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Q1 Determine order and degree(if defined) of differential equation \begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0\end{align} Ans: \begin{align} \frac{d^4y}{dx^4}\;+\;\sin(y^m)\;=0 \end{align}
\begin{align} \Rightarrow y^{m\;'}+\;\sin(y^m)\;=0 \end{align}
The highest order derivative present in the differential equation is ym '. Therefore, its order is four.
The given differential equation is not a polynomial equation in its derivatives. Hence, its degree is not defined.
Q2 Determine order and degree(if defined) of differential equation y' + 5y = 0 Ans: The given differential equation is:
y' + 5y = 0
The highest order derivative present in the differential equation isy'. 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.
Q3 Determine order and degree(if defined) of differential equation \begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align} Ans: \begin{align}\left(\frac{ds}{dt}\right)^4\;+\;3s\frac{d^2s}{dt^2}\;=\;0\end{align}
The highest order derivative present in the given differential equation is\begin{align}\frac{d^2s}{dt^2}.\end{align}
Therefore, its order is two. It is a polynomial equation in
\begin{align}\frac{d^2s}{dt^2} and \frac{ds}{dt}.\end{align}
The power raised to is 1. \begin{align} \frac{d^2s}{dt^2} \end{align}
Q4 Determine order and degree(if defined) of differential equation \begin{align}\left(\frac{d^2y}{dx^2}\right)^2\;+\;cos\left(\frac{dy}{dx}\right)\;=\;0\end{align} Ans: \begin{align}\left(\frac{d^2y}{dx^2}\right)^2\;+\;cos\left(\frac{dy}{dx}\right)\;=\;0\end{align}
The highest order derivative present in the given differential equation is \begin{align}\frac{d^2y}{dx^2}.\end{align}
Therefore, its order is 2. The given differential equation is not a polynomial equation in its derivatives.
Hence, its degree is not defined.
Q5 Determine order and degree(if defined) of differential equation \begin{align}\frac{d^2y}{dx^2}=\cos3x + sin3x\end{align} Ans: \begin{align}\frac{d^2y}{dx^2}=\cos3x + sin3x\end{align}
\begin{align}\Rightarrow\frac{d^2y}{dx^2} - \cos3x - sin3x = 0\end{align}
The highest order derivative present in the differential equation is\begin{align}\frac{d^2y}{dx^2}.\end{align}
Therefore, its order is two.It is a polynomial equation in \begin{align}\frac{d^2y}{dx^2}\end{align}
and the power raised to is 1.
\begin{align}\frac{d^2y}{dx^2}\end{align}
Hence, its degree is one.
Q6 Determine order and degree(if defined) of differential equation (ym)2 + (yn)3 + (y')4 + y5 =0 Ans: (ym)2 + (yn)3 + (y')4 + y5 =0
The highest order derivative present in the differential equation isym. Therefore, its order is three.
The given differential equation is a polynomial equation in ym , yn , y'.
The highest power raised to ym is 2. Hence, its degree is 2.
Q7 Determine order and degree(if defined) of differential equation ym + 2yn + y' =0 Ans: The highest order derivative present in the differential equation is ym. Therefore, its order is three.
It is a polynomial equation in ym , yn and y' . The highest power raised to ym is 1. Hence, its degree is 1.
Q8 Determine order and degree(if defined) of differential y' + y =ex Ans: y' + y =ex
y' + y - ex =0
The highest order derivative present in the differential equation is y'. Therefore, its order is one.
The given differential equation is a polynomial equation in y' and the highest power raised to y' is one. Hence, its degree is one.
Q9 Determine order and degree(if defined) of differential equation yn + (y')2 + 2y =0 Ans: yn + (y')2 + 2y =0
The highest order derivative present in the differential equation is yn. Therefore, its order is two.
The given differential equation is a polynomial equation in yn and y' and the highest power raised to yn is one.
Hence, its degree is one.
Q10 Determine order and degree(if defined) of differential equation yn + 2y' + siny = 0 Ans: yn + 2y' + siny = 0
The highest order derivative present in the differential equation is yn. Therefore, its order is two.
This is a polynomial equation in yn and y' and the highest power raised to yn is one. Hence, its degree is one.
Q11 The degree of the differential equation \begin{align}\left(\frac{d^2y}{dx^2}\right)^3\;+ \left(\frac{dy}{dx}\right)^2+\;sin\left(\frac{dy}{dx}\right)\;+ 1=\;0\end{align} is (A) 3 (B) 2 (C) 1 (D) not defined Ans: \begin{align}\left(\frac{d^2y}{dx^2}\right)^3\;+ \left(\frac{dy}{dx}\right)^2+\;sin\left(\frac{dy}{dx}\right)\;+ 1=\;0\end{align}
The given differential equation is not a polynomial equation in its derivatives. Therefore, its degree is not defined.
Hence, the correct answer is D.
