CBSE Class 12 Mathematics HOTs Differential Equations Set 02

Very Short Answer Type Questions

Question. Find the order and degree of the following differential equation \( x \sqrt{1-y^2} dx + y \sqrt{1-x^2} dy = 0 \).
Answer: The given equation can be written as \( x\sqrt{1-y^2}dx = -y\sqrt{1-x^2}dy \Rightarrow \frac{dy}{dx} = \frac{-x\sqrt{1-y^2}}{y\sqrt{1-x^2}} \)
Since the highest order derivative is \( \frac{dy}{dx} \), its order is 1.
It is a polynomial equation in \( \frac{dy}{dx} \) and the highest power of \( \frac{dy}{dx} \) is 1.
So, its degree is 1.

Question. Find the order and degree of the differential equation \( x - \sin\left(\frac{dy}{dx}\right) = 0 \).
Answer: Given differential equation is \( x - \sin\left(\frac{dy}{dx}\right) = 0 \).
Since the highest order derivative is \( \frac{dy}{dx} \), its order is 1.
Now, if possible, convert the given differential equation into a polynomial equation of derivatives.
Consider, \( x - \sin\left(\frac{dy}{dx}\right) = 0 \Rightarrow \sin\left(\frac{dy}{dx}\right) = x \)
\( \Rightarrow \frac{dy}{dx} = \sin^{-1} x \), which is a polynomial in \( \frac{dy}{dx} \).
Also, the highest power of \( \frac{dy}{dx} \) is 1.
So, its degree is 1.

Question. Find the degree of the differential equation \( 1 + \left(\frac{dy}{dx}\right)^2 = x \).
Answer: We have, \( 1 + \left(\frac{dy}{dx}\right)^2 = x \)
Since the highest order derivative is \( \frac{dy}{dx} \) and the highest power of \( \frac{dy}{dx} \) is 2,
\( \therefore \) Degree = 2.

Question. Find the order and the degree of the differential equation \( x^2 \frac{d^2y}{dx^2} = \left\{1 + \left(\frac{dy}{dx}\right)^2\right\}^4 \).
Answer: Given differential equation is \( x^2 \frac{d^2y}{dx^2} = \left\{1 + \left(\frac{dy}{dx}\right)^2\right\}^4 \).
Since the highest order derivative occurring in the differential equation is \( \frac{d^2y}{dx^2} \), the order is 2. As the given equation can be expressed as a polynomial in derivatives, its degree is 1, which is the power of \( \frac{d^2y}{dx^2} \).

Question. Find the order and degree (if defined) of the differential equation \( \frac{d^2y}{dx^2} + x \left(\frac{dy}{dx}\right)^2 = 2x^2 \log \left(\frac{d^2y}{dx^2}\right) \).
Answer: Since the highest order derivative occurring in the differential equation is \( \frac{d^2y}{dx^2} \), the order is 2. As the differential equation is not a polynomial in derivatives, its degree is not defined.

Question. Write the degree of the differential equation \( \left(\frac{dy}{dx}\right)^4 + 6x \frac{d^2y}{dx^2} = 0 \).
Answer: Given differential equation is \( \left(\frac{dy}{dx}\right)^4 + 6x \frac{d^2y}{dx^2} = 0 \).
Here, the highest order derivative is \( \frac{d^2y}{dx^2} \), whose degree is one. So, the degree of the differential equation is 1.

Question. Solve the differential equation \( \frac{dy}{dx} = \frac{1+y^2}{y^3} \).
Answer: We have, \( \frac{dy}{dx} = \frac{1+y^2}{y^3} \Rightarrow \frac{dx}{dy} = \frac{y^3}{1+y^2} \Rightarrow dx = \frac{y^3}{1+y^2} dy \).
On integrating both sides, we get
\( \int dx = \int \frac{y^3}{1+y^2} dy \)
\( \Rightarrow x = \int \left( y - \frac{y}{1+y^2} \right) dy \)
\( \Rightarrow x = \frac{y^2}{2} - \frac{1}{2} \log(1+y^2) + C \), which is the required solution.

Question. Solve the differential equation \( \frac{dy}{dx} = x^3 e^{-2y} \).
Answer: Given differential equation is \( \frac{dy}{dx} = x^3 e^{-2y} \Rightarrow e^{2y} dy = x^3 dx \).
On integrating both sides, we get
\( \int e^{2y} dy = \int x^3 dx \)
\( \Rightarrow \frac{e^{2y}}{2} = \frac{x^4}{4} + C \), which is the required solution.

Question. How many arbitrary constants are there in the particular solution of the differential equation \( \frac{dy}{dx} = -4xy^2; y(0)=1 \)?
Answer: In the particular solution of a differential equation, the number of arbitrary constants is always zero.

Question. Write the solution of the differential equation \( \frac{dy}{dx} = 2^{-y} \).
Answer: Given differential equation is \( \frac{dy}{dx} = 2^{-y} \).
On separating the variables, we get
\( 2^y dy = dx \)
On integrating both sides, we get
\( \int 2^y dy = \int dx \)
\( \Rightarrow \frac{2^y}{\log 2} = x + C_1 \)
\( \Rightarrow 2^y = x \log 2 + C_1 \log 2 \)
\( \therefore 2^y = x \log 2 + C \), where \( C = C_1 \log 2 \).

Question. For what value of \( n \), the following is a homogeneous differential equation \( \frac{dy}{dx} = \frac{x^3 - y^n}{x^2 y + xy^2} \)?
Answer: For the given differential equation to be homogeneous, the degree of \( (x^3 - y^n) \) and \( (x^2 y + xy^2) \) must be the same, and also the degree of \( x^3 \) and \( y^n \) must be the same.
\( \therefore n = 3 \).

Question. Find the integrating factor of the differential equation \( x \frac{dy}{dx} + 2y = x^2 \).
Answer: We have, \( x \frac{dy}{dx} + 2y = x^2 \Rightarrow \frac{dy}{dx} + \frac{2}{x}y = x \), which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \frac{2}{x} \) and \( Q = x \).
\( \therefore \) Integrating factor (IF) \( = e^{\int P dx} = e^{\int \frac{2}{x} dx} = e^{2 \log x} = e^{\log x^2} = x^2 \).

