Complete solutions for Class 12 Maths (NCERT) Exercise 9.2. Learn how to verify that a given function (explicit or implicit) is a solution to a corresponding differential equation (DE) by computing and substituting its derivatives ($y’$, $y”$) into the DE. Understand the difference between the number of arbitrary constants in the general solution (equal to order) and the particular solution (zero).
Table of Contents
To verify that a function is a solution to a differential equation, we must compute the necessary derivatives of the function and substitute them back into the differential equation. If the substitution results in a true statement ($0=0$ or $\text{LHS} = \text{RHS}$), the solution is verified.
Verification of Solutions (1-10)
1. $y = e^x + 1$ : $y” – y’ = 0$
- First Derivative: $y’ = \frac{d}{dx} (e^x + 1) = e^x$
- Second Derivative: $y” = \frac{d}{dx} (e^x) = e^x$
- Substitution: Substitute $y”$ and $y’$ into the DE:$$y” – y’ = e^x – e^x = 0$$LHS = RHS. (Verified)
2. $y = x^2 + 2x + C$ : $y’ – 2x – 2 = 0$
- First Derivative: $y’ = \frac{d}{dx} (x^2 + 2x + C) = 2x + 2$
- Substitution: Substitute $y’$ into the DE:$$y’ – 2x – 2 = (2x + 2) – 2x – 2 = 0$$LHS = RHS. (Verified)
3. $y = \cos x + C$ : $y’ + \sin x = 0$
- First Derivative: $y’ = \frac{d}{dx} (\cos x + C) = -\sin x$
- Substitution: Substitute $y’$ into the DE:$$y’ + \sin x = (-\sin x) + \sin x = 0$$LHS = RHS. (Verified)
4. $y = \sqrt{1 + x^2}$ : $y’ = \frac{xy}{1 + x^2}$
- First Derivative: $y’ = \frac{d}{dx} (1 + x^2)^{1/2}$$$y’ = \frac{1}{2} (1 + x^2)^{-1/2} \cdot (2x) = \frac{x}{\sqrt{1 + x^2}}$$
- Manipulation of RHS: Rewrite the expression in the DE using the function $y$:$$\text{RHS} = \frac{xy}{1 + x^2} = \frac{x \sqrt{1 + x^2}}{1 + x^2} = \frac{x}{\frac{1 + x^2}{\sqrt{1 + x^2}}} = \frac{x}{\sqrt{1 + x^2}}$$
- Comparison:$$\text{LHS} = y’ = \frac{x}{\sqrt{1 + x^2}}$$$$\text{RHS} = \frac{x}{\sqrt{1 + x^2}}$$LHS = RHS. (Verified)
5. $y = Ax$ : $xy’ = y$ ($x \neq 0$)
- First Derivative: $y’ = \frac{d}{dx} (Ax) = A$
- Substitution into LHS:$$\text{LHS} = xy’ = x(A) = Ax$$
- Substitution into RHS:$$\text{RHS} = y = Ax$$LHS = RHS. (Verified)
6. $y = x \sin x$ : $xy’ = y + x \sqrt{x^2 – y^2}$
- First Derivative: Use the product rule:$$y’ = \frac{d}{dx} (x \sin x) = 1 \cdot \sin x + x \cdot \cos x = \sin x + x \cos x$$
- Substitution into LHS:$$\text{LHS} = xy’ = x(\sin x + x \cos x) = x \sin x + x^2 \cos x$$
- Substitution into RHS: Substitute $y = x \sin x$:$$\text{RHS} = y + x \sqrt{x^2 – y^2} = x \sin x + x \sqrt{x^2 – (x \sin x)^2}$$$$\text{RHS} = x \sin x + x \sqrt{x^2 (1 – \sin^2 x)}$$$$\text{RHS} = x \sin x + x |x \cos x|$$Since the problem implies that $x$ and $\sqrt{x^2 – y^2}$ have the same sign (or the simplification assumes $x>0$), we take $|x|=x$:$$\text{RHS} = x \sin x + x (x \cos x) = x \sin x + x^2 \cos x$$LHS = RHS. (Verified)
7. $xy = \log y + C$ : $y’ = \frac{y^2}{1 – xy}$ ($xy \neq 1$)