Q12 The order of the differential equation \begin{align}2x^2\frac{d^2y}{dx^2}\;- \;3\frac{dy}{dx}\;+ y=\;0\end{align} is (A) 2 (B) 1 (C) 0 (D) not defined Ans: \begin{align}2x^2\frac{d^2y}{dx^2}\;- \;3\frac{dy}{dx}\;+ y=\;0\end{align}
The highest order derivative present in the given differential equation is
\begin{align}\frac{d^2y}{dx^2}\end{align}
Therefore, its order is two.
Hence, the correct answer is A.
Class 12 Mathematics Chapter 9: Differential Equations - NCERT Solutions
This page focuses on the complete NCERT solutions for Class 12 Mathematics Chapter 9: Differential Equations. This section provides detailed, easy-to-understand solutions for all the questions from this chapter. These Differential Equations question answers will offer you valuable insights and explanations.
In this chapter, we will shidy about differential equations and their solutions. Concepts of differential equations will help us in this class and also in higher studies. It has many applications in other branches like science, algebra, etc. Key topics of this chapter are - definition, order and degree, general and particular solutions, formation of differential equaĊon whose general solution is given, method of separation of variables, homogeneous differential equations of first order and degree.
Download pdf of NCERT Solutions for Class Mathematics Chapter 9 Differential Equations
Download pdf of NCERT Examplar with Solutions for Class Mathematics Chapter 9 Differential Equations
Exercise 1
Exercise 2
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Q1 y = ex +1 : yn -y' = 0 Ans: y = ex +1
Differentiating both sides of this equation with respect to x, we get:
\begin{align}\frac{dy}{dx}=\frac{d}{dx}(e^x + 1)\end{align}
=> y' = ex ...(1)
Now, differentiating equation (1) with respect to x, we get:
\begin{align}\frac{d}{dx}(y^{'})=\frac{d}{dx}(e^x)\end{align}
=> y'' = ex
Substituting the values of y' and y'' in the given differential equation, we get the L.H.S. as:
y'' - y' = ex - ex = 0 = R.H.S.
Thus, the given function is the solution of the corresponding differential equation.
Q2 y = x2 + 2x + C : y' - 2x - 2 = 0 Ans: y = x2 + 2x + C
Differentiating both sides of this equation with respect to x, we get:
\begin{align}y^{'}=\frac{d}{dx}(x^2 + 2x + C)\end{align}
=> y' = 2x + 2
Substituting the value of y' in the given differential equation, we get:
L.H.S. = y' - 2x - 2 = 2x + 2 - 2x - 2 = 0 = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Q3 y = cosx + C : y' + sinx = 0 Ans: y = cosx + C
Differentiating both sides of this equation with respect to x, we get:
\begin{align}y^{'}=\frac{d}{dx}(cosx + C)\end{align}
=> y' = - sinx
Substituting the value of y' in the given differential equation, we get:
L.H.S. = y' + sinx = - sinx + sinx = 0 = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Q4 \begin{align} y= \sqrt{1+x^2} : y^{'}=\frac{xy}{1+x^2}\end{align} Ans: \begin{align} y= \sqrt{1+x^2}\end{align}
Differentiating both sides of the equation with respect to x, we get:
\begin{align} y^{'}=\frac{d}{dx}\left(\sqrt{1+x^2} \right)\end{align}
\begin{align} y^{'}=\frac{1}{2\sqrt{1+x^2}}\frac{d}{dx}\left(1+x^2\right)\end{align}
\begin{align} y^{'}=\frac{2x}{2\sqrt{1+x^2}}\end{align}
\begin{align} y^{'}=\frac{x}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{\sqrt{1+x^2}}\frac{\sqrt{1+x^2}}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{\sqrt{1+x^2}}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{x}{1+x^2}{y}\end{align}
\begin{align}\Rightarrow y^{'}=\frac{xy}{1+x^2}\end{align}
∴ L.H.S. = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Q5 y = Ax : xy' = y (x ≠ 0) Ans: y = Ax
Differentiating both sides with respect to x, we get:
\begin{align}y^{'}=\frac{d}{dx}(Ax)\end{align}
⇒ y ' = A
Substituting the value of y' in the given differential equation, we get:
L.H.S. = xy' = xA = Ax = y = R.H.S.
Hence, the given function is the solution of the corresponding differential equation.
Q6 \begin{align} y = xsinx:xy{'}=y +x\sqrt{x^2 -y^2}(x\neq0\; and\; x>y\; or\; x<-y)\end{align} Ans: y= x.sinx
Differentiating both sides of this equation with respect to x, we get:
\begin{align} y^{'} =\frac{d}{dx}\left(x.sinx\right)\end{align}
\begin{align}\Rightarrow y^{'} =sinx. \frac{d}{dx}\left(x\right)+ x. \frac{d}{dx}\left(sinx\right)\end{align}
\begin{align} \Rightarrow y^{'} =sinx + x.cosx\end{align}
Differentiating both sides of this equation with respect to x, we get:
L.H.S. =xy' = x(sinx + xcosx)
\begin{align} =x.sinx + x^2.cosx\end{align}
\begin{align} =y + x^2.\sqrt{1-sin^2x}\end{align}
\begin{align} =y + x^2.\sqrt{1-\left(\frac{y}{x}\right)^2}\end{align}
\begin{align} =y + x^2.\sqrt{\frac{x^2-y^2}{x^2}}\end{align}
\begin{align} =y + x.\sqrt{x^2-y^2}\end{align}
R.H.S.