Question. Write the integrating factor of the following differential equation \( (1+y^2) + (4xy - \cot y) \frac{dy}{dx} = 0 \).
Answer: Given differential equation is \( (1+y^2) + (4xy - \cot y) \frac{dy}{dx} = 0 \).
The above equation can be written as
\( (\cot y - 4xy) \frac{dy}{dx} = 1 + y^2 \)
\( \Rightarrow \frac{\cot y - 4xy}{1+y^2} = \frac{dx}{dy} \)
\( \Rightarrow \frac{dx}{dy} = \frac{\cot y}{1+y^2} - \frac{4xy}{1+y^2} \)
\( \Rightarrow \frac{dx}{dy} + \frac{4y}{1+y^2}x = \frac{\cot y}{1+y^2} \), which is a linear differential equation of the form \( \frac{dx}{dy} + Px = Q \).
Here, \( P = \frac{4y}{1+y^2} \) and \( Q = \frac{\cot y}{1+y^2} \).
Now, integrating factor (IF) \( = e^{\int P dy} = e^{\int \frac{4y}{1+y^2} dy} \).
On putting \( 1+y^2 = t \Rightarrow 2y dy = dt \)
\( \therefore \text{IF} = e^{\int \frac{2 dt}{t}} = e^{2 \log |t|} = t^2 = (1+y^2)^2 \).

Short Answer Type Questions

Question. Find the order and degree of the differential equation \( y = x \frac{dy}{dx} + \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \).
Answer: Given differential equation is \( y = x \frac{dy}{dx} + \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \).
It has a radical sign in the power, which means it does not have positive integer powers. We first convert it into a differential equation having positive integer powers.
We can rewrite the given differential equation as
\( y - x \frac{dy}{dx} = \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \)
\( \Rightarrow y^2 + x^2 \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} = 1 + \left(\frac{dy}{dx}\right)^2 \) [squaring both sides]
\( \Rightarrow (x^2 - 1) \left(\frac{dy}{dx}\right)^2 - 2xy \frac{dy}{dx} + y^2 - 1 = 0 \), which represents a quadratic polynomial in \( \frac{dy}{dx} \).
Since the highest order derivative is \( \frac{dy}{dx} \), the order of the differential equation is 1, and the highest power of \( \frac{dy}{dx} \) is 2, so the degree of the differential equation is 2.

Question. Solve the following differential equation \( \frac{dy}{dx} = x^3 \csc y \), given that \( y(0) = 0 \).
Answer: We have, \( \frac{dy}{dx} = x^3 \csc y \)
\( \Rightarrow \frac{1}{\csc y} dy = x^3 dx \)
On integrating both sides, we get
\( \int \frac{1}{\csc y} dy = \int x^3 dx \)
\( \Rightarrow \int \sin y dy = \int x^3 dx \)
\( \Rightarrow -\cos y = \frac{x^4}{4} + C \).
It is given that \( y = 0 \) when \( x = 0 \).
\( -\cos 0 = 0 + C \Rightarrow C = -1 \).
\( \therefore -\cos y = \frac{x^4}{4} - 1 \Rightarrow \cos y = 1 - \frac{x^4}{4} \).

Question. Verify that \( ax^2 + by^2 = 1 \) is a solution of the differential equation \( x(y y_2 + y_1^2) = y y_1 \).
Answer: We have, \( ax^2 + by^2 = 1 \).
On differentiating both sides w.r.t. \( x \), we get
\( 2ax + 2by y_1 = 0 \Rightarrow 2(ax + by y_1) = 0 \Rightarrow ax + by y_1 = 0 \) ...(i) [since \( y_1 = \frac{dy}{dx} \)]
Again, on differentiating both sides w.r.t. \( x \), we get
\( a + b(y_1 y_1 + y y_2) = 0 \Rightarrow a + b(y_1^2 + y y_2) = 0 \) [since \( y_2 = \frac{d^2y}{dx^2} \)]
\( \Rightarrow a = -b(y y_2 + y_1^2) \) ...(ii)
On putting \( a = -b(y y_2 + y_1^2) \) in Eq. (i), we get
\( -b(y y_2 + y_1^2)x + by y_1 = 0 \Rightarrow b[-(y y_2 + y_1^2)x + y y_1] = 0 \Rightarrow x(y y_2 + y_1^2) = y y_1 \).
Hence verified.

Question. Verify that the function \( x + y = \tan^{-1} y \) is a solution of the differential equation \( y^2 y' + y^2 + 1 = 0 \).
Answer: Given function is \( x + y = \tan^{-1} y \) ...(i)
Here, the order of the differential equation is 1. So, we differentiate Eq. (i) one time.
On differentiating Eq. (i), w.r.t. \( x \), we get
\( 1 + y' = \frac{1}{1 + y^2} y' \)
\( \Rightarrow (1 + y^2)(1 + y') = y' \Rightarrow 1 + y' + y^2 + y^2 y' = y' \)
\( \Rightarrow 1 + y^2 + y^2 y' = 0 \), which is the given differential equation.
Hence verified.

Question. Find the general solution of the differential equation \( (x+2) \frac{dy}{dx} = x^2 + 5x - 3 \), \( x \neq -2 \).
Answer: Given differential equation is \( (x+2) \frac{dy}{dx} = x^2 + 5x - 3 \)
\( \Rightarrow dy = \frac{x^2 + 5x - 3}{x+2} dx \) [separating the variables]
On integrating both sides, we get
\( \int dy = \int \frac{x^2 + 5x - 3}{x+2} dx \)
\( \Rightarrow y = \int \left[ (x + 3) - \frac{9}{x + 2} \right] dx \) [dividing \( (x^2 + 5x - 3) \) by \( (x + 2) \)]
\( \Rightarrow y = \frac{x^2}{2} + 3x - 9 \log |x + 2| + C \), which is the required general solution.

Question. Find the general solution of the differential equation \( e^x \tan y dx + (1 - e^x) \sec^2 y dy = 0 \).
Answer: Given differential equation is \( e^x \tan y dx + (1 - e^x) \sec^2 y dy = 0 \)
\( \Rightarrow (1 - e^x) \sec^2 y dy = -e^x \tan y dx \)
\( \Rightarrow \frac{\sec^2 y}{\tan y} dy = \frac{-e^x}{1 - e^x} dx \)
On taking integration on both sides, we get
\( \int \frac{\sec^2 y}{\tan y} dy = \int \frac{-e^x}{1 - e^x} dx \)
\( \Rightarrow \log(\tan y) = \log(1 - e^x) + \log C \)
\( \Rightarrow \log(\tan y) = \log [C(1 - e^x)] \) [since \( \log m + \log n = \log mn \)]
\( \Rightarrow \tan y = C(1 - e^x) \Rightarrow y = \tan^{-1} [C(1 - e^x)] \), which is the required solution.

Question. Solve the differential equation \( xy \frac{dy}{dx} = (x + 2) (y + 2) \).
Answer: Given differential equation is \( xy \frac{dy}{dx} = (x + 2) (y + 2) \)
\( \Rightarrow \frac{y}{y+2} dy = \frac{x+2}{x} dx \)
\( \Rightarrow \int \frac{y}{y+2} dy = \int \frac{x+2}{x} dx \)
\( \Rightarrow \int \left( 1 - \frac{2}{y+2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx \)
\( \Rightarrow y - 2 \log |y + 2| = x + 2 \log |x| + C \), which is the required solution.

