Differential Equations: Chapter 9 Links
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9.1 | 9.3 | 9.4 | 9.5 | Misc. | MCQs
Differential Equations Exercise 9.2 introduces the concept of the solution of a differential equation. In this exercise, you will learn how to verify whether a given explicit or implicit function satisfies a differential equation. You will also understand the difference between a general solution and a particular solution, which are fundamental concepts in this chapter.
\displaystyle F(x,y,C)=0\qquad\Longrightarrow\qquad \text{General Solution}
A solution of a differential equation is a function that satisfies the given equation when its derivatives are substituted into it. A general solution contains one or more arbitrary constants, whereas a particular solution is obtained by assigning specific values to these constants, usually with the help of given initial or boundary conditions.
In this exercise, we will verify whether the given functions satisfy the corresponding differential equations and learn to distinguish between general and particular solutions. These concepts form the basis for solving differential equations in the remaining exercises of this chapter and are important for both CBSE board examinations and CUET.
Key Concepts
Before solving Exercise 9.2, it is important to understand what is meant by a solution of a differential equation. In this exercise, you will verify whether a given function (explicit or implicit) satisfies a differential equation and learn the difference between a general solution and a particular solution.
1. Solution of a Differential Equation
A function is called a solution of a differential equation if, after finding its required derivatives and substituting them into the equation, the left-hand side becomes equal to the right-hand side.
To verify a solution, simply differentiate the given function (if required) and substitute the values into the differential equation.
2. General Solution
A general solution contains one or more arbitrary constants such as C, C_1 or C_2. It represents an entire family of solutions of the differential equation.
For example, \displaystyle y=x^2+C is the general solution of \displaystyle \frac{dy}{dx}=2x.
3. Particular Solution
A particular solution is obtained by assigning specific values to the arbitrary constants in the general solution, usually using the given initial or boundary conditions.
For example, if \displaystyle y=x^2+C and y=1 when x=0, then C=1. Hence, the particular solution is \displaystyle y=x^2+1.
4. Explicit and Implicit Solutions
An explicit solution expresses y directly in terms of x, whereas an implicit solution contains both variables in a single equation.
- Explicit: \displaystyle y=x^2+1
- Implicit: \displaystyle x^2+y^2=1
5. Important Points
- A function is a solution only if it satisfies the given differential equation after substitution.
- A general solution always contains arbitrary constant(s).
- A particular solution does not contain any arbitrary constant.
- Both explicit and implicit functions can be solutions of a differential equation.
- Always simplify both sides of the differential equation before concluding whether the given function is a solution.
Before You Begin… Exercise 9.2 focuses on verifying whether a given function is a solution of a differential equation. First find the required derivative(s) of the given function. Then substitute these values into the differential equation and simplify both sides. If the left-hand side is equal to the right-hand side, the given function is a solution of the differential equation.
Let us now solve all the NCERT questions step by step of Differential Equations 9.2.
Question 1: Particular Solution of Differential Equations
1. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y=e^x+1\quad:\quad y''-y'=0
Solution
Think First… To verify a solution, first find the required derivative(s) of the given function. Then substitute them into the differential equation. If the left-hand side becomes equal to the right-hand side, the given function is a solution of the differential equation.
The given function is
\displaystyle y=e^x+1
Differentiating with respect to x, we get
\displaystyle y'=e^x
Differentiating again, we obtain
\displaystyle y''=e^x
Substituting the values of y' and y'' into the given differential equation, we get
\displaystyle \text{L.H.S.}=y''-y'=e^x-e^x=0=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Question 2: Differential Equations 9.2
2. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y=x^2+2x+C\quad:\quad y'-2x-2=0
Solution
Think First… While differentiating a general solution, remember that an arbitrary constant such as C is treated as a constant. Therefore, its derivative is 0.
The given function is
\displaystyle y=x^2+2x+C
Differentiating with respect to x, we get
\displaystyle y'=2x+2
Substituting the values of y' into the given differential equation, we get
\displaystyle \text{L.H.S.}=y'-2x-2=(2x+2)-2x-2=0=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
You may already be following Maths Better for the NCERT Solutions of the previous Class 12 Maths chapters, namely:
- Matrices
- Determinants
- Relations and Functions
- Inverse Trigonometric Functions
- Continuity and Differentiability
- Application of Derivatives
- Integrals
- Application of Integrals
Similarly, Differential Equations Exercise 9.2 helps you understand the concepts of a solution of a differential equation, general and particular solutions, and how to verify whether a given function satisfies a differential equation. A clear understanding of these concepts will make it easier to solve differential equations in the upcoming exercises. Now, let’s proceed to the next question.
