GSEB Class 12 Maths Solutions Chapter 9 Differential Equations Exercise 9.2

In each of the questions 1 to 10, verify that the given function (explicit or implicit) is a solution of the corresponding differential equation:

Question 1. \( y = e^x + 1 : y'' - y' = 0 \)
Answer: We are given the function \( y = e^x + 1 \).
First, we differentiate \( y \) with respect to \( x \):
\( y' = \frac{d}{dx} (e^x + 1) \)
\( y' = e^x + 0 \)
\( y' = e^x \) (1)
Next, we differentiate \( y' \) again with respect to \( x \) to find \( y'' \):
\( y'' = \frac{d}{dx} (e^x) \)
\( y'' = e^x \) (2)
Now, we subtract equation (1) from equation (2):
\( y'' - y' = e^x - e^x \)
\( y'' - y' = 0 \)
This result matches the given differential equation. Therefore, the function \( y = e^x + 1 \) is a solution.
In simple words: We find the first and second derivatives of the given function. When we subtract the first derivative from the second, the result is zero, which means the function solves the given equation.

Exam Tip: Remember to differentiate correctly step-by-step and then substitute the derivatives into the given differential equation to confirm if it holds true.

Question 2. \( y = x^2 + 2x + C : y' - 2x - 2 = 0 \)
Answer: We are given the function \( y = x^2 + 2x + C \).
To verify, we need to find the first derivative, \( y' \).
Differentiating \( y \) with respect to \( x \):
\( y' = \frac{d}{dx} (x^2 + 2x + C) \)
\( y' = 2x + 2 + 0 \)
\( y' = 2x + 2 \)
Now, we rearrange the terms to match the given differential equation:
\( y' - 2x - 2 = 0 \)
This shows that the given function satisfies the differential equation. Hence, it is a solution.
In simple words: We take the first derivative of the function. Then, we move the other terms to one side of the equation. If it matches the given differential equation, the function is a solution.

Exam Tip: For problems involving arbitrary constants (like C), differentiating once usually eliminates the constant, making it easier to match the differential equation.

Question 3. \( y = \cos x + C : y' + \sin x = 0 \)
Answer: We are given the function \( y = \cos x + C \).
To check this, we will differentiate \( y \) with respect to \( x \):
\( y' = \frac{d}{dx} (\cos x + C) \)
\( y' = -\sin x + 0 \)
\( y' = -\sin x \)
Now, substitute this value of \( y' \) into the given differential equation, \( y' + \sin x = 0 \):
\( (-\sin x) + \sin x = 0 \)
\( 0 = 0 \)
Since the equation holds true, the given function is indeed a solution to the differential equation.
In simple words: We find the derivative of the function. When we put this derivative into the equation and add sine x, it equals zero, which confirms it's a correct solution.

Exam Tip: Pay close attention to the signs when differentiating trigonometric functions, as a common error is to miss a negative sign.

Question 4. \( y = \sqrt{1+x^{2}} : y' = \frac{xy}{1+x^{2}} \)
Answer: We have the function \( y = \sqrt{1+x^{2}} \).
To find \( y' \), we differentiate \( y \) with respect to \( x \):
\( y' = \frac{d}{dx} (1+x^{2})^{1/2} \)
\( y' = \frac{1}{2} (1+x^{2})^{-1/2} \cdot (2x) \)
\( y' = \frac{x}{\sqrt{1+x^{2}}} \)
Now, we need to show that this matches the given differential equation, \( y' = \frac{xy}{1+x^{2}} \).
We know that \( y = \sqrt{1+x^{2}} \), so we can substitute this into our derived \( y' \):
\( y' = \frac{x}{y} \)
To match the target form, multiply the numerator and denominator by \( \sqrt{1+x^2} \):
\( y' = \frac{x}{\sqrt{1+x^2}} \cdot \frac{\sqrt{1+x^2}}{\sqrt{1+x^2}} \)
\( y' = \frac{x \sqrt{1+x^2}}{1+x^2} \)
Since \( y = \sqrt{1+x^2} \), we get:
\( y' = \frac{xy}{1+x^{2}} \)
This confirms that the given function is a solution to the differential equation.
In simple words: We take the derivative of the given function. Then, we substitute the original function back into the derivative. This shows that the derivative matches the given differential equation.

Exam Tip: Sometimes, after finding the derivative, you need to manipulate the expression by substituting the original function back in to match the required form of the differential equation.

