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Class 12 NCERT Mathematics Solutions- Chapter 9– Differential Equations Exercise 9.2
Chapter 9 of the Class 12 NCERT Mathematics Part II textbook, titled “Differential Equations,” explores methods for solving differential equations. This chapter includes techniques for handling first-order and higher-order differential equations. Exercise 9.2 focuses on solving specific differential equations using the concepts introduced in the chapter.
This section provides detailed solutions for Exercise 9.2 from Chapter 9 of the Class 12 NCERT Mathematics Part II textbook. The exercise involves solving problems related to differential equations, applying various methods to find solutions for given equations. Solutions are presented step-by-step to aid in understanding and applying differential equation techniques effectively.
NCERT Question 1 : Verify that the given function is a solution of the differential equation:
Function: $y = e^x + 1$
Differential Equation: $y” – y’ = 0$
Solution:
Given:
$$
y = e^x + 1
$$
Differentiate both sides with respect to $x$:
$$
y’ = \frac{dy}{dx} = \frac{d}{dx}(e^x + 1) = e^x
$$
Differentiate again:
$$
y” = \frac{d^2y}{dx^2} = \frac{d}{dx}(y’) = \frac{d}{dx}(e^x) = e^x
$$
Substitute $y’$ and $y”$ into the differential equation:
$$
y” – y’ = e^x – e^x = 0
$$
Conclusion
$$
\boxed{y = e^x + 1 \text{ is a solution of } y” – y’ = 0.}
$$
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NCERT Question 2 : Verify that the given function is a solution of the differential equation:
Function: $y = x^2 + 2x + C$
Differential Equation: $y’ – 2x – 2 = 0$
Solution
Given:
$$
y = x^2 + 2x + C
$$
Differentiate both sides with respect to $x$:
$$
y’ = \frac{dy}{dx} = \frac{d}{dx}(x^2 + 2x + C) = 2x + 2
$$
Substitute $y’$ into the differential equation:
$$
y’ – 2x – 2 = (2x + 2) – 2x – 2 = 0
$$
Conclusion
$$
\boxed{y = x^2 + 2x + C \text{ is a solution of } y’ – 2x – 2 = 0.}
$$
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NCERT Question 3 : Verify that the given function is a solution of the differential equation:
Function: $y = \cos x + C$
Differential Equation: $y’ + \sin x = 0$
Solution :
Given:
$$
y = \cos x + C
$$
Differentiate both sides with respect to $x$:
$$
y’ = \frac{dy}{dx} = \frac{d}{dx}(\cos x + C) = -\sin x
$$
Substitute $y’$ into the differential equation:
$$
y’ + \sin x = (-\sin x) + \sin x = 0
$$
Conclusion
$$
\boxed{y = \cos x + C \text{ is a solution of } y’ + \sin x = 0.}
$$
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NCERT Question 4 : Verify that the given function is a solution of the differential equation:
Function:
$$
y = \sqrt{1 + x^2}
$$
Differential Equation:
$$
y’ = \frac{x y}{1 + x^2}
$$
Solution :
Given:
$$
y = \sqrt{1 + x^2}
$$
Differentiating both sides with respect to $x$:
$$
y’ = \frac{d}{dx} \Big( (1 + x^2)^{1/2} \Big)
$$
Using the chain rule:
$$
y’ = \frac{1}{2} (1 + x^2)^{-1/2} \cdot 2x
$$
Simplifying:
$$
y’ = \frac{x}{\sqrt{1 + x^2}}
$$
Since $y = \sqrt{1 + x^2}$, we can rewrite:
$$
y’ = \frac{x}{1 + x^2} \cdot y
$$
$$
y’ = \frac{x y}{1 + x^2} \quad \text{(RHS)}
$$
Conclusion
$$
\boxed{y = \sqrt{1 + x^2} \text{ is a solution of } y’ = \frac{x y}{1 + x^2}}
$$
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NCERT Question 5 : Verify that the given function is a solution of the differential equation:
Function:
$$
y = A x
$$