Question. Solve the differential equation \( (x - 1) \frac{dy}{dx} = 2x^3 y \).
Answer: Given differential equation is \( (x - 1) \frac{dy}{dx} = 2x^3 y \)
\( \Rightarrow \frac{dy}{y} = \frac{2x^3}{x-1} dx \)
\( \Rightarrow \frac{dy}{y} = 2 \left( \frac{x^3}{x-1} \right) dx \)
\( \Rightarrow \frac{dy}{y} = 2 \left[ \frac{(x^3 - 1) + 1}{x - 1} \right] dx \)
\( \Rightarrow \frac{dy}{y} = 2 \left[ \frac{(x - 1)(x^2 + x + 1)}{x - 1} + \frac{1}{x - 1} \right] dx \) [since \( a^3 - b^3 = (a-b)(a^2+ab+b^2) \)]
\( \Rightarrow \frac{dy}{y} = 2 \left[ x^2 + x + 1 + \frac{1}{x-1} \right] dx \)
On integrating both sides, we get
\( \int \frac{dy}{y} = 2 \int \left[ x^2 + x + 1 + \frac{1}{x-1} \right] dx \)
\( \Rightarrow \log |y| = 2 \left[ \frac{x^3}{3} + \frac{x^2}{2} + x + \log |x-1| \right] + C \).

Question. Solve the differential equation \( x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy \).
Answer: Given differential equation is \( x^2 \frac{dy}{dx} = x^2 - 2y^2 + xy \)
\( \Rightarrow \frac{dy}{dx} = 1 - \frac{2y^2}{x^2} + \frac{y}{x} \) ...(i), which is a homogeneous differential equation.
On putting \( y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx} \) in Eq. (i), we get
\( v + x \frac{dv}{dx} = 1 - 2v^2 + v \)
\( \Rightarrow x \frac{dv}{dx} = 1 - 2v^2 \)
\( \Rightarrow \frac{dv}{1 - 2v^2} = \frac{dx}{x} \)
On integrating both sides, we get
\( \int \frac{dv}{1 - 2v^2} = \int \frac{dx}{x} \)
\( \Rightarrow \frac{1}{2} \int \frac{dv}{\frac{1}{2} - v^2} = \int \frac{dx}{x} \)
\( \Rightarrow \frac{1}{2} \times \frac{1}{2 \left( \frac{1}{\sqrt{2}} \right)} \log \left| \frac{\frac{1}{\sqrt{2}} + v}{\frac{1}{\sqrt{2}} - v} \right| = \log |x| + C \) [since \( \int \frac{dx}{a^2-x^2} = \frac{1}{2a}\log\left|\frac{a+x}{a-x}\right| \)]
\( \Rightarrow \frac{\sqrt{2}}{4} \log \left| \frac{1 + \sqrt{2}v}{1 - \sqrt{2}v} \right| = \log |x| + C \)
\( \Rightarrow \frac{1}{2\sqrt{2}} \log \left| \frac{x + \sqrt{2}y}{x - \sqrt{2}y} \right| = \log |x| + C \) [since \( v = \frac{y}{x} \)].

Question. Find the particular solution of the differential equation \( x \frac{dy}{dx} - y + x \csc\left(\frac{y}{x}\right) = 0 \), \( y = \frac{\pi}{3} \) when \( x = 1 \).
Answer: Given differential equation can be written as
\( x \frac{dy}{dx} - y + x \csc\left(\frac{y}{x}\right) = 0 \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \csc\left(\frac{y}{x}\right) \) ...(i), which is a homogeneous differential equation of the form \( \frac{dy}{dx} = f\left(\frac{y}{x}\right) \).
On putting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) in Eq. (i), we get
\( v + x \frac{dv}{dx} = v - \csc v \)
\( \Rightarrow x \frac{dv}{dx} = -\csc v \Rightarrow \sin v dv = -\frac{dx}{x} \)
On integrating both sides, we get
\( \int \sin v dv = -\int \frac{dx}{x} \)
\( \Rightarrow -\cos v = -\log |x| + C \Rightarrow \cos v = \log |x| - C \)
\( \Rightarrow \cos\left(\frac{y}{x}\right) = \log |x| - C \) ...(ii) [since \( v = \frac{y}{x} \)].
Also, given that \( y = \frac{\pi}{3} \) when \( x = 1 \).
Then, \( \cos \frac{\pi}{3} = \log 1 - C \Rightarrow \frac{1}{2} = 0 - C \Rightarrow C = -\frac{1}{2} \).
So, Eq. (ii) becomes \( \cos\left(\frac{y}{x}\right) = \log |x| + \frac{1}{2} \), which is the required equation.

Question. Find the general solution of the differential equation \( x \frac{dy}{dx} = y(\log y - \log x + 1) \).
Answer: Given differential equation is \( x \frac{dy}{dx} = y(\log y - \log x + 1) \)
\( \Rightarrow \frac{dy}{dx} = \frac{y}{x} \left( \log \frac{y}{x} + 1 \right) \) ...(i)
On putting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) in Eq. (i), we get
\( v + x \frac{dv}{dx} = v(\log v + 1) \)
\( \Rightarrow x \frac{dv}{dx} = v \log v \Rightarrow \frac{1}{v \log v} dv = \frac{1}{x} dx \).
On integrating both sides, we get
\( \int \frac{1}{v \log v} dv = \int \frac{1}{x} dx \)
\( \Rightarrow \log |\log v| = \log |x| + C \).
\( \therefore \) Required solution is \( \log \left| \log\left(\frac{y}{x}\right) \right| = \log |x| + C \).

Question. What will be the integrating factor of given differential equation \( y dx + (x - y^2) dy = 0 \)?
Answer: Given differential equation is \( y dx + (x - y^2) dy = 0 \)
\( \Rightarrow y \frac{dx}{dy} + (x - y^2) = 0 \Rightarrow y \frac{dx}{dy} + x = y^2 \)
\( \Rightarrow \frac{dx}{dy} + \frac{1}{y} x = y \) [dividing both sides by \( y \)], which is of the form \( \frac{dx}{dy} + Px = Q \).
Here, \( P = \frac{1}{y} \) and \( Q = y \).
\( \therefore \) Integrating factor (IF) \( = e^{\int P dy} = e^{\int \frac{1}{y} dy} = e^{\log y} = y \).

Question. Find integrating factor of the differential equation \( \left( \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right) \frac{dx}{dy} = 1 \).
Answer: The given differential equation can be written as
\( \frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \)
\( \Rightarrow \frac{dy}{dx} + \frac{y}{\sqrt{x}} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \), which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \frac{1}{\sqrt{x}} \) and \( Q = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \).
Now, Integrating factor (IF) \( = e^{\int P dx} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{2\sqrt{x}} \).