Question 5. \( y = Ax : xy' = y, (x \neq 0) \)
Answer: We are given the function \( y = Ax \) (1).
To verify, we need to find \( y' \).
Differentiating equation (1) with respect to \( x \):
\( y' = \frac{d}{dx} (Ax) \)
\( y' = A \) (2)
Now, we want to eliminate the arbitrary constant \( A \). From equation (1), we know that \( A = \frac{y}{x} \).
Substitute this value of \( A \) into equation (2):
\( y' = \frac{y}{x} \)
Multiply both sides by \( x \):
\( xy' = y \)
This matches the given differential equation. Hence, the function \( y = Ax \) is a solution.
In simple words: We differentiate the function to find y'. Then, we use the original equation to replace the constant 'A' with y/x. When we put this into the differentiated equation, it matches the given differential equation.

Exam Tip: When the differential equation does not contain the arbitrary constant, your strategy should be to differentiate the given function and then eliminate the constant using the original function or its derivatives.

Question 6. \( y = x \sin x : xy' = y + x \sqrt{x^2 - y^2} (x \neq 0 \text{ and } x > y \text{ or } x < -y) \)
Answer: We are given the function \( y = x \sin x \) (1).
First, differentiate \( y \) with respect to \( x \) using the product rule:
\( y' = \frac{d}{dx} (x \sin x) \)
\( y' = (1 \cdot \sin x) + (x \cdot \cos x) \)
\( y' = \sin x + x \cos x \) (2)
From equation (1), we can find \( \sin x = \frac{y}{x} \).
Also, we know that \( \sin^2 x + \cos^2 x = 1 \), so \( \cos x = \sqrt{1 - \sin^2 x} \).
Substitute \( \sin x = \frac{y}{x} \) into the expression for \( \cos x \):
\( \cos x = \sqrt{1 - \left(\frac{y}{x}\right)^2} = \sqrt{1 - \frac{y^2}{x^2}} = \sqrt{\frac{x^2 - y^2}{x^2}} = \frac{\sqrt{x^2 - y^2}}{|x|} \)
Since the problem specifies \( x \neq 0 \), we can use \( x \) instead of \( |x| \) in the context of the differential equation.
Substitute \( \sin x \) and \( \cos x \) into equation (2):
\( y' = \frac{y}{x} + x \left(\frac{\sqrt{x^2 - y^2}}{x}\right) \)
\( y' = \frac{y}{x} + \sqrt{x^2 - y^2} \)
Now, multiply both sides by \( x \):
\( xy' = y + x \sqrt{x^2 - y^2} \)
This matches the given differential equation. Thus, the function is a solution.
In simple words: We find the derivative of \( y \). Then we replace \( \sin x \) and \( \cos x \) using the original equation and trigonometric identities. After simplifying, the equation becomes the same as the given differential equation.

Exam Tip: When dealing with trigonometric functions in differential equations, it is often helpful to use fundamental trigonometric identities to simplify expressions or make substitutions.

Question 7. \( xy = \log y + C : y' = \frac{y^2}{1-xy} (xy \neq 1) \)
Answer: We are given the function \( xy = \log y + C \).
To verify, we will differentiate implicitly with respect to \( x \). Apply the product rule on the left side and chain rule on the right side:
\( \frac{d}{dx} (xy) = \frac{d}{dx} (\log y + C) \)
\( (1 \cdot y) + (x \cdot y') = \frac{1}{y} y' + 0 \)
\( y + xy' = \frac{y'}{y} \)
To isolate \( y' \), first multiply the entire equation by \( y \):
\( y^2 + xy y' = y' \)
Move all terms with \( y' \) to one side and terms without \( y' \) to the other:
\( y^2 = y' - xy y' \)
Factor out \( y' \) from the right side:
\( y^2 = y'(1 - xy) \)
Finally, divide by \( (1 - xy) \) to solve for \( y' \):
\( y' = \frac{y^2}{1-xy} \)
This result matches the given differential equation, confirming that the function is a solution.
In simple words: We differentiate both sides of the given equation, remembering to use implicit differentiation for y. We then rearrange the terms to solve for y', and the result matches the differential equation we were given.

Exam Tip: When using implicit differentiation, make sure to apply the chain rule correctly for terms involving \( y \) (e.g., \( \frac{d}{dx}(\log y) = \frac{1}{y} y' \)).

