Ex 9.2 class 12 maths ncert solutions | class 12 maths exercise 9.2 | class 12 maths ncert solutions chapter 9 exercise 9.2 | exercise 9.2 class 12 maths ncert solutions | chapter 9 class 12 maths ncert solutions | differential equations class 12 ncert solutions
The ex 9.2 Class 12 Maths NCERT solutions are a vital resource for mastering linear differential equations. This part of the Class 12 Maths Exercise 9.2 focuses on solving differential equations using variables separable methods. With step-by-step explanations, the Class 12 Maths NCERT Solutions Chapter 9 Exercise 9.2 help students clearly understand each concept. Whether you’re practicing for boards or aiming for a strong grasp on calculus, the Exercise 9.2 Class 12 Maths NCERT Solutions and overall Differential Equations Class 12 NCERT Solutions provide the clarity and confidence you need.
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EXERCISE 9.2
1. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y=e^{x}+1: y^{\prime \prime}-y^{\prime}=0\)
\(y=e^{x}+1: y^{\prime \prime}-y^{\prime}=0\)
Answer
It is given that \( \mathrm{y}=\mathrm{e}^{x}+1 \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(\frac{d y}{d x}=\frac{d}{d x}\left(e^{x}\right)\)
Now, Again, differentiating both sides w.r.t. \(x\), we get,
\(\frac{d}{d x}\left(y^{\prime}\right)=\frac{d}{d x}\left(e^{x}\right)\)
\(\Rightarrow \mathrm{y}^{\prime \prime}=\mathrm{e}^{x}\)
Now, Substituting the values of \( y^{\prime} \) and \( y^{\prime \prime} \) in the given differential equations, we get,
\(y^{\prime \prime}-y^{\prime}=e^{x}-e^{x}=\text { RHS }\)
Therefore, the given function is the solution of the corresponding differential equation.
2. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y=x^{2}+2 x+C: y^{\prime}-2 x-2=0\)
\(y=x^{2}+2 x+C: y^{\prime}-2 x-2=0\)
Answer
It is given that \( y=x^{2}+2 x+C \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(y=\frac{d}{d x}\left(x^{2}+2 x+C\right)\)
\(\Rightarrow y^{\prime}=2 x+2\)
Now, Substituting the values of \(y^{\prime}\) in the given differential equations, we get,
\(y^{\prime}-2 x-2=2 x+2-2 x-2=0=\text { RHS. }\)
Therefore, the given function is the solution of the corresponding differential equation.
3. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y=\cos x+C: y^{\prime}+\sin x=0\)
\(y=\cos x+C: y^{\prime}+\sin x=0\)
Answer
It is given that \( y=\cos x+C \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(y^{\prime}=\frac{d}{d x}(\cos x+C)\)
\(\Rightarrow y^{\prime}=-\sin x\)
Now, Substituting the values of \(y^{\prime}\) in the given differential equations, we get,
\(y^{\prime}+\sin x=-\sin x+\sin x=0=\text { RHS }\)
Therefore, the given function is the solution of the corresponding differential equation.
4. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y=\sqrt{1+x^{2}} ; y^{\prime}=\frac{x y}{1+x^{2}}\)
\(y=\sqrt{1+x^{2}} ; y^{\prime}=\frac{x y}{1+x^{2}}\)
Answer
It is given that \( y=\sqrt{1+x^{2}} \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(y^{\prime}=\frac{d}{d x}\left(\sqrt{1+x^{2}}\right)^{2}\)
\(\Rightarrow y^{\prime}=\frac{1}{2 \sqrt{1+x^{2}}} \cdot \frac{d}{d x}\left(1+x^{2}\right)\)
\(\Rightarrow y^{\prime}=\frac{2 x}{2 \sqrt{1+x^{2}}}\)
\(\Rightarrow y^{\prime}=\frac{x}{\sqrt{1+x^{2}}}\)
\(\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \times \sqrt{1+x^{2}}\)
\(\Rightarrow y^{\prime}=\frac{x}{1+x^{2}} \cdot y\)
\(\Rightarrow y^{\prime}=\frac{x y}{1+x^{2}}\)
Therefore, LHS \( = \) RHS
Therefore, the given function is the solution of the corresponding differential equation.
