CBSE Class 12 Mathematics Differential Equations VBQs Set A

BASIC CONCEPTS

Definition: An equation involving the independent variable \(x\) (say), dependent variable \(y\) (say) and the differential coefficients of dependent variable with respect to independent variable i.e., \(\frac{dy}{dx}\), \(\frac{d^2y}{dx^2}\), \dots, etc. is called a differential equation.
e.g., \(\frac{dy}{dx} + 4y = x\), \(\frac{d^2y}{dx^2} - 3\frac{dy}{dx} + 5y = x^2\) are differential equations.

Order and Degree of a Differential Equation:
The order of a differential equation is the order of the highest derivative occurring in the differential equation.
The degree of a differential equation is the degree of the highest order derivative occurring in the equation, when the differential coefficients are made free from radicals, fractions and it is written as a polynomial in differential co-efficient.

Question. Consider three differential equations:
(i) \(\frac{d^3y}{dx^3} + 2\left(\frac{d^2y}{dx^2}\right)^2 - \frac{dy}{dx} + y = 0\)
(ii) \(\frac{d^2y}{dx^2} = 1 + \sqrt{\frac{dy}{dx}}\)
(iii) \(\left(\frac{d^2y}{dx^2}\right)^3 + \sin\left(\frac{dy}{dx}\right) = 0\)
Answer:
(i) In this equation, the highest order derivative is 3 and its power is 1. Therefore, its order is 3 and degree 1.
(ii) In this equation, the differential co-efficient is not free from radical. Therefore, it is made free from radical as
\(\frac{d^2y}{dx^2} - 1 = \sqrt{\frac{dy}{dx}} \Rightarrow \left(\frac{d^2y}{dx^2}\right)^2 + 1 - 2\frac{d^2y}{dx^2} - \frac{dy}{dx} = 0\) [Squaring both sides]
Hence, order is 2 and degree is 2.
(iii) In this equation order of highest order derivative is 2 therefore, its order is 2, but this differential equation cannot be written in the form of polynomial in differential co-efficient.
Hence, its degree is not defined.
[Note: The order and degree of differential equations are always positive integers.]

Classification of Differential Equations

(A) Differential equations are classified according to their order:

• First order differential equations: First order differential equations are those in which only the first order derivative of the dependent variable occurs.
• Higher order differential equations: Differential equations of order two or more are referred as higher order differential equations.

(B) Another classification of differential equations refers to its linearity means linear and non linear differential equations:
Linear and non-linear differential equations: A differential equation, in which the dependent variable and its derivatives occur only in the 1st degree and are not multiplied together, is called a linear differential equation otherwise it is non linear.
Note: Every linear differential equation is always of the 1st degree but every differential equation of the 1st degree need not be the linear differential equation.

Solution of a differential equation: The solution of a differential equation is a relation between dependent and the independent variables which satisfies the given differential equation i.e., when this relation is substituted in given differential equation, makes left hand and right hand sides identically equal.
Note: If any relation contains \(n\) arbitrary constants, then the differential equation of \(n\)th order will be obtained after eliminating all the arbitrary constants.

General and particular solutions of differential equations: The general solution of a differential equation of \(n\)th order is a relation between dependent and independent variables having \(n\) arbitrary constants.
The solution obtained from the general solution by giving the particular values to these arbitrary constants is called the particular solution.

Forms of the solution of differential equations: The general solution may have more than one forms but the arbitrary constants must be same in the number.

Formation of differential equations: By forming a differential equation from a given equation representing family of curves, means finding a differential equation whose solution is the given equation. If an equation, representing a family of curves contains \(n\) arbitrary constants, then we differentiate the given equation \(n\) times to obtain \(n\) equations. Using all these equations, we eliminate the arbitrary constants. The equation so obtained is the differential equation of the \(n\)th order for the family of the given curves.
(i) If the given equation contains only one arbitrary constant then differentiate only one time and eliminate the constant then differential equation of the first order is obtained.
(ii) If the differential equation contains two arbitrary constant then differentiate only two times and eliminate the constants, then the differential equation of the second order is obtained.

Solution of differential equations: In this chapter, we shall only find the solutions of differential equations viz. differential equations with variables separable form, homogeneous and linear differential equations.