Question. Find the general solution of the differential equation \( e^{2x} \frac{dy}{dx} + 3e^{2x} y = 1 \).
Answer: Given differential equation is \( e^{2x} \frac{dy}{dx} + 3e^{2x} y = 1 \)
\( \Rightarrow e^{2x} \left( \frac{dy}{dx} + 3y \right) = 1 \Rightarrow \frac{dy}{dx} + 3y = e^{-2x} \).
This is of the form of \( \frac{dy}{dx} + Py = Q \).
Here, \( P = 3 \) and \( Q = e^{-2x} \).
Now, Integrating factor (IF) \( = e^{\int 3 dx} = e^{3x} \), and the solution of the differential equation is given by:
\( y \times \text{IF} = \int (Q \times \text{IF}) dx + C \)
\( \Rightarrow y \times e^{3x} = \int (e^{-2x} \times e^{3x}) dx + C = \int e^x dx + C \)
\( \Rightarrow y e^{3x} = e^x + C \), which is the required solution.

Question. Find the general solution of the differential equation \( y dx - x dy + (x \log x) dx = 0 \).
Answer: Given, \( y dx - x dy + (x \log x) dx = 0 \)
\( \Rightarrow (y + x \log x) dx - x dy = 0 \)
\( \Rightarrow \frac{dy}{dx} = \frac{y + x \log x}{x} \)
\( \Rightarrow \frac{dy}{dx} = \frac{y}{x} + \log x \Rightarrow \frac{dy}{dx} - \frac{y}{x} = \log x \).
On comparing with \( \frac{dy}{dx} + Py = Q \), we get \( P = -\frac{1}{x} \), and \( Q = \log x \).
\( \therefore \text{IF} = e^{\int P dx} = e^{\int -\frac{1}{x} dx} = e^{-\log x} = \frac{1}{x} \).
Now, the general solution of the differential equation is given by:
\( y \times \text{IF} = \int (\text{IF} \times Q) dx + C \)
\( \Rightarrow y \cdot \frac{1}{x} = \int \frac{1}{x} \cdot \log x dx + C \).
Let \( \log x = t \Rightarrow \frac{1}{x} dx = dt \).
\( \therefore y \cdot \frac{1}{x} = \int t dt + C \)
\( \Rightarrow \frac{y}{x} = \frac{t^2}{2} + C \)
\( \Rightarrow y = \frac{x}{2} (\log x)^2 + Cx \).

Question. Find the general solution of the following differential equation \( x \, dy - (y + 2x^2) dx = 0 \).
Answer: We have, \[ x \, dy - (y + 2x^2) dx = 0 \] \[ \Rightarrow x \frac{dy}{dx} - y - 2x^2 = 0 \] \[ \Rightarrow \frac{dy}{dx} - \frac{1}{x} y = 2x \quad \text{...(i)} \]
Eq. (i) is a linear differential equation.
On comparing Eq. (i) with \( \frac{dy}{dx} + Py = Q \), we get \[ P = -\frac{1}{x} \text{ and } Q = 2x \] \[ \therefore \text{IF} = e^{\int P \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\log x} = e^{\log(x)^{-1}} = \frac{1}{x} \]
The solution is \[ y \cdot \text{IF} = \int Q \cdot (\text{IF}) \, dx + C \] \[ \Rightarrow y \times \frac{1}{x} = \int \left( 2x \times \frac{1}{x} \right) dx + C \] \[ \Rightarrow \frac{y}{x} = \int 2 \, dx + C \] \[ \Rightarrow \frac{y}{x} = 2x + C \] \[ \Rightarrow y = 2x^2 + Cx \]

Question. Solve the differential equation \(\frac{dy}{dx} + 2xy = y\).
Answer: Given that \[ \frac{dy}{dx} + 2xy = y \] \[ \Rightarrow \frac{dy}{dx} + 2xy - y = 0 \] \[ \Rightarrow \frac{dy}{dx} + (2x - 1)y = 0 \]
which is a linear differential equation.
On comparing it with \( \frac{dy}{dx} + Py = Q \), we get \[ P = (2x - 1) \text{ and } Q = 0 \]
Now, \( \text{IF} = e^{\int P \, dx} = e^{\int (2x - 1) \, dx} = e^{x^2 - x} \)
and the solution is given by \[ y \cdot \text{IF} = \int Q \cdot (\text{IF}) \, dx + C \] \[ \Rightarrow y \cdot e^{x^2 - x} = \int (0 \times e^{x^2 - x}) dx + C \] \[ \Rightarrow y \cdot e^{x^2 - x} = 0 + C \] \[ \Rightarrow y = C e^{x - x^2} \]

Question. Solve the differential equation \( \cos x \frac{dy}{dx} + (2 \sin x) \cdot y = \sin x \cdot \cos x \).
Answer: Given differential equation is \[ \cos x \frac{dy}{dx} + (2 \sin x) \cdot y = \sin x \cdot \cos x \] \[ \Rightarrow \frac{dy}{dx} + \frac{2 \sin x}{\cos x} \cdot y = \frac{\sin x \cdot \cos x}{\cos x} \] \[ \therefore \frac{dy}{dx} + 2 \tan x \cdot y = \sin x \quad \text{...(i)} \]
which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = 2 \tan x \) and \( Q = \sin x \).
Now, \( \text{IF} = e^{\int P \, dx} = e^{\int 2 \tan x \, dx} = e^{2 \log \sec x} = e^{\log \sec^2 x} = \sec^2 x \quad [\because \int \tan x \, dx = \log \sec x] \)
and the required solution is given by \[ y \times \text{IF} = \int (Q \times \text{IF}) \, dx + C \] \[ \therefore y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x \, dx + C \] \[ \Rightarrow y \cdot \sec^2 x = \int \sec x \cdot \tan x \, dx + C \] \[ \Rightarrow y \cdot \sec^2 x = \sec x + C \quad [\because \int \sec x \tan x \, dx = \sec x] \] \[ \Rightarrow y = \cos x + C \cos^2 x \quad [\text{divide by } \sec^2 x] \]
which is the required solution.

