RBSE Solutions Class 12 Maths Chapter 12 Differential Equation Exercise 12.3

Question 1. Prove that \( y^2 = 4a (x+a) \) is the solution of: \( y \left[1-\left(\frac{dy}{dx}\right)^2\right] = 2x \frac{dy}{dx} \)
Answer: We are given the equation \( y^2 = 4a (x+a) \).
First, we differentiate both sides of the equation with respect to \( x \):
\( \frac{d}{dx} (y^2) = \frac{d}{dx} [4a(x+a)] \)
\( 2y \frac{dy}{dx} = 4a (1+0) \)
\( 2y \frac{dy}{dx} = 4a \)
Now, we can find the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{4a}{2y} \)
\( \frac{dy}{dx} = \frac{2a}{y} \)
Next, we square both sides of this derivative:
\( \left(\frac{dy}{dx}\right)^2 = \left(\frac{2a}{y}\right)^2 \)
\( \left(\frac{dy}{dx}\right)^2 = \frac{4a^2}{y^2} \)
Now, let's substitute this into the left side of the differential equation we need to prove, which is \( y \left[1-\left(\frac{dy}{dx}\right)^2\right] \):
\( y \left[1-\frac{4a^2}{y^2}\right] \)
\( y \left[\frac{y^2-4a^2}{y^2}\right] \)
\( \frac{y^2-4a^2}{y} \)
From the original given equation, \( y^2 = 4a(x+a) = 4ax + 4a^2 \).
This means \( y^2 - 4a^2 = 4ax \).
Substitute this back into the expression:
\( \frac{4ax}{y} \)
Now, let's look at the right side of the differential equation: \( 2x \frac{dy}{dx} \).
Substitute \( \frac{dy}{dx} = \frac{2a}{y} \) into this:
\( 2x \left(\frac{2a}{y}\right) \)
\( \frac{4ax}{y} \)
Since the left side \( \left(\frac{y^2-4a^2}{y}\right) \) equals \( \frac{4ax}{y} \) and the right side \( \left(2x \frac{dy}{dx}\right) \) also equals \( \frac{4ax}{y} \), both sides are equal. This confirms that the given equation is indeed the solution to the differential equation. The process of differentiation and substitution helps verify if a proposed solution fits the differential equation.
In simple words: First, we took the given equation and differentiated it to find what \( \frac{dy}{dx} \) is. Then, we put this value into the differential equation we needed to prove. After simplifying, we saw that both sides of the equation became the same, showing that the initial equation was indeed its solution.

🎯 Exam Tip: When proving a solution, always start by differentiating the given solution and then substitute the derivatives back into the differential equation. Ensure both sides match to confirm the proof.

Question 2. Prove that \( y = ae^{-2x} + be^x \) is the solution: \( \frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0 \).
Answer: We are given the equation \( y = ae^{-2x} + be^x \).
First, we find the first derivative, \( \frac{dy}{dx} \), by differentiating \( y \) with respect to \( x \):
\( \frac{dy}{dx} = \frac{d}{dx} (ae^{-2x} + be^x) \)
\( \frac{dy}{dx} = a(-2e^{-2x}) + b(e^x) \)
\( \frac{dy}{dx} = -2ae^{-2x} + be^x \) (This is equation (1))
Next, we find the second derivative, \( \frac{d^2y}{dx^2} \), by differentiating \( \frac{dy}{dx} \) with respect to \( x \):
\( \frac{d^2y}{dx^2} = \frac{d}{dx} (-2ae^{-2x} + be^x) \)
\( \frac{d^2y}{dx^2} = -2a(-2e^{-2x}) + b(e^x) \)
\( \frac{d^2y}{dx^2} = 4ae^{-2x} + be^x \) (This is equation (2))
Now, we substitute the expressions for \( y \), \( \frac{dy}{dx} \), and \( \frac{d^2y}{dx^2} \) into the left-hand side of the differential equation: \( \frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y \). This is a standard method to verify differential equations.
\( (4ae^{-2x} + be^x) + (-2ae^{-2x} + be^x) - 2(ae^{-2x} + be^x) \)
Open the brackets and group similar terms:
\( 4ae^{-2x} + be^x - 2ae^{-2x} + be^x - 2ae^{-2x} - 2be^x \)
Group the \( ae^{-2x} \) terms:
\( (4ae^{-2x} - 2ae^{-2x} - 2ae^{-2x}) + (be^x + be^x - 2be^x) \)
\( (4-2-2)ae^{-2x} + (1+1-2)be^x \)
\( 0 \cdot ae^{-2x} + 0 \cdot be^x \)
\( 0 + 0 \)
\( = 0 \)
Since the left-hand side equals 0, which is the right-hand side of the given differential equation, we have successfully proven that \( y = ae^{-2x} + be^x \) is the solution to the differential equation \( \frac{d^2y}{dx^2} + \frac{dy}{dx} - 2y = 0 \).
In simple words: We took the proposed answer and found its first and second derivatives. Then we put these derivatives and the original answer into the differential equation. When we added everything up, it all cancelled out to zero, which means the proposed answer is correct.

