RBSE Solutions Class 12 Maths Chapter 12 Differential Equation Exercise 12.4

Solve the Following Differential Equations:

Question 1. (e^x + 1) cos x dx + e^y sin x dy = 0.
Answer: The given equation is:
\( (e^x + 1) \cos x dx + e^y \sin x dy = 0 \)
We rearrange the terms to separate variables:
\( (e^x + 1) \cos x dx = - e^y \sin x dy \)
Now, divide both sides to group \( x \) terms with \( dx \) and \( y \) terms with \( dy \):
\( \frac{\cos x}{\sin x} dx = - \frac{e^y}{e^y + 1} dy \)
Next, integrate both sides of the equation:
\( \int \frac{\cos x}{\sin x} dx = - \int \frac{e^y}{e^y + 1} dy \)
This gives us:
\( \log |\sin x| = - \log |e^y + 1| + \log C \)
Combining the logarithm terms:
\( \log |\sin x| + \log |e^y + 1| = \log C \)
Using logarithm properties, \( \log A + \log B = \log (AB) \):
\( \log |\sin x (e^y + 1)| = \log C \)
Removing the logarithm from both sides:
\( \sin x (e^y + 1) = C \)
This is the general solution for the given differential equation, where C is the integration constant.
In simple words: We separate the parts with 'x' and 'dx' from the parts with 'y' and 'dy'. Then, we integrate both sides to find a solution. The goal is to get 'x' and 'y' on different sides of the equation before integrating.

🎯 Exam Tip: When solving differential equations, the first crucial step is always to successfully separate the variables. Any error in this separation will lead to an incorrect integration and solution.

Question 3. (x + 1) \( \frac{dy}{dx} \) = 2xy
Answer: The given differential equation is:
\( (x + 1) \frac{dy}{dx} = 2xy \)
First, we separate the variables by moving all \( y \) terms to one side and all \( x \) terms to the other:
\( \frac{dy}{y} = \frac{2x}{x + 1} dx \)
Now, integrate both sides:
\( \int \frac{1}{y} dy = \int \frac{2x}{x + 1} dx \)
For the right side integral, we can rewrite the numerator:
\( \int \frac{1}{y} dy = \int \frac{2(x + 1 - 1)}{x + 1} dx \)
\( \log |y| = \int \left( \frac{2(x+1)}{x+1} - \frac{2}{x+1} \right) dx \)
\( \log |y| = \int \left( 2 - \frac{2}{x+1} \right) dx \)
Integrating both terms on the right:
\( \log |y| = 2 \int 1 dx - 2 \int \frac{1}{x+1} dx \)
\( \log |y| = 2x - 2 \log |x + 1| + C \)
We can simplify this further:
\( \log |y| = 2(x - \log |x + 1|) + C \)
This is the general solution for the given differential equation.
In simple words: We separate the 'y' and 'dy' parts from the 'x' and 'dx' parts. Then, we integrate both sides of the equation. We rewrite the 'x' part to make it easier to integrate, and finally, we combine the terms with a constant 'C'.

🎯 Exam Tip: When integrating fractions like \( \frac{2x}{x+1} \), try to adjust the numerator to match the denominator plus a constant, which often simplifies the integration greatly.

Question 5. (sin x + cos x) dy + (cos x - sin x) dx = 0.
Answer: The given differential equation is:
\( (\sin x + \cos x) dy + (\cos x - \sin x) dx = 0 \)
First, we rearrange the equation to separate the variables:
\( (\sin x + \cos x) dy = - (\cos x - \sin x) dx \)
\( (\sin x + \cos x) dy = (\sin x - \cos x) dx \)
Now, we group terms with \( y \) and \( dy \) on one side and terms with \( x \) and \( dx \) on the other:
\( dy = \frac{\sin x - \cos x}{\sin x + \cos x} dx \)
Next, integrate both sides:
\( \int dy = \int \frac{\sin x - \cos x}{\sin x + \cos x} dx \)
Let \( t = \sin x + \cos x \). Then, \( dt = (\cos x - \sin x) dx = - (\sin x - \cos x) dx \).
So, \( -dt = (\sin x - \cos x) dx \).
Substituting these into the integral on the right:
\( y = \int \frac{-dt}{t} \)
\( y = - \int \frac{1}{t} dt \)
\( y = - \log |t| + \log C \)
Now, substitute back \( t = \sin x + \cos x \):
\( y = - \log |\sin x + \cos x| + \log C \)
This can be written as:
\( y = \log \left| \frac{C}{\sin x + \cos x} \right| \)
To remove the logarithm, we use the exponential function on both sides:
\( e^y = \frac{C}{\sin x + \cos x} \)
Therefore, the solution is:
\( e^y (\sin x + \cos x) = C \)
In simple words: We move the 'x' terms to one side with 'dx' and 'y' terms to the other side with 'dy'. Then, we integrate both sides. We use a substitution to make the integration easier, which helps us solve for 'y'.