Question. Solve the differential equation \( (1+x^2)\frac{dy}{dx} + y = e^{\tan^{-1} x} \).
Answer: Given differential equation is \[ (1+x^2)\frac{dy}{dx} + y = e^{\tan^{-1} x} \] \[ \Rightarrow \frac{dy}{dx} + \frac{y}{1 + x^2} = \frac{e^{\tan^{-1} x}}{1 + x^2} \]
which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \frac{1}{1 + x^2} \) and \( Q = \frac{e^{\tan^{-1} x}}{1 + x^2} \).
Now, \( \text{IF} = e^{\int P \, dx} = e^{\int \frac{1}{1 + x^2} \, dx} = e^{\tan^{-1} x} \)
The solution of the differential equation is given by \[ y \cdot e^{\tan^{-1} x} = \int \left( \frac{e^{\tan^{-1} x}}{1 + x^2} \times e^{\tan^{-1} x} \right) dx + C \] \[ \Rightarrow y e^{\tan^{-1} x} = \int \frac{e^{2\tan^{-1} x}}{1 + x^2} \, dx \] \[ \Rightarrow y e^{\tan^{-1} x} = \frac{e^{2\tan^{-1} x}}{2} + C \]

Question. Solve the differential equation \( x \frac{dy}{dx} + y = x \cos x + \sin x \), given that \( y = 1 \) when \( x = \frac{\pi}{2} \).
Answer: Given differential equation is \[ x \frac{dy}{dx} + y = x \cos x + \sin x \] \[ \Rightarrow \frac{dy}{dx} + \frac{y}{x} = \cos x + \frac{\sin x}{x} \quad \text{...(i)} \]
which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \frac{1}{x} \) and \( Q = \cos x + \frac{\sin x}{x} \). \[ \therefore \text{IF} = e^{\int P \, dx} = e^{\int \frac{1}{x} \, dx} = e^{\log x} = x \]
and the solution of the differential equation is given by \[ y \cdot (\text{IF}) = \int [Q \cdot (\text{IF})] \, dx + C \] \[ \Rightarrow yx = \int (x \cos x + \sin x) \, dx + C \] \[ \Rightarrow xy = \int x \cos x \, dx + \int \sin x \, dx + C \] \[ \Rightarrow xy = x \sin x - \int \sin x \, dx + \int \sin x \, dx + C \quad [\text{integrating by parts}] \] \[ \Rightarrow xy = x \sin x + C \quad \text{...(ii)} \]
On putting \( x = \frac{\pi}{2} \) and \( y = 1 \) in Eq. (ii), we get \[ \frac{\pi}{2} \times 1 = \frac{\pi}{2} \sin \frac{\pi}{2} + C \] \[ \Rightarrow \frac{\pi}{2} = \frac{\pi}{2} + C \Rightarrow C = 0 \]
Now, on putting the value of \( C \) in Eq. (ii), we get \[ xy = x \sin x + 0 \] \[ \Rightarrow y = \sin x \]

Long Answer Type Questions

Question. Prove that \( (x^2 - y^2) = c(x^2 + y^2)^2 \) is the general solution of the differential equation \( (x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy \), where \( c \) is a parameter.
Answer: The given differential equation is \[ (x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy \quad \text{...(i)} \]
and the given function is \[ (x^2 - y^2) = c(x^2 + y^2)^2 \quad \text{...(ii)} \]
On differentiating Eq. (ii) w.r.t. \( x \), we get \[ 2x - 2y \frac{dy}{dx} = 2c(x^2 + y^2) \left( 2x + 2y \frac{dy}{dx} \right) \] \[ \Rightarrow \left( x - y \frac{dy}{dx} \right) = 2c(x^2 + y^2) \left( x + y \frac{dy}{dx} \right) \quad \text{...(iii)} \] \[ \Rightarrow \left( x - y \frac{dy}{dx} \right) = \frac{2(x^2 - y^2)}{(x^2 + y^2)^2} (x^2 + y^2) \left( x + y \frac{dy}{dx} \right) \quad [\text{using Eq. (ii)}] \] \[ \Rightarrow (x^2 + y^2) \left( x - y \frac{dy}{dx} \right) = 2(x^2 - y^2) \left( x + y \frac{dy}{dx} \right) \] \[ \Rightarrow (x^2 + y^2)x - (x^2 + y^2)y \frac{dy}{dx} = 2(x^2 - y^2)x + 2(x^2 - y^2)y \frac{dy}{dx} \] \[ \Rightarrow \left\{ x(x^2 + y^2) - 2x(x^2 - y^2) \right\} = \left\{ 2y(x^2 - y^2) + y(x^2 + y^2) \right\} \frac{dy}{dx} \] \[ \Rightarrow (3xy^2 - x^3) = (3x^2y - y^3) \frac{dy}{dx} \] \[ \Rightarrow (x^3 - 3xy^2)dx = (y^3 - 3x^2y)dy \]
which is the given differential equation. Hence, the given function is the solution of the given differential equation.

Question. Show that the general solution of the differential equation \(\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0\) is given by \((x + y + 1) = A(1 - x - y - 2xy)\), where \(A\) is a parameter.
Answer: We have, \[ \frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0 \] \[ \Rightarrow \frac{dy}{y^2 + y + 1} + \frac{dx}{x^2 + x + 1} = 0 \] \[ \Rightarrow \int \frac{dy}{y^2 + y + 1} + \int \frac{dx}{x^2 + x + 1} = C \] \[ \Rightarrow \int \frac{dy}{y^2 + y + 1 + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2} + \int \frac{dx}{x^2 + x + 1 + \left(\frac{1}{2}\right)^2 - \left(\frac{1}{2}\right)^2} = C \] \[ \Rightarrow \int \frac{dy}{\left(y + \frac{1}{2}\right)^2 + \left(1 - \frac{1}{4}\right)} + \int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(1 - \frac{1}{4}\right)} = C \] \[ \Rightarrow \int \frac{dy}{\left(y + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} + \int \frac{dx}{\left(x + \frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = C \] \[ \Rightarrow \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{y + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right) = C \] \[ \Rightarrow \tan^{-1} \left( \frac{2y + 1}{\sqrt{3}} \right) + \tan^{-1} \left( \frac{2x + 1}{\sqrt{3}} \right) = \frac{\sqrt{3}C}{2} = k \quad \text{(say)} \] \[ \Rightarrow \tan^{-1} \left[ \frac{\frac{2y + 1}{\sqrt{3}} + \frac{2x + 1}{\sqrt{3}}}{1 - \left(\frac{2y + 1}{\sqrt{3}}\right)\left(\frac{2x + 1}{\sqrt{3}}\right)} \right] = k \quad \left[ \because \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right) \right] \] \[ \Rightarrow \tan^{-1} \left[ \frac{\frac{2y + 1 + 2x + 1}{\sqrt{3}}}{1 - \frac{4xy + 2x + 2y + 1}{3}} \right] = k \] \[ \Rightarrow \tan^{-1} \left[ \frac{\frac{2(x + y + 1)}{\sqrt{3}}}{\frac{3 - (4xy + 2x + 2y + 1)}{3}} \right] = k \] \[ \Rightarrow \frac{2\sqrt{3}(x + y + 1)}{2(1 - x - y - 2xy)} = \tan k \] \[ \Rightarrow \frac{\sqrt{3}(x + y + 1)}{1 - x - y - 2xy} = \tan k \] \[ \Rightarrow x + y + 1 = \frac{1}{\sqrt{3}} \tan k (1 - x - y - 2xy) \]
Ans. \( x + y + 1 = A(1 - x - y - 2xy) \), where \( A = \frac{1}{\sqrt{3}} \tan k \) is an arbitrary constant.