🎯 Exam Tip: Always calculate the derivatives one by one (first then second) to avoid mistakes. Be careful with signs and constants when differentiating exponential terms.

Question 3. Prove that \( y = \frac{c-x}{1+cx} \) is the solution of \( (1+x^2)\frac{dy}{dx} + (1+y^2) = 0 \).
Answer: We are given the equation \( y = \frac{c-x}{1+cx} \).
First, we differentiate \( y \) with respect to \( x \) using the quotient rule, \( \frac{d}{dx} \left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \).
Here, \( u = c-x \) and \( v = 1+cx \). So, \( u' = -1 \) and \( v' = c \).
\( \frac{dy}{dx} = \frac{(-1)(1+cx) - (c-x)(c)}{(1+cx)^2} \)
\( \frac{dy}{dx} = \frac{-1-cx - (c^2-cx)}{(1+cx)^2} \)
\( \frac{dy}{dx} = \frac{-1-cx - c^2+cx}{(1+cx)^2} \)
\( \frac{dy}{dx} = \frac{-(1+c^2)}{(1+cx)^2} \)
Now, we need to show that \( (1+x^2)\frac{dy}{dx} + (1+y^2) = 0 \).
Let's substitute \( \frac{dy}{dx} \) into the left-hand side of the differential equation:
\( (1+x^2) \left[ \frac{-(1+c^2)}{(1+cx)^2} \right] + (1+y^2) \)
We also know that \( y = \frac{c-x}{1+cx} \). We can rearrange this to express \( c \) in terms of \( x \) and \( y \).
\( y(1+cx) = c-x \)
\( y+cxy = c-x \)
\( x+y = c-cxy \)
\( x+y = c(1-xy) \)
\( c = \frac{x+y}{1-xy} \)
Now, let's substitute this expression for \( c \) into \( 1+c^2 \):
\( 1+c^2 = 1 + \left(\frac{x+y}{1-xy}\right)^2 \)
\( 1+c^2 = 1 + \frac{(x+y)^2}{(1-xy)^2} \)
\( 1+c^2 = \frac{(1-xy)^2 + (x+y)^2}{(1-xy)^2} \)
Expand the squares:
\( 1+c^2 = \frac{1-2xy+x^2y^2 + x^2+2xy+y^2}{(1-xy)^2} \)
\( 1+c^2 = \frac{1+x^2+y^2+x^2y^2}{(1-xy)^2} \)
\( 1+c^2 = \frac{(1+x^2) + y^2(1+x^2)}{(1-xy)^2} \)
\( 1+c^2 = \frac{(1+x^2)(1+y^2)}{(1-xy)^2} \)
Now we substitute \( \frac{dy}{dx} = \frac{-(1+c^2)}{(1+cx)^2} \) and \( 1+c^2 = \frac{(1+x^2)(1+y^2)}{(1-xy)^2} \) into the differential equation.
The term \( (1+cx)^2 \) can also be related to \( (1-xy)^2 \).
We have \( y(1+cx) = c-x \), so \( 1+cx = \frac{c-x}{y} \).
Then \( (1+cx)^2 = \frac{(c-x)^2}{y^2} \).
This requires more complex substitutions. A simpler way is to use the expression for \( c \) in the \( \frac{dy}{dx} \) formula directly.
Let's substitute \( c = \frac{x+y}{1-xy} \) directly into \( \frac{dy}{dx} = \frac{-(1+c^2)}{(1+cx)^2} \).
First, substitute \( c \) into \( 1+cx \):
\( 1+cx = 1 + \left(\frac{x+y}{1-xy}\right)x \)
\( 1+cx = \frac{1-xy + x^2+xy}{1-xy} \)
\( 1+cx = \frac{1+x^2}{1-xy} \)
So, \( (1+cx)^2 = \frac{(1+x^2)^2}{(1-xy)^2} \).
Now substitute \( 1+c^2 \) and \( (1+cx)^2 \) back into the expression for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = \frac{-\left(\frac{(1+x^2)(1+y^2)}{(1-xy)^2}\right)}{\left(\frac{(1+x^2)^2}{(1-xy)^2}\right)} \)
\( \frac{dy}{dx} = \frac{-(1+x^2)(1+y^2)}{(1+x^2)^2} \)
\( \frac{dy}{dx} = \frac{-(1+y^2)}{1+x^2} \)
Now substitute this into the differential equation \( (1+x^2)\frac{dy}{dx} + (1+y^2) = 0 \):
\( (1+x^2) \left[ \frac{-(1+y^2)}{1+x^2} \right] + (1+y^2) \)
\( -(1+y^2) + (1+y^2) \)
\( = 0 \)
Thus, the given equation \( y = \frac{c-x}{1+cx} \) is a solution to the differential equation. The process of finding 'c' in terms of 'x' and 'y' is a key step in simplifying the expression.
In simple words: We first found the derivative of \( y \) from the given equation. This derivative included the constant 'c'. We then found a way to write 'c' using 'x' and 'y', and substituted it back into the derivative. Finally, we put this simplified derivative into the main equation we wanted to prove. It all added up to zero, showing that the original equation is a correct solution.