🎯 Exam Tip: When the numerator is a derivative of the denominator (or a multiple of it), a simple substitution can quickly solve the integral. Always look for this pattern!

Question 6. \( \frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}} \)
Answer: The given differential equation is:
\( \frac{dy}{dx} = \frac{3e^{2x} + 3e^{4x}}{e^x + e^{-x}} \)
First, we simplify the right-hand side. Factor out \( 3e^{2x} \) from the numerator:
\( \frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{e^x + e^{-x}} \)
To make the denominator look like \( e^x(1 + e^{2x}) \), multiply the denominator by \( e^x \) and the numerator by \( e^x \):
\( \frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{e^x(1 + e^{-2x})} \)
This step seems wrong. Let's restart the simplification by factoring \( e^{-x} \) from the denominator:
\( e^x + e^{-x} = e^{-x} (e^{2x} + 1) \)
So, the equation becomes:
\( \frac{dy}{dx} = \frac{3e^{2x}(1 + e^{2x})}{e^{-x}(1 + e^{2x})} \)
We can cancel out \( (1 + e^{2x}) \) from the numerator and denominator:
\( \frac{dy}{dx} = \frac{3e^{2x}}{e^{-x}} \)
Using the rule \( \frac{a^m}{a^n} = a^{m-n} \):
\( \frac{dy}{dx} = 3e^{2x - (-x)} \)
\( \frac{dy}{dx} = 3e^{3x} \)
Now, we separate the variables:
\( dy = 3e^{3x} dx \)
Next, integrate both sides:
\( \int dy = \int 3e^{3x} dx \)
\( y = 3 \frac{e^{3x}}{3} + C \)
\( y = e^{3x} + C \)
This is the general solution to the differential equation. Exponential functions are commonly used in modeling growth or decay processes.
In simple words: First, we simplify the right side of the equation by factoring common parts and canceling them out. This makes the equation much simpler. Then, we move 'dx' to the right side and integrate both sides to find 'y'.

🎯 Exam Tip: Always look for opportunities to simplify complex expressions before integration, as this can turn a difficult problem into a straightforward one. Factoring is key here.

Question 7. \( \sec^2 x \tan y dy + \sec^2 y \tan x dx = 0. \)
Answer: The given differential equation is:
\( \sec^2 x \tan y dy + \sec^2 y \tan x dx = 0 \)
First, rearrange the terms to separate variables:
\( \sec^2 x \tan y dy = - \sec^2 y \tan x dx \)
Divide both sides by \( \tan y \sec^2 y \tan x \sec^2 x \) (or similar terms) to group \( y \) terms with \( dy \) and \( x \) terms with \( dx \):
\( \frac{\sec^2 x}{\tan x} dx + \frac{\sec^2 y}{\tan y} dy = 0 \)
Now, integrate both sides:
\( \int \frac{\sec^2 x}{\tan x} dx + \int \frac{\sec^2 y}{\tan y} dy = \int 0 dx \)
For both integrals, we can use a simple substitution. Let \( u = \tan x \), then \( du = \sec^2 x dx \). Let \( v = \tan y \), then \( dv = \sec^2 y dy \).
So, the integrals become:
\( \int \frac{1}{u} du + \int \frac{1}{v} dv = C_1 \)
\( \log |\tan x| + \log |\tan y| = C_1 \)
Using the logarithm property \( \log A + \log B = \log (AB) \):
\( \log |\tan x \tan y| = C_1 \)
To remove the logarithm, we take the exponential of both sides:
\( |\tan x \tan y| = e^{C_1} \)
Let \( C = e^{C_1} \) (which is a positive constant).
\( \tan x \tan y = C \)
This is the general solution. This type of equation is common in physics problems involving fields.
In simple words: We separate the 'x' parts with 'dx' and 'y' parts with 'dy'. Then, we integrate both sides. We use a trick where we notice that the top part of each fraction is the derivative of the bottom part, which makes integration very easy with logarithms.