Question 3: Differential Equations
3. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y=\cos x+C\quad:\quad y'+\sin x=0
Solution
Think First… Differentiate the given function carefully. Remember that the derivative of \cos x is -\sin x and the derivative of the arbitrary constant C is 0.
The given function is
\displaystyle y=\cos x+C
Differentiating with respect to x, we get
\displaystyle y'=-\sin x
Substituting the value of y' into the given differential equation, we get
\displaystyle \text{L.H.S.}=y'+\sin x=-\sin x+\sin x=0=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Question 4: Differential Equations 9.2
4. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y=\sqrt{1+x^2}\quad:\quad y'=\frac{xy}{1+x^2}
Solution
Think First… When the given function contains a square root, first differentiate it using the chain rule. Then substitute both y and y' into the given differential equation and simplify.
The given function is
\displaystyle y=\sqrt{1+x^2}=(1+x^2)^{1/2}
Differentiating with respect to x, we get
\displaystyle y'=\frac12(1+x^2)^{-1/2}\cdot2x=\frac{x}{\sqrt{1+x^2}}
Since \displaystyle y=\sqrt{1+x^2}, therefore
\displaystyle y'=\frac{x}{y}
Also,
\displaystyle \frac{xy}{1+x^2}=\frac{xy}{y^2}=\frac{x}{y}
Hence,
\displaystyle \text{L.H.S.}=y'=\frac{x}{y}=\frac{xy}{1+x^2}=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Question 5: Differential Equations 9.2
5. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y=Ax\quad:\quad xy'=y\;(x\ne0)
Solution
Think First… Here, A is an arbitrary constant. While differentiating, treat A as a constant and then substitute the values into the given differential equation.
The given function is
\displaystyle y=Ax
Differentiating with respect to x, we get
\displaystyle y'=A
Substituting the values of y and y' into the given differential equation, we get
\displaystyle \text{L.H.S.}=xy'=x(A)=Ax
\displaystyle \text{R.H.S.}=y=Ax
Thus,
\displaystyle \text{L.H.S.}=Ax=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
By now, you have already completed the NCERT Solutions for Matrices, Determinants, Relations and Functions, Inverse Trigonometric Functions, Continuity and Differentiability, Application of Derivatives, Integrals and Application of Integrals. Likewise, every exercise of Differential Equations is explained here with detailed, step-by-step solutions to help you build strong concepts and prepare confidently for your CBSE board examinations.
Many of these questions are also available in video format on my YouTube Channel, @MathsBetter, where each solution is explained in a simple and student-friendly manner. Now, let’s proceed to the next question.
Question 6: Exercise 9.2
6. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y=x\sin x\;:\; xy'=y+x\sqrt{x^2-y^2}\;\;(x\ne0,\;x>y\;\mathrm{or}\;x\lt-y)
Solution
Think First… Use the product rule to differentiate the given function. Then simplify the expression under the square root using the identity \displaystyle 1-\sin^2x=\cos^2x. Finally, use the given condition to remove the modulus and verify that the left-hand side is equal to the right-hand side.
The given function is
\displaystyle y=x\sin x
Differentiating with respect to x using the product rule, we get
\displaystyle y'=x\cos x+\sin x
Therefore,
\displaystyle xy'=x(x\cos x+\sin x)=x^2\cos x+x\sin x
Also,
\displaystyle x^2-y^2=x^2-x^2\sin^2x=x^2(1-\sin^2x)=x^2\cos^2x
Since x>y or x\lt-y, we have
\displaystyle \sqrt{x^2-y^2}=\sqrt{x^2\cos^2x}=|x\cos x|=x\cos x
Hence,
\displaystyle y+x\sqrt{x^2-y^2}=x\sin x+x(x\cos x)=x\sin x+x^2\cos x
Thus,
\displaystyle \text{L.H.S.}=xy'=x^2\cos x+x\sin x=y+x\sqrt{x^2-y^2}=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Think First… Use the product rule to differentiate the given function. Then simplify the expression under the square root using the identity \displaystyle 1-\sin^2x=\cos^2x. Finally, use the given condition to remove the modulus and verify that the left-hand side is equal to the right-hand side.