- Differentiate Implicitly with respect to $x$:$$\frac{d}{dx} (xy) = \frac{d}{dx} (\log y + C)$$$$1 \cdot y + x \cdot y’ = \frac{1}{y} \cdot y’ + 0$$$$y + xy’ = \frac{y’}{y}$$
- Isolate $y’$: Collect $y’$ terms and solve:$$xy’ – \frac{y’}{y} = -y$$$$y’ \left( x – \frac{1}{y} \right) = -y$$$$y’ \left( \frac{xy – 1}{y} \right) = -y$$$$y’ = \frac{-y^2}{xy – 1} = \frac{-y^2}{-(1 – xy)} = \frac{y^2}{1 – xy}$$LHS = RHS. (Verified)
8. $y – \cos y = x$ : $(y \sin y + \cos y + x) y’ = y$
- Differentiate Implicitly with respect to $x$:$$\frac{d}{dx} (y) – \frac{d}{dx} (\cos y) = \frac{d}{dx} (x)$$$$y’ – (-\sin y \cdot y’) = 1$$$$y’ (1 + \sin y) = 1 \implies y’ = \frac{1}{1 + \sin y}$$
- Substitution into LHS of DE: Substitute $y’$ and use the given function $x = y – \cos y$:$$\text{LHS} = (y \sin y + \cos y + x) y’$$$$\text{LHS} = (y \sin y + \cos y + (y – \cos y)) \cdot \left( \frac{1}{1 + \sin y} \right)$$$$\text{LHS} = (y \sin y + y) \cdot \frac{1}{1 + \sin y}$$$$\text{LHS} = y ( \sin y + 1 ) \cdot \frac{1}{1 + \sin y} = y$$LHS = RHS ($y$). (Verified)
9. $x + y = \tan^{-1} y$ : $y^2 y’ + y^2 + 1 = 0$
- Differentiate Implicitly with respect to $x$:$$\frac{d}{dx} (x) + \frac{d}{dx} (y) = \frac{d}{dx} (\tan^{-1} y)$$$$1 + y’ = \frac{1}{1 + y^2} \cdot y’$$
- Isolate $y’$:$$1 = y’ \left( \frac{1}{1 + y^2} – 1 \right)$$$$1 = y’ \left( \frac{1 – (1 + y^2)}{1 + y^2} \right)$$$$1 = y’ \left( \frac{-y^2}{1 + y^2} \right) \implies y’ = -\frac{1 + y^2}{y^2}$$
- Substitution into DE:$$\text{LHS} = y^2 y’ + y^2 + 1$$$$\text{LHS} = y^2 \left( -\frac{1 + y^2}{y^2} \right) + y^2 + 1$$$$\text{LHS} = -(1 + y^2) + y^2 + 1 = -1 – y^2 + y^2 + 1 = 0$$LHS = RHS. (Verified)
10. $y = \sqrt{a^2 – x^2}$, $x \in (-a, a)$ : $x + y \frac{dy}{dx} = 0$ ($y \neq 0$)
- Differentiate:$$\frac{dy}{dx} = \frac{d}{dx} (a^2 – x^2)^{1/2} = \frac{1}{2} (a^2 – x^2)^{-1/2} (-2x)$$$$\frac{dy}{dx} = -\frac{x}{\sqrt{a^2 – x^2}}$$
- Substitution into DE: Substitute $\frac{dy}{dx}$ and use $y = \sqrt{a^2 – x^2}$:$$\text{LHS} = x + y \frac{dy}{dx}$$$$\text{LHS} = x + (\sqrt{a^2 – x^2}) \left( -\frac{x}{\sqrt{a^2 – x^2}} \right)$$$$\text{LHS} = x – x = 0$$LHS = RHS. (Verified)
Multiple Choice Questions (11-12)
11. The number of arbitrary constants in the general solution of a differential equation of fourth order are:
The number of arbitrary constants in the general solution of a differential equation is equal to the order of the differential equation.
Since the order is four, the number of arbitrary constants is 4.
The correct answer is (D) 4.
12. The number of arbitrary constants in the particular solution of a differential equation of third order are:
A particular solution is obtained from the general solution by assigning specific values to the arbitrary constants (usually using initial or boundary conditions). Therefore, the particular solution contains no arbitrary constants.
The correct answer is (D) 0.