Type 1: Variables separable form

(A) Variables separable form: If in the given equation, it is possible to get all the terms containing \(x\) and \(dx\) to one side and all the terms containing \(y\) and \(dy\) to the other, the variables are said to be separable.
Procedure to solve the differential equations with variables separable form:
Consider the equation \(\frac{dy}{dx} = X \cdot Y\) where \(X\) is a function of \(x\) only and \(Y\) is a function of \(y\) only.
(i) Put the equation in the form \(\frac{1}{Y} \cdot dy = X \cdot dx\)
(ii) Integrating both the sides, we get \(\int \frac{dy}{Y} = \int X dx + C\), where \(C\) is an arbitrary constant.
Thus, the required solution is obtained.

(B) Equations Reducible to Variables Separable Form: Equations of the form \(\frac{dy}{dx} = f(ax + by + c)\) can be reduced to form in which the variables are separable form.
Procedure to solve an equation reducible to variables separable form:
(i) Write the given equation in form \(\frac{dy}{dx} = f(ax + by + c)\).
(ii) Put \(ax + by + c = z\), so that \(\frac{dy}{dx} = \frac{1}{b} \left( \frac{dz}{dx} - a \right)\).
(iii) Putting this \(\frac{dy}{dx}\) in the given equation, we get \(\frac{1}{b} \left( \frac{dz}{dx} - a \right) = f(z)\). This equation is reduced in the form : \(\frac{dz}{a + bf(z)} = dx\). After integrating, we get the required result.

Type 2 : Homogeneous Function and Homogeneous Differential Equation

Homogeneous function: A function \(F(x, y)\) is called homogeneous function of degree \(n\) if \(F(\lambda x, \lambda y) = \lambda^n F(x, y)\), where \(\lambda\) is non-zero real number.
A differential equation of the form \(\frac{dy}{dx} = F(x, y)\) is called homogeneous differential equation, if \(F(x, y)\) is a homogeneous function of degree zero, i.e., \(F(\lambda x, \lambda y) = \lambda^0 F(x, y)\).
Example: \((x^2 + xy)dy = (x^2 + y^2)dx \Rightarrow \frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}\) is homogeneous differential equation because
Here, \(F(x, y) = \frac{x^2 + y^2}{x^2 + xy}\)
\(\therefore F(\lambda x, \lambda y) = \frac{\lambda^2x^2 + \lambda^2y^2}{\lambda^2x^2 + \lambda x \cdot \lambda y} = \frac{\lambda^2(x^2 + y^2)}{\lambda^2(x^2 + xy)} = \lambda^0 F(x, y)\)
Hence, \(F(x, y)\) is homogeneous function of degree zero. Therefore, \(\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}\) is a homogeneous differential equation.
To solve this type of equation we proceed as follows:
(i) Suppose \(y = vx\) and so \(\frac{dy}{dx} = v + x\frac{dv}{dx}\).
(ii) The value \(y = vx\) and \(\frac{dy}{dx} = v + x\frac{dv}{dx}\) is substituted in given equation. The equation reduces to variable separable form, which can be solved by integrating both sides.
(iii) Finally, \(v\) is replaced by \(\frac{y}{x}\) to get the required solution.
Note: If the homogeneous differential equation is in the form \(\frac{dx}{dy} = F(x, y)\) then we substitute \(x = vy\) and so \(\frac{dx}{dy} = v + y\frac{dv}{dy}\) and proceed as above.

Type 3: Linear Differential Equations Form

A linear differential equation is that in which the dependent variable and its differential co-efficient occur in the first degree and not multiplied together.
Thus, the standard form of a linear differential equation of the first order is \(\frac{dy}{dx} + Py = Q\), where \(P\) and \(Q\) are functions of \(x\) or constants.
Now, we find a function \(F\) of \(x\), by which we can multiply both sides of the given equation so that the LHS becomes a complete differential. Such a function \(F\) is called the integrating factor (IF).
In this case \(\text{IF} = e^{\int P dx}\) and solution is given by \(y \cdot \text{IF} = \int (Q \cdot \text{IF}) dx + C\).

Sometimes the Equation can be Made Linear Differential as Follows:
\(\frac{dx}{dy} + Px = Q\) in which \(x\) is treated as dependent variable while \(y\) is treated as independent variable and \(P, Q\) are function of \(y\) or constant.
In this case \(\text{IF} = e^{\int P dy}\) and solution is given by, \(x \cdot \text{IF} = \int (Q \cdot \text{IF}) dy + C\).