Question 7: Differential Equations
7. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle xy=\log y+C\;:\; y'=\frac{y^2}{1-xy}\;\;(xy\ne1)
Solution
Think First… Since the given equation is an implicit function, differentiate both sides with respect to x. Use the product rule for xy and the chain rule for \log y, then solve for y'.
The given function is
\displaystyle xy=\log y+C
Differentiating both sides with respect to x, we get
\displaystyle x\frac{dy}{dx}+y=\frac1y\frac{dy}{dx}
Rearranging the terms,
We get, \displaystyle x\frac{dy}{dx}-\frac1y\frac{dy}{dx}=-y
\displaystyle \frac{dy}{dx}\left(x-\frac1y\right)=-y
\displaystyle \frac{dy}{dx}=\frac{-y}{x-\frac1y}=\frac{-y^2}{xy-1}=\frac{y^2}{1-xy}
This is exactly the right-hand side of the given differential equation.
Hence,
\displaystyle \text{L.H.S.}=y'=\frac{y^2}{1-xy}=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Question 8: Differential Equations 9.2
8. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y-\cos y=x\;:\; (y\sin y+\cos y+x)\,y'=y
Solution
Think First… The given equation is an implicit function. Differentiate both sides with respect to x. While differentiating \cos y, use the chain rule, then substitute the given relation to simplify the result.
The given function is
\displaystyle y-\cos y=x
Differentiating both sides with respect to x, we get
\displaystyle y'+\sin y\,y'=1
\displaystyle y'(1+\sin y)=1
Also, from the given function,
\displaystyle x=y-\cos y
Therefore,
\displaystyle y\sin y+\cos y+x=y\sin y+\cos y+y-\cos y=y(1+\sin y)
Hence,
\displaystyle (y\sin y+\cos y+x)\,y'=y(1+\sin y)\cdot\frac1{1+\sin y}=y
Thus,
\displaystyle \text{L.H.S.}=y=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Question 9: Differential Equations
9. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle x+y=\tan^{-1}y\quad:\quad y^2y'+y^2+1=0
Solution
Think First… The given equation is an implicit function. Differentiate both sides with respect to x. While differentiating \tan^{-1}y, use the chain rule, then simplify the resulting expression.
The given function is
\displaystyle x+y=\tan^{-1}y
Differentiating both sides with respect to x, we get
\displaystyle 1+y'=\frac{1}{1+y^2}\,y'
Multiplying both sides by \displaystyle 1+y^2, we obtain
\displaystyle (1+y^2)(1+y')=y'
Expanding and simplifying,
\displaystyle 1+y^2+y'+y^2y'=y'
\displaystyle y^2y'+y^2+1=0
This is exactly the given differential equation.
Hence,
\displaystyle \text{L.H.S.}=y^2y'+y^2+1=0=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Question 10: Differential Equations Exercise 9.2
10. Verify that the given function is a solution of the corresponding differential equation:
\displaystyle y=\sqrt{a^2-x^2},\;x\in(-a,a)\;:\; x+y\frac{dy}{dx}=0\;(y\ne0)
Solution
Think First… Differentiate the given function using the chain rule. Then substitute the values of y and \displaystyle \frac{dy}{dx} into the differential equation and simplify.
The given function is
\displaystyle y=\sqrt{a^2-x^2}=(a^2-x^2)^{1/2}
Differentiating with respect to x, we get
\displaystyle \frac{dy}{dx}=\frac12(a^2-x^2)^{-1/2}(-2x)=\frac{-x}{\sqrt{a^2-x^2}}
Since \displaystyle y=\sqrt{a^2-x^2}, therefore
\displaystyle \frac{dy}{dx}=-\frac{x}{y}
Substituting the values of y and \displaystyle \frac{dy}{dx} into the given differential equation, we get
\displaystyle \text{L.H.S.}=x+y\frac{dy}{dx}=x+y\left(-\frac{x}{y}\right)=x-x=0=\text{R.H.S.}
Therefore, the given function is a solution of the given differential equation.