Question. Show that given differential equation \((x - y)dy = (x + y)dx\) is homogeneous and solve it.
Answer: Given differential equation is \[ (x - y) dy = (x + y) dx \] or \[ \frac{dy}{dx} = \frac{x + y}{x - y} \quad \text{...(i)} \]
Let \[ F(x, y) = \frac{x + y}{x - y} \quad \text{...(ii)} \]
On replacing \( x \) by \( \lambda x \) and \( y \) by \( \lambda y \) in Eq. (ii), we get \[ F(\lambda x, \lambda y) = \frac{\lambda x + \lambda y}{\lambda x - \lambda y} = \frac{\lambda(x+y)}{\lambda(x-y)} = \lambda^0 F(x, y) \]
Thus, \( F(x, y) \) is a homogeneous function of degree zero.
Now, on putting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) in Eq. (i), we get \[ v + x \frac{dv}{dx} = \frac{x + vx}{x - vx} \] \[ \Rightarrow v + x \frac{dv}{dx} = \frac{1 + v}{1 - v} \] \[ \Rightarrow x \frac{dv}{dx} = \frac{1 + v}{1 - v} - v \] \[ \Rightarrow x \frac{dv}{dx} = \frac{1 + v - v + v^2}{1 - v} \] \[ \Rightarrow x \frac{dv}{dx} = \frac{1 + v^2}{1 - v} \]
On separating the variables, we get \[ \frac{1 - v}{1 + v^2} dv = \frac{1}{x} dx \]
On integrating both sides, we get \[ \int \frac{1 - v}{1 + v^2} dv = \int \frac{1}{x} dx \] \[ \Rightarrow \int \frac{1}{1 + v^2} dv - \frac{1}{2} \int \frac{2v}{1 + v^2} dv = \int \frac{1}{x} dx \] \[ \Rightarrow \tan^{-1} v - \frac{1}{2} \log |1 + v^2| = \log |x| + C \] \[ \Rightarrow \tan^{-1} \left( \frac{y}{x} \right) - \frac{1}{2} \log \left| 1 + \frac{y^2}{x^2} \right| = \log |x| + C \quad \left[ \text{put } v = \frac{y}{x} \right] \] \[ \Rightarrow 2 \tan^{-1} \left( \frac{y}{x} \right) - \log \left| 1 + \frac{y^2}{x^2} \right| = 2(\log |x| + C) \] \[ \Rightarrow 2 \tan^{-1} \left( \frac{y}{x} \right) = \log \left( \frac{x^2 + y^2}{x^2} \right) + \log |x|^2 + 2C \] \[ \Rightarrow 2 \tan^{-1} \left( \frac{y}{x} \right) = \log \left[ \left( \frac{x^2 + y^2}{x^2} \right) |x|^2 \right] + 2C \] \[ \Rightarrow 2 \tan^{-1} \left( \frac{y}{x} \right) - \log |x^2 + y^2| = 2C \quad [\because \log m + \log n = \log mn] \] \[ \Rightarrow \tan^{-1} \left( \frac{y}{x} \right) - \frac{1}{2} \log |x^2 + y^2| = C_1 \]
which is the required solution.

Question. Show that the differential equation \( x \sin \frac{y}{x} \frac{dy}{dx} + x - y \sin \frac{y}{x} = 0 \) is homogeneous and find the particular solution, when \( x = 1 \) and \( y = \frac{\pi}{2} \).
Answer: We have, \[ x \sin \frac{y}{x} \frac{dy}{dx} + x - y \sin \frac{y}{x} = 0 \]
On dividing by \( x \sin \frac{y}{x} \), we get \[ \frac{dy}{dx} + \frac{x}{x \sin \frac{y}{x}} - \frac{y \sin \frac{y}{x}}{x \sin \frac{y}{x}} = 0 \] \[ \Rightarrow \frac{dy}{dx} + \text{cosec} \frac{y}{x} - \frac{y}{x} = 0 \] \[ \Rightarrow \frac{dy}{dx} = \frac{y}{x} - \text{cosec} \frac{y}{x} \quad \text{...(i)} \]
Let \( f(x, y) = \frac{y}{x} - \text{cosec} \frac{y}{x} \)
and \( f(\lambda x, \lambda y) = \frac{\lambda y}{\lambda x} - \text{cosec} \frac{\lambda y}{\lambda x} = \lambda^0 f(x, y) \).
Therefore, \( f(x, y) \) is a homogeneous function of degree zero, so the given differential equation is homogeneous.
Let \( y = vx \)
On differentiating both sides w.r.t. \( x \), we get \[ \frac{dy}{dx} = v + x \frac{dv}{dx} \quad \text{...(ii)} \]
On substituting the values of \( y \) and \( \frac{dy}{dx} \) in Eq. (i), we get \[ v + x \frac{dv}{dx} = v - \text{cosec} v \] \[ \Rightarrow x \frac{dv}{dx} = -\text{cosec} v \] \[ \Rightarrow \frac{dv}{\text{cosec} v} = -\frac{dx}{x} \] \[ \Rightarrow \sin v \, dv = -\frac{dx}{x} \]
On integrating both sides, we get \[ \int \sin v \, dv = -\int \frac{dx}{x} \] \[ \Rightarrow -\cos v = -\log x + C \] \[ \Rightarrow \cos v = \log x + C \]
On putting \( v = \frac{y}{x} \), we get \[ \cos \frac{y}{x} = \log x + C \quad \text{...(iii)} \]
On substituting \( x = 1 \) and \( y = \frac{\pi}{2} \) in Eq. (iii), we get \[ \cos \frac{\pi}{2} = \log 1 + C \] \[ \Rightarrow C = 0 \]
On substituting \( C = 0 \) in Eq. (iii), we get \[ \cos \frac{y}{x} = \log x \]
which is the particular solution of the differential equation.