Questions

Question. Verify that the given function is a solution of the corresponding differential equation: \(xy = \log y + C : y' = \frac{y^2}{1 - xy} (xy \neq 1)\).
Answer: The given function is \(xy = \log y + C\). Now,
\(x \cdot \frac{dy}{dx} + y = \frac{1}{y} \frac{dy}{dx} \Rightarrow \left(\frac{1}{y} - x\right) \frac{dy}{dx} = y \Rightarrow \frac{dy}{dx} = \frac{y^2}{1 - xy}\)
which is given differential equation.
Thus, \(xy = \log y + C\) is a solution of the given differential equation.

Question. Verify that the given function is a solution of the corresponding differential equation: \(y - \cos y = x : (y \sin y + \cos y + x) y' = y\).
Answer: The given function is \(y - \cos y = x\).
Now, \(\frac{dy}{dx} + \sin y \frac{dy}{dx} = 1 \Rightarrow \frac{dy}{dx} = \frac{1}{1 + \sin y}\)
Putting values of \(y'\) and \(x\) in given differential equation, we have
LHS \( = (y \sin y + \cos y + y - \cos y) \cdot \frac{1}{(1 + \sin y)} = y(1 + \sin y) \cdot \frac{1}{(1 + \sin y)} = y = \text{RHS}\)
Thus, \(y - \cos y = x\) is a solution of the given differential equation.

Question. Form the differential equation of the family of hyperbolas having foci on x-axis and centre at origin.
Answer: We know that the differential equation of the family of hyperbolas having foci on \(x\)-axis and centre at origin is
\(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \Rightarrow \frac{2x}{a^2} - \frac{2y}{b^2} \frac{dy}{dx} = 0 \Rightarrow \frac{y}{x} \frac{dy}{dx} = \frac{b^2}{a^2}\)
Again differentiating both sides w.r.t. \(x\), we have
\(\frac{y}{x} \cdot \frac{d^2y}{dx^2} + \frac{dy}{dx} \cdot \left[ \frac{x \frac{dy}{dx} - y}{x^2} \right] = 0 \Rightarrow \frac{y}{x} \frac{d^2y}{dx^2} + \frac{1}{x} \left( \frac{dy}{dx} \right)^2 - \frac{y}{x^2} \frac{dy}{dx} = 0\)
\(\Rightarrow xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \left( \frac{dy}{dx} \right) = 0\)
which is the required differential equation.

Question. Write the solution of the differential equation \((e^x + e^{-x}) dy = (e^x - e^{-x}) dx\)
Answer: We have, \(dy = \frac{e^x - e^{-x}}{e^x + e^{-x}} \cdot dx\)
Integrating both sides, we get
\(y = \int \frac{e^x - e^{-x}}{e^x + e^{-x}} dx = \log | e^x + e^{-x} | + C\)

Question. Find the general solution of the following differential equation: \(e^x \tan y dx + (1 - e^x) \sec^2 y dy = 0\) 
Answer: Given differential equation,
\(e^x \tan y dx + (1 - e^x) \sec^2 y dy = 0 \Rightarrow e^x \tan y dx = -(1 - e^x) \sec^2 y dy\)
\(\therefore dy = \frac{e^x}{e^x - 1} \cdot \frac{\tan y}{\sec^2 y} dx \Rightarrow \frac{\sec^2 y dy}{\tan y} = \frac{e^x}{e^x - 1} dx\)
Integrating both sides, we get
\(\int \frac{\sec^2 y dy}{\tan y} = \int \frac{e^x}{e^x - 1} dx \Rightarrow \log | \tan y | = \log | e^x - 1 | + \log C\)
\(\Rightarrow \log | \tan y | = \log | (e^x - 1) C |\)
\(\therefore \tan y = (e^x - 1) C\)

Question. For the differential equation \(xy \frac{dy}{dx} = (x + 2) (y + 2)\), find the solution curve passing through the point \((1, -1)\).
Answer: The given equation is \(xy \frac{dy}{dx} = (x + 2) (y + 2) \Rightarrow \frac{y}{y + 2} dy = \frac{x + 2}{x} dx\)
\(\Rightarrow \int \frac{y}{y + 2} dy = \int \left( \frac{x + 2}{x} \right) dx \Rightarrow \int \left( 1 - \frac{2}{y + 2} \right) dy = \int \left( 1 + \frac{2}{x} \right) dx\)
\(\Rightarrow y - 2 \log | y + 2 | = x + 2 \log | x | + C\)
\(\Rightarrow y = x + 2 \log | x | + 2 \log | y + 2 | + C\)
\(\Rightarrow y = x + 2 \log | x (y + 2) | + C\)
Since the line passes through the point \((1, -1)\). So, putting \(x = 1, y = -1\).
We have, \(-1 = 1 + 2 \log | 1 (-1 + 2) | + C \Rightarrow C = -2\)
\(\therefore y = x + 2 \log | x (y + 2) | - 2\), which is the required equation of the curve.