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5. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y=A x: x y^{\prime}=y(x \neq 0)\)
\(y=A x: x y^{\prime}=y(x \neq 0)\)
Answer
It is given that \( \mathrm{y}=\mathrm{A}x \)
Now, differentiating both sides w.r.t. \(x\), we get,
\(y^{\prime}=\frac{d}{d x}(A x)\)
\(\Rightarrow \mathrm{y}^{\prime}=\mathrm{A}\)
Now, Substituting the values of \(\mathrm{y}^{\prime}\) in the given differential equations, we get,
\(x y^{\prime}=x . A=A x=y=R H S\)
Therefore, the given function is the solution of the corresponding differential equation.
6. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y=x \sin x: x y^{\prime}=y+x \sqrt{x^{2}-y^{2}}(x \neq 0 \text { and } x > y \text { or } x < -y)\)
\(y=x \sin x: x y^{\prime}=y+x \sqrt{x^{2}-y^{2}}(x \neq 0 \text { and } x > y \text { or } x < -y)\)
Answer
It is given that \( y=x \sin x \)
Now, differentiating both sides w.r.t. \(x\), we get,
\(y^{\prime}=\frac{d}{d x}(x \sin x)\)
\(\Rightarrow y^{\prime}=\sin x \frac{d}{d x}(x)+x \cdot \frac{d}{d x}(\sin x)\)
\(\Rightarrow y^{\prime}=\sin x+x \cos x\)
Now, Substituting the values of \(y^{\prime}\) in the given differential equations, we get,
\(\text {LHS }=xy^{\prime}=x(\sin x+x \cos x)\)
\(=x \sin x+x^{2} \cos x\)
\(=y+x^{2} \cdot \sqrt{1-\sin ^{2} x}\)
\(=y+x^{2} \sqrt{1-\left(\frac{y}{x}\right)^{2}}\)
\(=y+x \sqrt{(y)^{2}-(x)^{2}}\)
\(=\text {RHS }\)
Therefore, the given function is the solution of the corresponding differential equation.
7. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(xy=\log y+C: y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)\)
\(xy=\log y+C: y^{\prime}=\frac{y^{2}}{1-x y}(x y \neq 1)\)
Answer
It is given that \( x y=\log y+C \)
Now, differentiating both sides w.r.t. \( x \), we get,
\(\frac{d}{d x}(x y)=\frac{d}{d x}(\log y)\)
\(\Rightarrow y \cdot \frac{d}{d x}(x)+x \cdot \frac{d y}{d x}=\frac{1}{y} \frac{d y}{d x}\)
\(\Rightarrow y+xy=\frac{1}{y} \frac{d y}{d x}\)
\(\Rightarrow y^{2}+xyy^{\prime}=y^{\prime}\)
\(\Rightarrow(xy-1) y^{\prime}=-y^{2}\)
\(\Rightarrow y^{\prime}=\frac{y^{2}}{1-x y}\)
Thus, LHS \( = \) RHS
Therefore, the given function is the solution of the corresponding differential equation.
8. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y-\cos y=x:(y \sin y+\cos y+x) y^{\prime}=y\)
\(y-\cos y=x:(y \sin y+\cos y+x) y^{\prime}=y\)
Answer
It is given that \( y-\cos y=x \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(\frac{d y}{d x}-\frac{d}{d x} \cos y=\frac{d}{d x}(x)\)
\(\Rightarrow y^{\prime}+\sin y \cdot y^{\prime}=1\)
\(\Rightarrow y^{\prime}(1+\sin y)=1\)
\(\Rightarrow y^{\prime}=\frac{1}{1+\sin y}\)
Now, Substituting the values of \(y^{\prime}\) in the given differential equations, we get,
\(\text {LHS }=(y \sin y+\cos y+x) y^{\prime}\)
\(=(y \sin y+\cos y+y-\cos y) \times \frac{1}{1+\sin y}\)
\(=y(1+\sin y) \times \frac{1}{1+\sin y}\)
\(=y=\text {RHS }\)
Therefore, the given function is the solution of the corresponding differential equation.