Question. Find the solution of the differential equation \(\left\{x \cos \left(\frac{y}{x}\right) + y \sin \left(\frac{y}{x}\right)\right\} y \, dx = \left\{y \sin \left(\frac{y}{x}\right) - x \cos \left(\frac{y}{x}\right)\right\} x \, dy\).
Answer: Given, \[ \left\{x \cos \left(\frac{y}{x}\right) + y \sin \left(\frac{y}{x}\right)\right\} y \, dx = \left\{y \sin \left(\frac{y}{x}\right) - x \cos \left(\frac{y}{x}\right)\right\} x \, dy \] \[ \Rightarrow \frac{dy}{dx} = \frac{y \left\{ x \cos\left(\frac{y}{x}\right) + y \sin\left(\frac{y}{x}\right) \right\}}{x \left\{ y \sin\left(\frac{y}{x}\right) - x \cos\left(\frac{y}{x}\right) \right\}} \quad \text{...(i)} \]
Clearly, the given differential equation is homogeneous.
On putting \( y = vx \) and \( \frac{dy}{dx} = v + x \frac{dv}{dx} \) in Eq. (i), we get \[ v + x \frac{dv}{dx} = \frac{vx(x \cos v + vx \sin v)}{x(vx \sin v - x \cos v)} \] \[ \Rightarrow v + x \frac{dv}{dx} = \frac{v(\cos v + v \sin v)}{v \sin v - \cos v} \] \[ \Rightarrow x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v} - v \] \[ \Rightarrow x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} \] \[ \Rightarrow x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v} \] \[ \Rightarrow \left( \frac{v \sin v - \cos v}{v \cos v} \right) dv = \frac{2}{x} dx \] \[ \Rightarrow \left( \tan v - \frac{1}{v} \right) dv = \frac{2}{x} dx \]
On integrating both sides, we get \[ \int \left( \tan v - \frac{1}{v} \right) dv = \int \frac{2}{x} dx \] \[ \Rightarrow \int \tan v \, dv - \int \frac{1}{v} dv = 2 \int \frac{1}{x} dx \] \[ \Rightarrow -\log |\cos v| - \log |v| = 2 \log |x| + C \] \[ \Rightarrow \log |v \cos v| + 2 \log |x| = -C \quad [\because \log m + \log n = \log mn] \] \[ \Rightarrow \log |(v \cos v) x^2| = -C \] \[ \Rightarrow (v \cos v) x^2 = e^{-C} \] \[ \Rightarrow x^2 v \cos v = A \quad [\text{consider } A = e^{-C}] \] \[ \Rightarrow x^2 \frac{y}{x} \cos \frac{y}{x} = A \quad \left[ \text{put } v = \frac{y}{x} \right] \] \[ \Rightarrow xy \cos \frac{y}{x} = A \]
which is the required solution.

Question. Find the particular solution of the differential equation \( ( \tan^{-1} y - x ) dy = ( 1 + y^2 ) dx \), given that \( x = 1 \) when \( y = 0 \).
Answer: Given, \[ ( \tan^{-1} y - x ) dy = ( 1 + y^2 ) dx \] \[ \Rightarrow (1 + y^2) \frac{dx}{dy} + x = \tan^{-1} y \] \[ \Rightarrow \frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1} y}{1 + y^2} \]
This is of the form \( \frac{dx}{dy} + Px = Q \).
Here, \( P = \frac{1}{1 + y^2} \) and \( Q = \frac{\tan^{-1} y}{1 + y^2} \).
Integrating factor (IF) \( = e^{\int P \, dy} = e^{\int \frac{1}{1 + y^2} \, dy} = e^{\tan^{-1} y} \).
The solution is given by: \[ x \times \text{IF} = \int (Q \times \text{IF}) \, dy + C \] \[ \Rightarrow x \times e^{\tan^{-1} y} = \int \left( \frac{\tan^{-1} y}{1 + y^2} \times e^{\tan^{-1} y} \right) dy + C \]
On putting \( \tan^{-1} y = t \Rightarrow \frac{1}{1+y^2} dy = dt \), we get \[ x \cdot e^{\tan^{-1} y} = \int t \cdot e^t \, dt + C \] \[ \Rightarrow x \cdot e^{\tan^{-1} y} = t e^t - e^t + C \] \[ \Rightarrow x e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + C \]
Also, it is given that \( x = 1 \) when \( y = 0 \).
Therefore, we have \[ 1 \cdot e^0 = e^0(0 - 1) + C \Rightarrow 1 = -1 + C \Rightarrow C = 2 \]
Hence, the required solution is \[ x e^{\tan^{-1} y} = e^{\tan^{-1} y} (\tan^{-1} y - 1) + 2 \]

Question. Find the particular solution of the differential equation \( (1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1+x^2} \), given that \( y = 0 \) when \( x = 1 \).
Answer: Given differential equation is \[ (1 + x^2) \frac{dy}{dx} + 2xy = \frac{1}{1+x^2} \]
On dividing both sides by \( (x^2 + 1) \), we get \[ \frac{dy}{dx} + \frac{2x}{x^2 + 1} y = \frac{1}{(x^2 + 1)^2} \]
which is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \frac{2x}{x^2 + 1} \) and \( Q = \frac{1}{(x^2 + 1)^2} \).
Now, integrating factor \( \text{(IF)} = e^{\int P \, dx} = e^{\int \frac{2x}{x^2 + 1} \, dx} \).
Put \( x^2 + 1 = t \Rightarrow 2x \, dx = dt \), then: \[ \int \frac{2x}{x^2 + 1} \, dx = \int \frac{dt}{t} = \log |t| = \log |x^2 + 1| \] \[ \therefore \text{IF} = e^{\log |x^2 + 1|} = x^2 + 1 \]
So, the required general solution is \[ y \times \text{IF} = \int (Q \times \text{IF}) \, dx + C \] \[ y(x^2 + 1) = \int \frac{1}{(x^2 + 1)^2} \times (x^2 + 1) \, dx + C \] \[ \Rightarrow y(x^2 + 1) = \int \frac{1}{x^2 + 1} \, dx + C \] \[ \Rightarrow y(x^2 + 1) = \tan^{-1} x + C \quad \text{...(i)} \]
When \( x = 1 \), then \( y = 0 \): \[ \therefore 0 = \tan^{-1} 1 + C \Rightarrow C = -\frac{\pi}{4} \]
Now, putting this in Eq. (i), we get: \[ y(x^2 + 1) = \tan^{-1} x - \frac{\pi}{4} \]
which is the required differential equation.

Question. Solve the following differential equation \( (\cot^{-1} y + x) dy = (1 + y^2) dx \).
Answer: We have, \[ (\cot^{-1} y + x) dy = (1 + y^2) dx \] \[ \Rightarrow \frac{dx}{dy} = \frac{\cot^{-1} y + x}{1+y^2} \] \[ \Rightarrow \frac{dx}{dy} - \frac{1}{1+y^2} x = \frac{\cot^{-1} y}{1+y^2} \]
This is a linear differential equation of the form \( \frac{dx}{dy} + Px = Q \).
Here, \( P = \frac{-1}{1+y^2} \) and \( Q = \frac{\cot^{-1} y}{1+y^2} \).
\[ \therefore \text{IF} = e^{\int P \, dy} = e^{\int -\frac{1}{1+y^2} \, dy} = e^{-\cot^{-1} y} \]
Now, the solution of the linear differential equation is given by: \[ x \cdot \text{IF} = \int (Q \times \text{IF}) \, dy + C \] \[ \therefore x e^{-\cot^{-1} y} = \int \left( \frac{\cot^{-1} y}{1+y^2} \right) e^{-\cot^{-1} y} \, dy + C \quad \text{...(i)} \]
On putting \( \cot^{-1} y = t \Rightarrow -\frac{1}{1+y^2} dy = dt \), Eq. (i) becomes: \[ x e^{-\cot^{-1} y} = -\int t e^{-t} \, dt + C \] \[ \Rightarrow x e^{-\cot^{-1} y} = -[ -t e^{-t} - e^{-t} ] + C \] \[ \Rightarrow x e^{-\cot^{-1} y} = e^{-\cot^{-1} y}(t + 1) + C \] \[ \Rightarrow x e^{-\cot^{-1} y} = e^{-\cot^{-1} y}(\cot^{-1} y + 1) + C \]
which is the required solution.