Question. Show that the differential equation \(2ye^{x/y} dx + (y - 2xe^{x/y}) dy = 0\) is homogeneous and find its particular solution, given that \(x = 0\) when \(y = 1\). [CBSE (North) 2016; 
Answer: Given: \(2ye^{x/y} dx + (y - 2xe^{x/y}) dy = 0\)
\(\Rightarrow \frac{dx}{dy} = -\frac{y - 2xe^{x/y}}{2ye^{x/y}} \Rightarrow \frac{dx}{dy} = \frac{2xe^{x/y} - y}{2ye^{x/y}}\)
Let \(F(x, y) = \frac{2xe^{x/y} - y}{2ye^{x/y}}\)
\(\therefore F(\lambda x, \lambda y) = \frac{2\lambda xe^{\lambda x/\lambda y} - \lambda y}{2\lambda ye^{\lambda x/\lambda y}} = \lambda^0 \frac{2xe^{x/y} - y}{2ye^{x/y}} = \lambda^0 F(x, y)\)
Hence, given differential equation is homogeneous.
Now, \(\frac{dx}{dy} = \frac{2xe^{x/y} - y}{2ye^{x/y}}\) ...(i)
Let \(x = vy \Rightarrow \frac{dx}{dy} = v + y \frac{dv}{dy}\)
\(\therefore (i) \Rightarrow v + y \frac{dv}{dy} = \frac{2vye^{vy/y} - y}{2ye^{vy/y}}\)
\(\Rightarrow v + y \frac{dv}{dy} = \frac{2ve^v - 1}{2e^v} \Rightarrow y \frac{dv}{dy} = \frac{2ve^v - 1}{2e^v} - v\)
\(\Rightarrow y \frac{dv}{dy} = \frac{2ve^v - 1 - 2ve^v}{2e^v} \Rightarrow y \frac{dv}{dy} = -\frac{1}{2e^v}\)
\(\Rightarrow 2e^v dv = - \frac{dy}{y}\)
\(\Rightarrow \int 2e^v dv = - \int \frac{dy}{y} \Rightarrow 2e^v = -\log y + C \Rightarrow 2e^{x/y} + \log y = C\)
When \(x = 0, y = 1\)
\(\therefore 2e^0 + \log 1 = C \text{ or } C = 2\)
Hence, the required solution is
\(2e^{x/y} + \log y = 2\)

Question. Show that the given differential equation is homogeneous and solve it. \(\{x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})\} y dx = \{y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})\} x dy\).
Answer: The given differential equation can be expressed as
\(\frac{dy}{dx} = \frac{\{x \cos (\frac{y}{x}) + y \sin (\frac{y}{x})\} y}{\{y \sin (\frac{y}{x}) - x \cos (\frac{y}{x})\} x} = f(x, y) \text{ (say)}\)
Now \(f(\lambda x, \lambda y) = \frac{\{\lambda x \cos (\frac{\lambda y}{\lambda x}) + \lambda y \sin (\frac{\lambda y}{\lambda x})\} \lambda y}{\{\lambda y \sin (\frac{\lambda y}{\lambda x}) - \lambda x \cos (\frac{\lambda y}{\lambda x})\} \lambda x} = \lambda^0 f(x, y)\)
Therefore, \(f(x, y)\) is a homogeneous function of degree zero. So, the given differential equation is a homogeneous differential equation.
Put \(y = vx\) so that \(\frac{dy}{dx} = v + x \frac{dv}{dx}\)
Putting values of \(\frac{dy}{dx}\) and \(y\) in the given equation, we have
\(v + x \frac{dv}{dx} = \frac{\{x \cos v + vx \sin v\} vx}{\{vx \sin v - x \cos v\} x} \Rightarrow x \frac{dv}{dx} = \frac{(v \cos v + v^2 \sin v) }{v \sin v - \cos v} - v\)
\(\Rightarrow x \frac{dv}{dx} = \frac{v \cos v + v^2 \sin v - v^2 \sin v + v \cos v}{v \sin v - \cos v} \Rightarrow x \frac{dv}{dx} = \frac{2v \cos v}{v \sin v - \cos v}\)
\(\Rightarrow \frac{v \sin v - \cos v}{v \cos v} dv = \frac{2}{x} dx \Rightarrow \left( \tan v - \frac{1}{v} \right) dv = \frac{2}{x} dx\)
\(\Rightarrow \int \left( \tan v - \frac{1}{v} \right) dv = 2 \int \frac{1}{x} dx \Rightarrow \log | \sec v | - \log v = 2 \log x + \log C\)
\(\therefore \log | \sec v | - \log | v | - \log x^2 = \log C \Rightarrow \log \frac{\sec v}{vx^2} = \log C\)
\(\Rightarrow \frac{\sec v}{vx^2} = C \Rightarrow \frac{\sec (\frac{y}{x})}{\frac{y}{x} \cdot x^2} = C \Rightarrow \sec (\frac{y}{x}) = Cxy\)
which is the required solution.