Differential Equation:
$$
x y’ = y, \quad x \neq 0
$$
Solution :
Given:
$$
y = A x
$$
Differentiating both sides with respect to $x$:
$$
y’ = \frac{d}{dx}(A x) = A
$$
Substituting into the differential equation:
$$
x y’ = x \cdot A = A x = y
$$
Thus,
$$
x y’ = y \quad \text{(RHS)}
$$
Conclusion
$$
\boxed{y = A x \text{ is a solution of } x y’ = y, , x \neq 0}
$$
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NCERT Question 6 : Verify that the given function is a solution of the differential equation:
Function:
$$
y = x \sin x
$$
Differential Equation:
$$
x y’ = y + x \sqrt{x^2 – y^2}, \quad x \neq 0 \text{ and } x > y \text{ or } x < -y
$$
Solution :
Given:
$$
y = x \sin x
$$
Differentiating both sides with respect to $x$:
$$
y’ = \frac{d}{dx}(x \sin x) = \frac{d}{dx}(x) \cdot \sin x + x \cdot \frac{d}{dx}(\sin x) = \sin x + x \cos x
$$
Multiplying both sides by $x$:
$$
x y’ = x(\sin x + x \cos x) = x \sin x + x^2 \cos x
$$
Since $y = x \sin x$, we have:
$$
x y’ = y + x^2 \cos x
$$
Also, compute the square root term:
$$
\sqrt{x^2 – y^2} = \sqrt{x^2 – (x \sin x)^2} = \sqrt{x^2 (1 – \sin^2 x)} = \sqrt{x^2 \cos^2 x} = x \cos x
$$
Thus,
$$
y + x \sqrt{x^2 – y^2} = y + x \cdot x \cos x = y + x^2 \cos x
$$
This matches the left-hand side:
$$
x y’ = y + x \sqrt{x^2 – y^2}
$$
Conclusion
$$
\boxed{y = x \sin x \text{ is a solution of the given differential equation.}}
$$
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NCERT Question 7 : Verify that
$$
xy = \log y + C
$$
is a solution of the differential equation
$$
y’ = \frac{y^2}{1 – xy}.
$$
Solution :
Given:
$$
xy = \log y + C
$$
Differentiate both sides with respect to $x$:
$$
\frac{d}{dx}(xy) = \frac{d}{dx}(\log y + C)
$$
Using the product rule on the left-hand side:
$$
y + x y’ = \frac{1}{y} y’ + 0
$$
(since $C$ is constant)
Simplify:
$$
y + x y’ = \frac{y’}{y} \implies y + x y’ = \frac{y’}{y}
$$
Multiply through by $y$:
$$
y^2 + xy y’ = y’ \implies xy y’ – y’ = -y^2 \implies y'(xy – 1) = -y^2
$$
Finally, solve for $y’$:
$$
y’ = \frac{-y^2}{xy – 1} = \frac{y^2}{1 – xy}
$$
This matches the given differential equation.
Conclusion
$$
\boxed{y = \text{solution of } xy = \log y + C \text{ satisfies } y’ = \frac{y^2}{1-xy}}
$$
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NCERT Question 8 : Verify that
$$
y – \cos y = x
$$
is a solution of the differential equation
$$
(y \sin y + \cos y + x) y’ = y.
$$
Solution :
Given:
$$
y – \cos y = x
$$
Differentiate both sides with respect to $x$:
$$
\frac{dy}{dx} – \frac{d}{dx}(\cos y) = \frac{d}{dx}(x)
$$
Using the chain rule on $\cos y$:
$$
\frac{dy}{dx} – (-\sin y \cdot y’) = 1
$$
Simplify:
$$
y’ + y’ \sin y = 1 \implies y'(1 + \sin y) = 1
$$
So we have:
$$
y’ = \frac{1}{1 + \sin y}
$$
Substitute $y’$ into the given differential equation:
$$
(y \sin y + \cos y + x) y’ = (y \sin y + \cos y + x) \cdot \frac{1}{1 + \sin y}
$$
From the original equation $y – \cos y = x$, we have $\cos y + x = y$. So:
$$
(y \sin y + \cos y + x) y’ = (y \sin y + y) \cdot \frac{1}{1 + \sin y} = y \cdot \frac{1 + \sin y}{1 + \sin y} = y
$$
Hence,
$$
(y \sin y + \cos y + x) y’ = y = \text{RHS}
$$
Conclusion
$$
\boxed{y – \cos y = x \text{ is a solution of } (y \sin y + \cos y + x) y’ = y}
$$
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NCERT Question 9 : Verify that
$$
x + y = \tan^{-1}y
$$
is a solution of the differential equation
$$
y^2 y’ + y^2 + 1 = 0.
