differential equation is $y”’$ and its order is 3.

The given differential equation is a polynomial equation in derivatives and the highest power raised to highest order $y”’$ is two, so its degree is 2.

Hence, order is 3 and degree is 2.

**Question 7. **$y”’+2y”+y’=0$

### Answer:

Given: $y”’+2y”+y’=0$

The highest order derivative present in the differential equation is $y”’$ and its order is 3.

The given differential equation is a polynomial equation in derivatives $y”’,y”$ and $y’$ and the highest power raised to highest order $y”’$ is two, so its degree is 1.

Hence, order is 3 and degree is 1.

**Question 8. **$y’+y={{e}^{x}}$

### Answer:

Given: $y’+y={{e}^{x}}$

The highest order derivative present in the differential equation is $y’$ and its order is

The given differential equation is a polynomial equation in derivative $y’$. It may be noted that ${{e}^{x}}$ is an exponential function and not a polynomial function but is not an exponential function of derivatives and the highest power raised to highest order derivative $y’$ is one so its degree is one.

Hence, order is 1 and degree is 1.

**Question 9.** $y”+{{\left( y’ \right)}^{2}}+2y=0$

### Answer:

Given: $y”+{{\left( y’ \right)}^{2}}+2y=0$

The highest order derivative present in the differential equation is $y”$ and its order is

The given differential equation is a polynomial equation in derivatives $y”$ and $y’$ and the highest power raised to highest order $y”$ is one, so its degree is 1.

Hence, order is 2 and degree is 1.

**Question 10.** $y”+2y’+\sin y=0$

### Answer:

Given: $y”+2y’+\sin y=0$

The highest order derivative present in the differential equation is $y”$ and its order is

The given differential equation is a polynomial equation in derivative $y”$and $y’$. It may be noted that $\sin y$ is not a polynomial function of $y$, it is a T-function of $y$ but is not a T-function of derivatives and the highest power raised to highest order derivative $y”$ is one so its degree is one.

Hence, order is 2 and degree is 1.

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

**(A) 3 **

**(B) 2 **

**(C) 1 **

**(D) Not defined**

### Answer:

Given: ${{\left( \frac{{{d}^{2}}y}{d{{x}^{2}}} \right)}^{3}}+{{\left( \frac{dy}{dx} \right)}^{2}}+\sin \left( \frac{dy}{dx} \right)+1=0$ ……….(i)

This equation is not a polynomial in derivatives as $\sin \left( \frac{dy}{dx} \right)$ is a T-function of derivative $\frac{dy}{dx}.$

Therefore, degree of given equation is not defined.

Hence, option (D) is correct.

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

**(A) 2 **

**(B) 1 **

**(C) 0 **

**(D) Not defined**

### Answer:

Given: $2{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}-3\frac{dy}{dx}+y=0$

The highest order derivative present in the differential equation is $\frac{{{d}^{2}}y}{d{{x}^{2}}}$ and its order is 2.

Therefore, option (A) is correct.