Question. Form the differential equation representing the family of ellipses foci on x-axis and centre at the origin.
Answer: The family of ellipses having foci on x-axis and centre at the origin, is given by \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)
Differentiating with respect to \(x\), we get
\(\frac{2x}{a^2} + \frac{2y}{b^2} \left( \frac{dy}{dx} \right) = 0 \Rightarrow \frac{2y \frac{dy}{dx}}{b^2} = - \frac{2x}{a^2}\)
\(\Rightarrow \frac{y \frac{dy}{dx}}{x} = - \frac{b^2}{a^2}\)
Again by differentiating with respect to \(x\), we get
\(\frac{x \left[ y \frac{d^2y}{dx^2} + \left( \frac{dy}{dx} \right)^2 \right] - \left( y \frac{dy}{dx} \right)}{x^2} = 0\)
\(\therefore\) The required equation is \(xy \frac{d^2y}{dx^2} + x \left( \frac{dy}{dx} \right)^2 - y \frac{dy}{dx} = 0\).

Question. Form the differential equation of the family of parabolas having vertex at the origin and axis along positive y-axis.
Answer: The equation of parabola having vertex at origin and axis along +ve \(y\)-axis is
\(x^2 = 4ay\) ...(i), where \(a\) is parameter.
Differentiating w.r.t. \(x\) we get,
\(2x = 4a \cdot \frac{dy}{dx}\) i.e., \(x = 2ay' \Rightarrow a = \frac{x}{2y'}\) where \(y' = \frac{dy}{dx}\)
Putting \(a = \frac{x}{2y'}\) in (i), we get
\(x^2 = 4 \cdot \frac{x}{2y'} \cdot y \Rightarrow y' = \frac{2y}{x} \Rightarrow xy' = 2y \Rightarrow xy' - 2y = 0\)
It is required differential equation.

Question. Solve: \(x \frac{dy}{dx} + y - x + xy \cot x = 0 (x \neq 0)\) 
Answer: The given differential equation \(x \frac{dy}{dx} + y - x + xy \cot x = 0\)
\(\Rightarrow \frac{dy}{dx} + \left( \cot x + \frac{1}{x} \right) y = 1\) (Dividing both sides by \(x\)) ...(i)
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\), where \(P = \cot x + \frac{1}{x}\) and \(Q = 1\).
So, \(\text{IF} = e^{\int (\cot x + \frac{1}{x}) dx} = e^{\log | \sin x | + \log | x |}\)
\(= e^{\log | x \sin x |} = x \sin x\) (\(x \sin x\) is always +ve for any \(x\))
Multiplying both sides by IF in equation (i), we get
\(x \sin x \frac{dy}{dx} + x \sin x \left( \cot x + \frac{1}{x} \right) y = x \sin x\)
\(\Rightarrow x \sin x \frac{dy}{dx} + (x \cos x + \sin x) y = x \sin x \Rightarrow \frac{d}{dx} (yx \sin x) = x \sin x\) [By product rule]
On integrating both sides, we get
\(y (x \sin x) = \int x \sin x dx + C\) ...(ii)
Let \(I = \int x \sin x dx = x \cdot (-\cos x) - \int 1 \cdot (-\cos x) dx\) (Using by parts)
\(I = -x \cos x + \sin x\)
Putting the value of \(I\) in (ii), we get
\(y (x \sin x) = -x \cos x + \sin x + C \Rightarrow y (x \sin x) = \sin x - x \cos x + C\)
Hence, \(y = \frac{1}{x} - \cot x + \frac{C}{x \sin x}\) is the required solution.