Remember! A general solution always contains one or more arbitrary constants such as C, C₁ or C₂. A particular solution is obtained by assigning specific values to these constants. Therefore, a particular solution contains no arbitrary constants.
Question 11: Differential Equations 9.2 – MCQ
11. The number of arbitrary constants in the general solution of a fourth order differential equation is
- (A) 1
- (B) 2
- (C) 3
- (D) 4
Solution
Think First… The general solution of an nth-order differential equation contains exactly n arbitrary constants. The number of arbitrary constants is equal to the order of the differential equation.
The given differential equation is of fourth order.
Therefore, its general solution contains 4 arbitrary constants.
✅ Hence, the correct answer is (D)
Quick Tip! The general solution of an nth-order differential equation contains exactly n arbitrary constants. Since a particular solution is obtained by fixing the values of these constants, it contains zero arbitrary constants.
Question 12: Differential Equations 9.2 – MCQ
12. The number of arbitrary constants in the particular solution of a third order differential equation is
- (A) 3
- (B) 2
- (C) 1
- (D) 0
Solution
Think First… A particular solution is obtained by assigning specific values to all the arbitrary constants in the general solution. Therefore, a particular solution contains no arbitrary constants.
A particular solution is obtained from the general solution after replacing all the arbitrary constants with specific values.
Hence, a particular solution does not contain any arbitrary constant.
✅ Hence, the correct answer is (D)
Observation! Do not confuse numerical constants such as 2, 3, −5 with arbitrary constants such as C, C₁, C₂. A particular solution may contain numerical constants, but it contains no arbitrary constants.
Common Mistakes to Avoid
- Not differentiating the given function correctly: Before substituting into the differential equation, carefully find the required derivative(s). A small mistake in differentiation can lead to an incorrect conclusion.
- Forgetting that the derivative of an arbitrary constant is zero: In a general solution, constants such as C or A are treated as constants while differentiating. Their derivatives are always 0.
- Not using implicit differentiation: When the given function is in implicit form, such as F(x,y)=0, differentiate both sides with respect to x. Remember that y is a function of x.
- Ignoring the chain rule: While differentiating expressions like \log y, \sin y or \tan^{-1}y, always multiply by \displaystyle \frac{dy}{dx}.
- Forgetting the product rule: Expressions such as xy or x\sin x must be differentiated using the product rule.
- Not simplifying before comparing both sides: After substitution, simplify the left-hand side and right-hand side completely before deciding whether the given function is a solution.
- Confusing general and particular solutions: A general solution contains arbitrary constants, whereas a particular solution does not contain any arbitrary constant.
- Ignoring the given conditions: Conditions such as x\ne0, y\ne0 or xy\ne1 are necessary for the validity of the given differential equation and should not be overlooked.
Continue Learning
Exercise 9.2 introduces the concept of the solution of a differential equation. In this exercise, you verify whether a given function satisfies a differential equation by finding its derivative(s) and substituting them into the given equation. You also learn the difference between a general solution and a particular solution.
Before starting Exercise 9.3, revise the following concepts:
- What is meant by a solution of a differential equation.
- How to verify a solution by differentiating the given function and substituting it into the differential equation.
- The difference between an explicit and an implicit solution.
- How to apply the product rule and the chain rule while differentiating implicit functions.
- The difference between a general solution and a particular solution.
- That the derivative of an arbitrary constant is always zero.
- How to simplify both sides of the differential equation before concluding whether the given function is a solution.
- The importance of the given conditions such as x\ne0 or y\ne0 while verifying a solution.
Explore More
A clear understanding of the solution of a differential equation is essential before learning how to solve differential equations. Whenever you are given a function, first find the required derivative(s) carefully and substitute them into the given differential equation. If the left-hand side becomes equal to the right-hand side, the function is a solution of the differential equation. Also remember that the derivative of an arbitrary constant is always zero, and implicit functions should be differentiated using the product rule and the chain rule wherever required. With regular practice, verifying solutions will become much easier and prepare you for solving differential equations in the remaining exercises of this chapter.
All the best and keep learning 👍