$$
Solution :
Given:
$$
x + y = \tan^{-1}y
$$
Differentiate both sides with respect to $x$:
$$
\frac{d}{dx}(x + y) = \frac{d}{dx}(\tan^{-1}y)
$$
This gives:
$$
1 + y’ = \frac{y’}{1 + y^2}
$$
Rewriting, we get:
$$
1 + y’ = \frac{y’}{1 + y^2} \implies y’ \left(1 – \frac{1}{1 + y^2}\right) = -1
$$
Simplify the term in brackets:
$$
1 – \frac{1}{1 + y^2} = \frac{(1 + y^2) – 1}{1 + y^2} = \frac{y^2}{1 + y^2}
$$
Hence:
$$
y’ \cdot \frac{y^2}{1 + y^2} = -1 \implies y’ = -\frac{1 + y^2}{y^2}
$$
Verification
Substitute $y’ = -\frac{1 + y^2}{y^2}$ into the differential equation:
$$
y^2 y’ + y^2 + 1 = y^2 \cdot \left(-\frac{1 + y^2}{y^2}\right) + y^2 + 1 = -(1 + y^2) + y^2 + 1 = 0
$$
Conclusion
$$
\boxed{x + y = \tan^{-1}y \text{ is a solution of } y^2 y’ + y^2 + 1 = 0}
$$
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NCERT Question 10 : Verify that
$$
y=\sqrt{a^2-x^2},\qquad |x|\le a,\ a>0.
$$
is a solution of the differential equation
$$
x + y\frac{dy}{dx}=0.
$$
Solution :
Given:
$$
y=\sqrt{a^2-x^2},\qquad |x|\le a,\ a>0.
$$
Differentiate with respect to $x$ using the chain rule:
$$
y=(a^2-x^2)^{1/2}\quad\Rightarrow\quad
y’=\frac{1}{2}(a^2-x^2)^{-1/2}\cdot(-2x)=\frac{-x}{\sqrt{a^2-x^2}}.
$$
Since $y=\sqrt{a^2-x^2}$, we can rewrite the derivative as
$$
y’=-\frac{x}{y}.
$$
Now substitute $y’$ into the left-hand side of the differential equation:
$$
x + y\frac{dy}{dx}=x + y\left(-\frac{x}{y}\right)=x-x=0.
$$
Thus the equation
$$
x + y\frac{dy}{dx}=0
$$
is satisfied by $y=\sqrt{a^2-x^2}$ (for $|x|\le a$). The same computation holds for the negative branch $y=-\sqrt{a^2-x^2}$ (the derivative again equals $-x/y$), so both
$$
y=\pm\sqrt{a^2-x^2}
$$
are solutions on the domain $|x|\le a$.
$$
\boxed{\text{Hence }y=\pm\sqrt{a^2-x^2}\text{ satisfy }x+y,\dfrac{dy}{dx}=0.}
$$
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NCERT Question 11 : The number of arbitrary constants in the general solution of a differential equation of fourth order are:
(A) 0 (B) 2 (C) 3 (D) 4
Solution:
We know that the number of arbitrary constants in the general solution of a differential equation is equal to the order of the differential equation.
Since the given differential equation is of fourth order,
the general solution will contain four arbitrary constants.
Hence,
$$
\text{Number of arbitrary constants} = 4
$$
$$
\boxed{\text{Correct Option: (D)}}
$$
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NCERT Question 12 : The number of arbitrary constants in the particular solution of a differential equation of third order are:
(A) 3 (B) 2 (C) 1 (D) 0
Solution:
In a particular solution, all arbitrary constants are determined using the given initial or boundary conditions.
Therefore, unlike a general solution (which contains constants equal to the order of the equation), the particular solution contains no arbitrary constants.
Hence,
$$
\text{Number of arbitrary constants} = 0
$$
$$
\boxed{\text{Correct Option: (D)}}
$$
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Summary
Chapter 9 of the Class 12 NCERT Mathematics Part II textbook, “Differential Equations,” deals with solving differential equations using various methods. Exercise 9.2 involves applying these techniques to specific problems. Detailed solutions are provided to help students understand and solve differential equations effectively.