Question. Find the particular solution of the differential equation \((1 + e^{2x}) dy + (1 + y^2) e^x dx = 0\) given that \(y = 1\) when \(x = 0\). 
Answer: We have, \((1 + e^{2x}) dy + (1 + y^2) e^x dx = 0\) and given that \(y = 1\), when \(x = 0\)
\(\therefore \frac{dy}{1 + y^2} = - \frac{e^x}{1 + e^{2x}} dx \Rightarrow \int \frac{dy}{1 + y^2} = - \int \frac{e^x}{1 + e^{2x}} dx\)
Integrating both sides, we get
\(\int \frac{dy}{1 + y^2} = - \int \frac{e^x dx}{1 + (e^x)^2}\)
\(\Rightarrow \tan^{-1} y = - \int \frac{dt}{1 + t^2}\) [Putting \(e^x = t \Rightarrow e^x dx = dt\)]
\(\Rightarrow \tan^{-1} y = - \tan^{-1} (t) + C \Rightarrow \tan^{-1} y = - \tan^{-1} (e^x) + C\) ...(i)
Put \(x = 0, y = 1\) in (i), we get
\(\tan^{-1} 1 = - \tan^{-1} (e^0) + C \Rightarrow \frac{\pi}{4} = - \frac{\pi}{4} + C \Rightarrow C = \frac{\pi}{2}\)
Putting the value of \(C\) in (i), we get
\(\tan^{-1} y = - \tan^{-1} (e^x) + \frac{\pi}{2} \Rightarrow \frac{\pi}{2} = \tan^{-1} (e^x) + \tan^{-1} y\)
Hence, \(\tan^{-1}(e^x) + \tan^{-1} y = \frac{\pi}{2}\) is the required solution.

Question. Show that the differential equation is homogeneous and solve it. \((1 + e^{x/y}) dx + e^{x/y} (1 - \frac{x}{y}) dy = 0\)
Answer: We have, \((1 + e^{x/y}) dx + e^{x/y} (1 - \frac{x}{y}) dy = 0 \Rightarrow (1 + e^{x/y}) dx = - e^{x/y} (1 - \frac{x}{y}) dy\)
\(\therefore \frac{dx}{dy} = \frac{- e^{x/y} (1 - \frac{x}{y})}{(1 + e^{x/y})} = g(\frac{x}{y})\) ...(i)
Here, RHS of differential equation is of the form \(g(\frac{x}{y})\), so it is a homogeneous function of degree zero.
Now, we put \(x = vy\) and \(\frac{dx}{dy} = v + y \frac{dv}{dy}\)
From (i), we get \(v + y \frac{dv}{dy} = \frac{- e^v (1 - v)}{1 + e^v}\)
\(\Rightarrow y \frac{dv}{dy} = \frac{- e^v (1 - v)}{1 + e^v} - v = \frac{- e^v + ve^v - v - ve^v}{1 + e^v} = \frac{-(v + e^v)}{1 + e^v}\)
\(\Rightarrow \frac{1 + e^v}{-(v + e^v)} dv = \frac{dy}{y}\)
On integrating both sides, we get
\(-\log | v + e^v | + \log C = \log | y | \Rightarrow \log C = \log | y | + \log | v + e^v |\)
\(\Rightarrow \log C = \log | y (v + e^v) | \Rightarrow \log C = \log | y (\frac{x}{y} + e^{x/y}) |\)
\(\Rightarrow C = y (\frac{x}{y} + e^{x/y})\) or \(C = x + ye^{x/y}\)
Hence, \(x + ye^{x/y} = C\) is the required solution.

Question. Solve the following differential equation: \(\frac{dy}{dx} + 2y \tan x = \sin x\), given that \(y = 0\), when \(x = \frac{\pi}{3}\). 
Answer: Given differential equation is \(\frac{dy}{dx} + 2y \tan x = \sin x\).
Comparing it with \(\frac{dy}{dx} + Py = Q\), we get \(P = 2 \tan x, Q = \sin x\)
\(\text{IF} = e^{\int 2 \tan x dx} = e^{2 \log \sec x} = e^{\log \sec^2 x} = \sec^2 x\) [\(\because e^{\log z} = z\)]
Hence, general solution is \(y \cdot \sec^2 x = \int \sin x \cdot \sec^2 x dx + C\).
\(y \cdot \sec^2 x = \int \sec x \cdot \tan x dx + C \Rightarrow y \cdot \sec^2 x = \sec x + C \Rightarrow y = \cos x + C \cos^2 x\)
Putting \(y = 0\) and \(x = \frac{\pi}{3}\), we get \(0 = \cos \frac{\pi}{3} + C \cdot \cos^2 \frac{\pi}{3}\)
\(\Rightarrow 0 = \frac{1}{2} + \frac{C}{4} \Rightarrow C = -2\)
\(\therefore\) Required solution is \(y = \cos x - 2 \cos^2 x\).

