Exercise 9.2 Practice
Chapter 9: Differential Equations – NCERT Solutions
Q1
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Verify that the function $y = e^x + 1$ is a solution of the differential equation $y'' - y' = 0$.
Given $y = e^x + 1$.
Differentiating w.r.t $x$: $y' = e^x$.
Differentiating again: $y'' = e^x$.
Substitute in DE: $y'' - y' = e^x - e^x = 0$.
$\boxed{\text{Verified}}$
Q2
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Verify that $y = x^2 + 2x + C$ is a solution of $y' - 2x - 2 = 0$.
Given $y = x^2 + 2x + C$.
$y' = 2x + 2$.
Substitute in DE: $(2x + 2) - 2x - 2 = 0$.
$\boxed{\text{Verified}}$
Q3
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Verify that $y = \cos x + C$ is a solution of $y' + \sin x = 0$.
Given $y = \cos x + C$.
$y' = -\sin x$.
Substitute in DE: $-\sin x + \sin x = 0$.
$\boxed{\text{Verified}}$
Q4
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Verify that $y = \sqrt{1+x^2}$ is a solution of $y' = \frac{xy}{1+x^2}$.
$y = (1+x^2)^{1/2}$.
$y' = \frac{1}{2}(1+x^2)^{-1/2} \cdot 2x = \frac{x}{\sqrt{1+x^2}}$.
RHS of DE: $\frac{xy}{1+x^2} = \frac{x\sqrt{1+x^2}}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$.
LHS = RHS.
$\boxed{\text{Verified}}$
Q5
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Verify that $y = Ax$ is a solution of $xy' = y$ ($x \ne 0$).
$y = Ax \Rightarrow y' = A$.
LHS: $xy' = x(A) = Ax$.
RHS: $y = Ax$. LHS = RHS.
$\boxed{\text{Verified}}$
Q6
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Verify that $y = x \sin x$ is a solution of $xy' = y + x\sqrt{x^2-y^2}$.
$y' = x \cos x + \sin x$. LHS: $x(x \cos x + \sin x) = x^2 \cos x + x \sin x$.
RHS: $x \sin x + x\sqrt{x^2 - x^2 \sin^2 x} = x \sin x + x|x \cos x|$.
Assuming $x>0, \cos x>0$, RHS = $x \sin x + x^2 \cos x$. LHS = RHS.
$\boxed{\text{Verified}}$
Q7
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Verify that $xy = \log y + C$ is a solution of $y' = \frac{y^2}{1-xy}$.
Differentiate $xy = \log y + C$: $x y' + y = \frac{1}{y} y'$.
$y = y'(\frac{1}{y} - x) = y'(\frac{1-xy}{y})$.
$y' = \frac{y^2}{1-xy}$.
$\boxed{\text{Verified}}$
Q8
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Verify that $y - \cos y = x$ is a solution of $(y \sin y + \cos y + x)y' = y$.
Differentiate: $y' + \sin y \cdot y' = 1 \Rightarrow y' = \frac{1}{1+\sin y}$.
LHS of DE: $(y \sin y + \cos y + x) \frac{1}{1+\sin y}$.
Put $x = y - \cos y$: $(y \sin y + \cos y + y - \cos y) \frac{1}{1+\sin y} = y(1+\sin y)\frac{1}{1+\sin y} = y$.
$\boxed{\text{Verified}}$
Q9
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Verify that $x + y = \tan^{-1} y$ is a solution of $y^2 y' + y^2 + 1 = 0$.
Differentiate: $1 + y' = \frac{1}{1+y^2} y'$.
$1 = y'(\frac{1}{1+y^2} - 1) = y'(\frac{-y^2}{1+y^2})$.
$1+y^2 = -y^2 y' \Rightarrow y^2 y' + y^2 + 1 = 0$.
$\boxed{\text{Verified}}$
Q10
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Verify that $y = \sqrt{a^2 - x^2}$ is a solution of $x + y \frac{dy}{dx} = 0$.
$y' = \frac{1}{2\sqrt{a^2-x^2}}(-2x) = \frac{-x}{y}$.
LHS: $x + y(\frac{-x}{y}) = x - x = 0$.
$\boxed{\text{Verified}}$
Q11
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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
(A) 0 (B) 2 (C) 3 (D) 4
The number of arbitrary constants in the general solution of a differential equation of order $n$ is equal to its order $n$.
$\boxed{\text{(D) 4}}$
Q12
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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
(A) 3 (B) 2 (C) 1 (D) 0
A particular solution is obtained by giving particular values to the arbitrary constants. Thus, it contains no arbitrary constants.
$\boxed{\text{(D) 0}}$