class 12 maths ncert solutions chapter 9 exercise 9.2 || exercise 9.2 class 12 maths ncert solutions || chapter 9 class 12 maths ncert solutions || differential equations class 12 ncert solutions || class 12 maths exercise 9.2 || ex 9.2 class 12 maths ncert solutions
9. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(x+y=\tan ^{-1} y: y^{2} y^{\prime}+y^{2}+1=0\)
\(x+y=\tan ^{-1} y: y^{2} y^{\prime}+y^{2}+1=0\)
Answer
It is given that \( x+y=\tan ^{-1} y \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(\frac{d}{d x}(x+y)=\frac{d}{d x}\left(\tan ^{-1} y\right)\)
\(\Rightarrow 1+y^{\prime}=\left[\frac{1}{1+y^{2}}\right] y^{\prime}\)
\(\Rightarrow y^{\prime}\left[\frac{1}{1+y^{2}}-1\right]=1\)
\(\Rightarrow y^{\prime}\left[\frac{1-\left(1+y^{2}\right)}{1+y^{2}}\right]=1\)
\(\Rightarrow y^{\prime}\left[\frac{-y^{2}}{1+y^{2}}\right]=1\)
\(\Rightarrow y^{\prime}=\frac{-\left(1+y^{2}\right)}{y^{2}}\)
Now, Substituting the values of \(y^{\prime}\) in the given differential equations, we get,
\(\text {LHS }=y^{2} y^{\prime}+y^{2}+1=y^{2}\left[\frac{-\left(1+y^{2}\right)}{y^{2}}\right]+y^{2}+1\)
\(=-1-y^{2}+y^{2}+1\)
\(=0=\text {RHS }\)
Therefore, the given function is the solution of the corresponding differential equation.
10. In each of the question verify that the given functions (explicit or implicit) is a solution of the corresponding differential equation:
\(y=\sqrt{a^{2}-x^{2}} x \in(-a, a): x+y \frac{d y}{d x}=0(y \neq 0)\)
\(y=\sqrt{a^{2}-x^{2}} x \in(-a, a): x+y \frac{d y}{d x}=0(y \neq 0)\)
Answer
It is given that \( \mathrm{y}=\sqrt{a^{2}-x^{2}} \)
Now, differentiating both sides w.r.t. \(x \), we get,
\(\frac{d y}{d x}=\frac{d}{d x}\left(\sqrt{a^{2}-x^{2}}\right)\)
\(\Rightarrow \frac{d y}{d x}=\frac{1}{2 \sqrt{a^{2}-x^{2}}} \cdot \frac{d}{d x} a^{2}-x^{2}\)
\(=\frac{1}{2 \sqrt{a^{2}-x^{2}}}(-2 x)\)
Now, Substituting the values of \(y^{\prime}\) in the given differential equations, we get,
\(\text {LHS }=x+y \frac{d y}{d x}\)
\(=x+\sqrt{a^{2}-x^{2}} \times-\frac{x}{2 \sqrt{a^{2}-x^{2}}}\)
\(=x-x=0=\text {RHS. }\)
Therefore, the given function is the solution of the corresponding differential equation.
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
A. 0 B. 2 C. 3 D. 4
Answer
As we know that the number of constant in the general solution of a differential equation of order n is equal to its order.
Thus, the number of constant in the general equation of fourth order differential equation is four.
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
A. 3 B. 2 C. 1 D. 0
Answer
In a particular solution of a differential equation, there is no arbitrary constant.