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: The given equation is \(\frac{dy}{dx} + \frac{y^2 + y + 1}{x^2 + x + 1} = 0\).
\(\therefore \frac{dy}{y^2 + y + 1} = -\frac{dx}{x^2 + x + 1} \Rightarrow \int \frac{dy}{y^2 + y + 1} = -\int \frac{dx}{x^2 + x + 1}\)
\(\Rightarrow \int \frac{dy}{y^2 + y + \frac{1}{4} + (1 - \frac{1}{4})} = -\int \frac{dx}{x^2 + x + \frac{1}{4} + (1 - \frac{1}{4})}\)
\(\Rightarrow \int \frac{dy}{(y + \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = -\int \frac{dx}{(x + \frac{1}{2})^2 + \left(\frac{\sqrt{3}}{2}\right)^2}\)
\(\therefore \frac{1}{\sqrt{3}/2} \tan^{-1} \left[ \frac{y + 1/2}{\sqrt{3}/2} \right] = -\frac{1}{\sqrt{3}/2} \tan^{-1} \left[ \frac{x + 1/2}{\sqrt{3}/2} \right] + C\)
\(\Rightarrow \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2y + 1}{\sqrt{3}} \right) + \frac{2}{\sqrt{3}} \tan^{-1} \left( \frac{2x + 1}{\sqrt{3}} \right) = C\)
\(\Rightarrow \frac{2}{\sqrt{3}} \left[ \tan^{-1} \left( \frac{2y + 1}{\sqrt{3}} \right) + \tan^{-1} \left( \frac{2x + 1}{\sqrt{3}} \right) \right] = C\)
\(\Rightarrow \frac{2}{\sqrt{3}} \tan^{-1} \left[ \frac{\frac{2y + 1}{\sqrt{3}} + \frac{2x + 1}{\sqrt{3}}}{1 - \left( \frac{2y + 1}{\sqrt{3}} \right) \left( \frac{2x + 1}{\sqrt{3}} \right)} \right] = C\)
\(\Rightarrow \tan^{-1} \left[ \frac{\sqrt{3} (2y + 1 + 2x + 1)}{3 - (2y + 1)(2x + 1)} \right] = \frac{\sqrt{3}}{2} C\)
\(\Rightarrow \left[ \frac{2\sqrt{3} (x + y + 1)}{2(1 - x - y - 2xy)} \right] = \tan \left( \frac{\sqrt{3}}{2} C \right)\)
\(\Rightarrow \frac{x + y + 1}{1 - x - y - 2xy} = \frac{1}{\sqrt{3}} \tan \left( \frac{\sqrt{3}}{2} C \right) = A\)
\(\Rightarrow (x + y + 1) = A (1 - x - y - 2xy)\).

Question. Solve the differential equation \(ye^{x/y} dx = (xe^{x/y} + y^2) dy (y \neq 0)\).
Answer: The given equation is \(ye^{x/y} dx = (xe^{x/y} + y^2) dy\).
\(\therefore \frac{dx}{dy} = \frac{xe^{x/y} + y^2}{ye^{x/y}}\), is a homogeneous differential equation. ...(i)
Put \(x = vy\) so that \(\frac{dx}{dy} = v + y \frac{dv}{dy}\)
\(\therefore v + y \frac{dv}{dy} = \frac{vye^{v} + y^2}{ye^{v}} \Rightarrow v + y \frac{dv}{dy} = \frac{y(ve^{v} + y)}{ye^{v}} \Rightarrow y \frac{dv}{dy} = \frac{ve^v + y}{e^v} - v\)
\(\Rightarrow y \frac{dv}{dy} = \frac{ve^v + y - ve^v}{e^v} \Rightarrow y \frac{dv}{dy} = \frac{y}{e^v} \Rightarrow e^v dv = dy\)
\(\Rightarrow \int e^v dv = \int dy \Rightarrow e^v = y + C \Rightarrow e^{x/y} = y + C\)