Question 8. \( y - \cos y = x : (y \sin y + \cos y + x)y' = y \)
Answer: We are given the function \( y - \cos y = x \) (1).
To verify, we differentiate implicitly with respect to \( x \):
\( \frac{d}{dx} (y - \cos y) = \frac{d}{dx} (x) \)
\( y' - (-\sin y \cdot y') = 1 \)
\( y' + y' \sin y = 1 \)
Factor out \( y' \):
\( y'(1 + \sin y) = 1 \)
From equation (1), we know that \( x = y - \cos y \). We need to achieve the form \( (y \sin y + \cos y + x)y' = y \).
Let's try to manipulate the derived \( y'(1 + \sin y) = 1 \).
Multiply both sides by \( y \):
\( y y'(1 + \sin y) = y \)
\( (y + y \sin y)y' = y \)
We also know \( x = y - \cos y \), which means \( y = x + \cos y \).
Substitute \( y \) in the left side's first term: \( (x + \cos y + y \sin y)y' = y \)
This matches the given differential equation. Hence, the function is a solution.
In simple words: We differentiate the function implicitly. Then, we multiply by y and use the original equation to substitute for y. This helps us get the same form as the required differential equation.

Exam Tip: Implicit differentiation often requires careful algebraic manipulation after differentiation to match the target differential equation. Substituting the original function back into the derived expression can be a key step.

Question 9. \( x + y = \tan^{-1}y : y^2y' + y^2 + 1 = 0 \)
Answer: We have the function \( x + y = \tan^{-1}y \).
Differentiate implicitly with respect to \( x \):
\( \frac{d}{dx} (x) + \frac{d}{dx} (y) = \frac{d}{dx} (\tan^{-1}y) \)
\( 1 + y' = \frac{1}{1+y^2} y' \)
To clear the denominator, multiply the entire equation by \( (1+y^2) \):
\( (1+y^2)(1 + y') = y' \)
\( 1 + y^2 + y'(1+y^2) = y' \)
\( 1 + y^2 + y' + y^2y' = y' \)
Now, subtract \( y' \) from both sides:
\( 1 + y^2 + y^2y' = 0 \)
Rearrange the terms to match the required form:
\( y^2y' + y^2 + 1 = 0 \)
This result matches the given differential equation. Therefore, the function is a solution.
In simple words: We differentiate the equation implicitly, making sure to handle the inverse tangent function correctly. Then, we rearrange the terms algebraically until they match the given differential equation.

Exam Tip: Remember the derivative of \( \tan^{-1}y \) with respect to \( x \) is \( \frac{1}{1+y^2} \frac{dy}{dx} \). Clear denominators early in implicit differentiation to simplify the algebra.

Question 10. \( y = \sqrt{a^2-x^2} \quad x \in (-a, a) : x + y \frac{dy}{dx} = 0 \)
Answer: We are given the function \( y = \sqrt{a^2-x^2} \).
To simplify differentiation, square both sides of the equation:
\( y^2 = a^2 - x^2 \)
Rearrange the terms to get:
\( x^2 + y^2 = a^2 \)
Now, differentiate this equation implicitly with respect to \( x \):
\( \frac{d}{dx} (x^2) + \frac{d}{dx} (y^2) = \frac{d}{dx} (a^2) \)
\( 2x + 2y \frac{dy}{dx} = 0 \)
Divide the entire equation by 2:
\( x + y \frac{dy}{dx} = 0 \)
This matches the given differential equation. Therefore, the function \( y = \sqrt{a^2-x^2} \) is a solution.
In simple words: We square the function to get rid of the square root. Then we differentiate this new equation indirectly. After simplifying, we get the same differential equation, proving that the function is a solution.

Exam Tip: Squaring both sides before differentiating can simplify expressions involving square roots, making implicit differentiation easier to perform.

Choose the correct answers in the following questions 11 and 12:

Question 11. The number of arbitrary constants in the general solution of a differential equation of fourth order is
(a) 0
(b) 2
(c) 3
(d) 4
Answer: (d) 4
In simple words: The number of unknown constants in a differential equation's general solution is always the same as the highest order of the derivative in that equation. Since this is a fourth-order equation, there will be four constants.

Exam Tip: Remember the fundamental rule: the order of a differential equation directly corresponds to the number of arbitrary constants in its general solution.

Question 12. The number of arbitrary constants in the particular solution of a differential equation of third order is
(a) 3
(b) 2
(c) 1
(d) 0
Answer: (d) 0
In simple words: A particular solution is when all the arbitrary constants in the general solution have been given specific values. Because of this, a particular solution will not have any arbitrary constants remaining.

Exam Tip: Differentiate between general solutions (which contain arbitrary constants) and particular solutions (where these constants have been determined by initial or boundary conditions and thus have specific numerical values, meaning no arbitrary constants remain).