Question. Solve the differential equation \( (x - \sin y) dy + (\tan y) dx = 0 \), given \( y(0) = 0 \).
Answer: Given differential equation is \[ (x - \sin y) dy + (\tan y) dx = 0 \] \[ \Rightarrow \frac{dx}{dy} = -\frac{x - \sin y}{\tan y} \] \[ \Rightarrow \frac{dx}{dy} = -\frac{x}{\tan y} + \frac{\sin y}{\tan y} \] \[ \Rightarrow \frac{dx}{dy} + \cot y \cdot x = \cos y \quad \left[ \because \frac{1}{\tan \theta} = \cot \theta = \frac{\cos \theta}{\sin \theta} \right] \]
It is a linear differential equation of the form \( \frac{dx}{dy} + Px = Q \).
Here, \( P = \cot y \) and \( Q = \cos y \). \[ \therefore \text{IF} = e^{\int P \, dy} = e^{\int \cot y \, dy} = e^{\log \sin y} = \sin y \]
The general solution is given by: \[ x \times \text{IF} = \int (\text{IF} \times Q) \, dy + C \] \[ \Rightarrow x \cdot \sin y = \int \sin y \cdot \cos y \, dy + C \] \[ \Rightarrow x \cdot \sin y = \frac{1}{2} \int 2 \sin y \cdot \cos y \, dy + C \] \[ \Rightarrow x \cdot \sin y = \frac{1}{2} \int \sin 2y \, dy + C \quad [\because 2\sin \theta \cos \theta = \sin 2\theta] \] \[ \Rightarrow x \cdot \sin y = -\frac{1}{4} \cos 2y + C \quad \text{...(i)} \]
Also, given that \( y = 0 \) when \( x = 0 \).
On putting \( x = 0 \) and \( y = 0 \) in Eq. (i), we get: \[ 0 = -\frac{1}{4} \cos 0 + C \Rightarrow C = \frac{1}{4} \quad [\because \cos 0 = 1] \]
On putting the value of \( C \) in Eq. (i), we get: \[ x \cdot \sin y = -\frac{1}{4} \cos 2y + \frac{1}{4} \] \[ \Rightarrow x \cdot \sin y = \frac{1}{4}(1 - \cos 2y) = \frac{1}{2} \sin^2 y \quad [\because \cos 2\theta = 1 - 2\sin^2 \theta] \] \[ \therefore 2x = \sin y \]
which is the required solution.

Question. Find the particular solution of differential equation \( \frac{dy}{dx} + y \cot x = 2x + x^2 \cot x \), \( x \neq 0 \), given that \( y = 0 \) when \( x = \frac{\pi}{2} \).
Answer: We have, \[ \frac{dy}{dx} + y \cot x = 2x + x^2 \cot x, \quad (x \neq 0) \]
This is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \).
Here, \( P = \cot x \) and \( Q = 2x + x^2 \cot x \). \[ \therefore \text{IF} = e^{\int P \, dx} = e^{\int \cot x \, dx} = e^{\log |\sin x|} = \sin x \]
The general solution is given by: \[ y \cdot \text{IF} = \int (\text{IF} \times Q) \, dx + C \] \[ \Rightarrow y \cdot \sin x = \int (2x + x^2 \cot x) \sin x \, dx + C \] \[ \Rightarrow y \cdot \sin x = 2 \int x \sin x \, dx + \int x^2 \cos x \, dx + C \] \[ \Rightarrow y \cdot \sin x = 2 \int x \sin x \, dx + \left[ x^2 \sin x - 2 \int x \sin x \, dx \right] + C \] \[ \Rightarrow y \cdot \sin x = x^2 \sin x + C \quad \text{...(i)} \]
On putting \( x = \frac{\pi}{2} \) and \( y = 0 \) in Eq. (i), we get: \[ (0)\sin \frac{\pi}{2} = \left(\frac{\pi}{2}\right)^2 \sin \frac{\pi}{2} + C \] \[ \Rightarrow 0 = \frac{\pi^2}{4} + C \Rightarrow C = -\frac{\pi^2}{4} \]
On putting \( C = -\frac{\pi^2}{4} \) in Eq. (i), we get: \[ y \cdot \sin x = x^2 \sin x - \frac{\pi^2}{4} \] \[ \Rightarrow y = x^2 - \frac{\pi^2}{4} \text{cosec} x \quad [\text{dividing both sides by } \sin x] \]

Question. Find the particular solution of the differential equation \( y e^y dx = (y^3 + 2x e^y) dy \), given \( y(0) = 1 \).
Answer: We have, \[ y e^y dx = (y^3 + 2x e^y) dy \] \[ \Rightarrow \frac{dx}{dy} = \frac{y^3 + 2x e^y}{y e^y} \] \[ \Rightarrow \frac{dx}{dy} = y^2 e^{-y} + \frac{2x}{y} \] \[ \Rightarrow \frac{dx}{dy} - \frac{2}{y} x = y^2 e^{-y} \quad \text{...(i)} \]
This is a linear differential equation of the form \( \frac{dx}{dy} + Px = Q \).
Here, \( P = -\frac{2}{y} \) and \( Q = y^2 e^{-y} \). \[ \therefore \text{IF} = e^{\int P \, dy} = e^{\int -\frac{2}{y} \, dy} = e^{-2 \log y} = y^{-2} = \frac{1}{y^2} \]
The solution is given by: \[ x \times \text{IF} = \int (Q \times \text{IF}) \, dy + C \] \[ \Rightarrow x \left( \frac{1}{y^2} \right) = \int \left( y^2 e^{-y} \times \frac{1}{y^2} \right) dy + C \] \[ \Rightarrow \frac{x}{y^2} = \int e^{-y} \, dy + C \] \[ \Rightarrow \frac{x}{y^2} = -e^{-y} + C \quad \text{...(ii)} \]
It is given that \( y(0) = 1 \), i.e., \( y = 1 \) when \( x = 0 \).
On putting \( x = 0 \) and \( y = 1 \) in Eq. (ii), we get: \[ 0 = -e^{-1} + C \Rightarrow C = e^{-1} = \frac{1}{e} \]
On putting \( C = e^{-1} \) in Eq. (ii), we get: \[ \frac{x}{y^2} = -e^{-y} + e^{-1} \] \[ \Rightarrow x = y^2(e^{-1} - e^{-y}) \]
which is the required solution.