🎯 Exam Tip: For problems involving implicit constants (like 'c'), it's often helpful to express the constant in terms of x and y from the original equation before substituting back into the derivative, simplifying the algebra significantly.

Question 4. Prove that \( y = a \cos (\log x) + b \sin (\log x) \) is the solution of \( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \).
Answer: We are given the equation \( y = a \cos (\log x) + b \sin (\log x) \).
First, we find the first derivative, \( \frac{dy}{dx} \), by differentiating \( y \) with respect to \( x \). Remember the chain rule for derivatives.
\( \frac{dy}{dx} = \frac{d}{dx} [a \cos (\log x) + b \sin (\log x)] \)
\( \frac{dy}{dx} = a \left( -\sin(\log x) \cdot \frac{1}{x} \right) + b \left( \cos(\log x) \cdot \frac{1}{x} \right) \)
\( \frac{dy}{dx} = \frac{1}{x} [-a \sin(\log x) + b \cos(\log x)] \)
Now, multiply by \( x \) to get \( x \frac{dy}{dx} \):
\( x \frac{dy}{dx} = -a \sin(\log x) + b \cos(\log x) \)
Next, we find the second derivative, \( \frac{d^2y}{dx^2} \). We differentiate \( x \frac{dy}{dx} \) using the product rule: \( (uv)' = u'v + uv' \).
Let \( u = x \) and \( v = \frac{dy}{dx} \). Then \( u' = 1 \) and \( v' = \frac{d^2y}{dx^2} \).
So, \( \frac{d}{dx} \left( x \frac{dy}{dx} \right) = 1 \cdot \frac{dy}{dx} + x \frac{d^2y}{dx^2} \).
Also, differentiate the right side: \( \frac{d}{dx} [-a \sin(\log x) + b \cos(\log x)] \)
\( = -a \left( \cos(\log x) \cdot \frac{1}{x} \right) + b \left( -\sin(\log x) \cdot \frac{1}{x} \right) \)
\( = -\frac{a}{x} \cos(\log x) - \frac{b}{x} \sin(\log x) \)
So, we have:
\( \frac{dy}{dx} + x \frac{d^2y}{dx^2} = -\frac{1}{x} [a \cos(\log x) + b \sin(\log x)] \)
We know that \( y = a \cos (\log x) + b \sin (\log x) \). Substitute \( y \) into the right side:
\( \frac{dy}{dx} + x \frac{d^2y}{dx^2} = -\frac{y}{x} \)
Now, multiply the entire equation by \( x \):
\( x \frac{dy}{dx} + x^2 \frac{d^2y}{dx^2} = -y \)
Rearrange the terms to match the differential equation:
\( x^2 \frac{d^2y}{dx^2} + x \frac{dy}{dx} + y = 0 \)
This matches the given differential equation, proving that \( y = a \cos (\log x) + b \sin (\log x) \) is its solution. Understanding the chain rule for derivatives involving \( \log x \) is crucial here.
In simple words: We took the given equation and found its first and second derivatives. We then multiplied by 'x' and 'x squared' as needed by the problem. When we added all these parts together, they cancelled out perfectly to zero, which means the original equation is the correct solution.