🎯 Exam Tip: Recognizing that \( \frac{f'(x)}{f(x)} \) integrates to \( \log |f(x)| \) is a powerful shortcut for many differential equation problems. Practice identifying this pattern.

Question 8. \( \frac{dy}{dx} = \frac{x(2\log x + 1)}{\sin y + y \cos y} \)
Answer: The given differential equation is:
\( \frac{dy}{dx} = \frac{x(2\log x + 1)}{\sin y + y \cos y} \)
First, we separate the variables by cross-multiplying:
\( (\sin y + y \cos y) dy = x(2\log x + 1) dx \)
Now, integrate both sides:
\( \int (\sin y + y \cos y) dy = \int (2x\log x + x) dx \)
Consider the left-hand side integral \( \int (\sin y + y \cos y) dy \):
We can see that \( \frac{d}{dy}(y \sin y) = 1 \cdot \sin y + y \cdot \cos y = \sin y + y \cos y \).
So, \( \int (\sin y + y \cos y) dy = y \sin y \).
Now, consider the right-hand side integral \( \int (2x\log x + x) dx \):
We can also see that \( \frac{d}{dx}(x^2 \log x) = 2x \log x + x^2 \cdot \frac{1}{x} = 2x \log x + x \).
So, \( \int (2x\log x + x) dx = x^2 \log x \).
Combining both sides with the constant of integration \( C \):
\( y \sin y = x^2 \log x + C \)
This is the general solution for the differential equation. Recognizing patterns for derivatives can greatly simplify integration.
In simple words: We separate the 'y' and 'dy' terms from the 'x' and 'dx' terms. Then, we integrate both sides. For this problem, both sides are special because they are exact derivatives of simple functions, which makes them very easy to integrate.

🎯 Exam Tip: Always be on the lookout for expressions that are exact derivatives of common functions. Sometimes, a complex-looking sum can be the derivative of a simple product, like \( \frac{d}{dx}(uv) = u'v + uv' \).

Question 9. (1 + cos x) dy = (1 - cos x) dx
Answer: The given differential equation is:
\( (1 + \cos x) dy = (1 - \cos x) dx \)
First, we separate the variables:
\( dy = \frac{1 - \cos x}{1 + \cos x} dx \)
Now, we use trigonometric identities to simplify the right-hand side. We know that \( 1 - \cos x = 2\sin^2 \left( \frac{x}{2} \right) \) and \( 1 + \cos x = 2\cos^2 \left( \frac{x}{2} \right) \).
So, the expression becomes:
\( dy = \frac{2\sin^2 \left( \frac{x}{2} \right)}{2\cos^2 \left( \frac{x}{2} \right)} dx \)
\( dy = \tan^2 \left( \frac{x}{2} \right) dx \)
We also know the identity \( \tan^2 \theta = \sec^2 \theta - 1 \). Applying this:
\( dy = \left( \sec^2 \left( \frac{x}{2} \right) - 1 \right) dx \)
Next, integrate both sides:
\( \int dy = \int \left( \sec^2 \left( \frac{x}{2} \right) - 1 \right) dx \)
\( y = \int \sec^2 \left( \frac{x}{2} \right) dx - \int 1 dx \)
For \( \int \sec^2 \left( \frac{x}{2} \right) dx \), let \( u = \frac{x}{2} \), then \( du = \frac{1}{2} dx \implies dx = 2 du \).
So, \( \int \sec^2 u (2du) = 2 \int \sec^2 u du = 2 \tan u = 2 \tan \left( \frac{x}{2} \right) \).
Therefore, the full solution is:
\( y = 2 \tan \left( \frac{x}{2} \right) - x + C \)
This is the general solution for the differential equation. Using trigonometric identities often simplifies complex expressions before integration.
In simple words: We first separate the 'dy' and 'dx' parts. Then, we use special math rules for sine and cosine to make the 'x' part much simpler. This simpler form is easier to integrate. Finally, we integrate both sides to find 'y'.

🎯 Exam Tip: Always be aware of common trigonometric identities. They are often essential for simplifying expressions in calculus, especially before integration, and can transform a difficult problem into an easy one.