Key Features of NCERT Class 12 Mathematics Chapter 'Differential Equations' question answers :
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Popular Questions of Class 12 Mathematics
- Q:- Given an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive. - Q:- Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3,13, 14} defined as
R = {(x, y): 3x − y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x − y is as integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y} - Q:- Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
- Q:- Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. - Q:- Show that the relation R in the set R of real numbers, defined as R = {(a, b): a ≤ b2} is neither reflexive nor symmetric nor transitive.
- Q:-
In each of the following cases, state whether the function is one-one, onto or bijective. Justify your answer.
(i) f : R → R defined by f(x) = 3 – 4x
(ii) f : R → R defined by f(x) = 1 + x2
- Q:- Let L be the set of all lines in XY plane and R be the relation in L defined as R = {(L1, L2): L1 is parallel to L2}. Show that R is an equivalence relation. Find the set of all lines related to the line y = 2x + 4.
- Q:-
Show that the Modulus Function f : R → R, given by f(x) = |x|, is neither oneone nor onto, where | x | is x, if x is positive or 0 and |x| is – x, if x is negative.
- Q:-
Prove that the Greatest Integer Function f : R → R, given by f(x) = [x], is neither one-one nor onto, where [x] denotes the greatest integer less than or equal to x.
- Q:- Show that the relation R in R defined as R = {(a, b): a ≤ b}, is reflexive and transitive but not symmetric.
Recently Viewed Questions of Class 12 Mathematics
- Q:- Determine whether each of the following relations are reflexive, symmetric and transitive:
(i) Relation R in the set A = {1, 2, 3,13, 14} defined as
R = {(x, y): 3x − y = 0}
(ii) Relation R in the set N of natural numbers defined as
R = {(x, y): y = x + 5 and x < 4}
(iii) Relation R in the set A = {1, 2, 3, 4, 5, 6} as
R = {(x, y): y is divisible by x}
(iv) Relation R in the set Z of all integers defined as
R = {(x, y): x − y is as integer}
(v) Relation R in the set A of human beings in a town at a particular time given by
(a) R = {(x, y): x and y work at the same place}
(b) R = {(x, y): x and y live in the same locality}
(c) R = {(x, y): x is exactly 7 cm taller than y}
(d) R = {(x, y): x is wife of y}
(e) R = {(x, y): x is father of y} - Q:-
Let f : {1, 3, 4} → {1, 2, 5} and g : {1, 2, 5} → {1, 3} be given by f = {(1, 2), (3, 5), (4, 1)} and g = {(1, 3), (2, 3), (5, 1)}. Write down gof.
- Q:- Check whether the relation R defined in the set {1, 2, 3, 4, 5, 6} as
R = {(a, b): b = a + 1} is reflexive, symmetric or transitive. - Q:-
Answer the following as true or false.
\begin{align}(i) \overrightarrow{a}\; and\; \overrightarrow{-a}\; are\; collinear.\end{align}
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
- Q:- Show that the relation R in the set {1, 2, 3} given by R = {(1, 2), (2, 1)} is symmetric but neither reflexive nor transitive.
- Q:-
A stone is dropped into a quiet lake and waves move in circles at the speed of 5 cm/s. At the instant when the radius of the circular wave is 8 cm, how fast is the enclosed area increasing?
- Q:- Given an example of a relation. Which is
(i) Symmetric but neither reflexive nor transitive.
(ii) Transitive but neither reflexive nor symmetric.
(iii) Reflexive and symmetric but not transitive.
(iv) Reflexive and transitive but not symmetric.
(v) Symmetric and transitive but not reflexive. - Q:-
State with reason whether following functions have inverse
(i) f : {1, 2, 3, 4} → {10} with
f = {(1, 10), (2, 10), (3, 10), (4, 10)}
(ii) g : {5, 6, 7, 8} → {1, 2, 3, 4} with
g = {(5, 4), (6, 3), (7, 4), (8, 2)}
(iii) h : {2, 3, 4, 5} → {7, 9, 11, 13} with
h = {(2, 7), (3, 9), (4, 11), (5, 13)}
- Q:- If A=\(\begin{bmatrix}1 & 2\\4 & 2\end{bmatrix}\), then show that |2A| = 4|A|
- Q:-
Determine order and degree(if defined) of differential equation (ym)2 + (yn)3 + (y')4 + y5 =0
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