🎯 Exam Tip: Remember to apply the chain rule correctly when differentiating functions of \( \log x \). Pay close attention to the product rule when differentiating terms like \( x \frac{dy}{dx} \).

Question 5. Prove that \( xy = \log y + c \) is the solution of \( \frac{dy}{dx} = \frac{y^2}{1-xy} (xy \neq 1) \).
Answer: We are given the equation \( xy = \log y + c \).
To prove this is the solution, we need to find \( \frac{dy}{dx} \) from the given equation by implicitly differentiating both sides with respect to \( x \).
\( \frac{d}{dx} (xy) = \frac{d}{dx} (\log y + c) \)
For the left side, \( \frac{d}{dx} (xy) \), use the product rule \( (uv)' = u'v + uv' \).
Here, \( u = x \) and \( v = y \). So, \( u' = 1 \) and \( v' = \frac{dy}{dx} \).
\( 1 \cdot y + x \cdot \frac{dy}{dx} = y + x \frac{dy}{dx} \)
For the right side, \( \frac{d}{dx} (\log y + c) \), differentiate term by term.
\( \frac{d}{dx} (\log y) = \frac{1}{y} \cdot \frac{dy}{dx} \) (using the chain rule)
\( \frac{d}{dx} (c) = 0 \) (since \( c \) is a constant)
So, combining both sides:
\( y + x \frac{dy}{dx} = \frac{1}{y} \frac{dy}{dx} \)
Now, we need to isolate \( \frac{dy}{dx} \). Move all terms with \( \frac{dy}{dx} \) to one side and other terms to the other side.
\( x \frac{dy}{dx} - \frac{1}{y} \frac{dy}{dx} = -y \)
Factor out \( \frac{dy}{dx} \) from the left side:
\( \frac{dy}{dx} \left( x - \frac{1}{y} \right) = -y \)
Simplify the term inside the parenthesis:
\( \frac{dy}{dx} \left( \frac{xy-1}{y} \right) = -y \)
Now, solve for \( \frac{dy}{dx} \):
\( \frac{dy}{dx} = -y \cdot \frac{y}{xy-1} \)
\( \frac{dy}{dx} = \frac{-y^2}{xy-1} \)
We can also write this as:
\( \frac{dy}{dx} = \frac{y^2}{-(xy-1)} \)
\( \frac{dy}{dx} = \frac{y^2}{1-xy} \)
This matches the differential equation we were asked to prove, given the condition \( (xy \neq 1) \). This condition is important because it prevents division by zero. The implicit differentiation method is very useful for equations where y is not easily isolated.
In simple words: We started with the given equation and used implicit differentiation to find \( \frac{dy}{dx} \). This means we treated \( y \) as a function of \( x \) and applied the chain rule. After rearranging the terms, we got the exact differential equation we wanted to prove, showing the original equation is its solution.

🎯 Exam Tip: Remember to use the product rule for terms like \( xy \) and the chain rule for terms like \( \log y \) when performing implicit differentiation. Carefully rearrange the equation to isolate \( \frac{dy}{dx} \).