Question. Find the particular solution of the differential equation: \(\frac{dy}{dx} + y \cot x = 4x \csc x (x \neq 0)\), given that \(y = 0\) when \(x = \frac{\pi}{2}\).
Answer: Given differential equation is \(\frac{dy}{dx} + y \cot x = 4x \csc x\).
It is of the type \(\frac{dy}{dx} + Py = Q\), where \(P = \cot x\), \(Q = 4x \csc x\).
\(\therefore \text{IF} = e^{\int P dx} = e^{\int \cot x dx} = e^{\log | \sin x |} = \sin x\)
Its solution is given by
\(y \sin x = \int (4x \csc x \cdot \sin x) dx\)
\(\Rightarrow y \sin x = \int 4x dx = \frac{4x^2}{2} + C \Rightarrow y \sin x = 2x^2 + C\) ... (i)
Now, put \(y = 0\) when \(x = \frac{\pi}{2}\) in (i), we get
\(0 = 2 \times \frac{\pi^2}{4} + C \Rightarrow C = -\frac{\pi^2}{2}\)
Hence, the particular solution of given differential equation is \(y \sin x = 2x^2 - \frac{\pi^2}{2}\)

Question. Write the integrating factor of the following differential equation: \((1 + y^2) + (2xy - \cot y)\frac{dy}{dx} = 0\) 
Answer: \((1 + y^2) + (2xy - \cot y)\frac{dy}{dx} = 0\)
\(\Rightarrow (2xy - \cot y)\frac{dy}{dx} = -(1 + y^2) \Rightarrow \frac{dy}{dx} = -\frac{1 + y^2}{2xy - \cot y}\)
\(\Rightarrow \frac{dx}{dy} = -\frac{(2xy - \cot y)}{1 + y^2} \Rightarrow \frac{dx}{dy} + \frac{2y}{1 + y^2} \cdot x = \frac{\cot y}{1 + y^2}\)
It is in the form \(\frac{dx}{dy} + Px = Q\), where \(P\) and \(Q\) are functions of \(y\).
\(\Rightarrow \text{IF} = e^{\int P dy} = e^{\int \frac{2y}{1 + y^2} dy} = e^{\log |1 + y^2|} = 1 + y^2\)

Short Answer Questions-II

Question. Solve the following differential equation: \((1 + e^{y/x})dy + e^{y/x} \left(1 - \frac{y}{x}\right) dx = 0, (x \neq 0)\). 
Answer: Given differential equation
\((1 + e^{y/x})dy + e^{y/x} \left(1 - \frac{y}{x}\right) dx = 0\)
\(\Rightarrow (1 + e^{y/x})dy = \left(\frac{y}{x} - 1\right) e^{y/x} dx\)
\(\Rightarrow \frac{dy}{dx} = \frac{\left(\frac{y}{x} - 1\right) e^{y/x}}{(1 + e^{y/x})}\)
It is a homogeneous differential equation
Put \(y = vx\) and \(\frac{dy}{dx} = v + x \frac{dv}{dx}\)
We have,
\(v + x \frac{dv}{dx} = \frac{(v - 1) e^v}{1 + e^v} = \frac{ve^v - e^v}{1 + e^v}\)
\(\Rightarrow x \frac{dv}{dx} = \frac{ve^v - e^v}{1 + e^v} - v = \frac{ve^v - e^v - v - ve^v}{1 + e^v}\)
\(\Rightarrow x \frac{dv}{dx} = -\frac{(v + e^v)}{1 + e^v}\)
\(\Rightarrow \frac{1 + e^v}{v + e^v} dv = -\frac{dx}{x}\)
On integrating both sides, we have
\(\int \frac{1 + e^v}{v + e^v} dv = -\int \frac{dx}{x}\)
\(\Rightarrow \log |v + e^v| = -\log |x| + \log |C|\)
\(\Rightarrow \log |v + e^v| + \log |x| = \log |C|\)
\(\Rightarrow \log |x(v + e^v)| = \log |C|\)
\(\Rightarrow x(v + e^v) = C \Rightarrow x\left(\frac{y}{x} + e^{y/x}\right) = C\)
\(\Rightarrow y + x e^{y/x} = C\)

Question. Find the particular solution of the differential equation \(\log\left(\frac{dy}{dx}\right) = 3x + 4y\), given that \(y = 0\) when \(x = 0\). 
Answer: Given differential equation is \(\log\left(\frac{dy}{dx}\right) = 3x + 4y\)
\(\Rightarrow \frac{dy}{dx} = e^{3x+4y} \Rightarrow \frac{dy}{dx} = e^{3x} \cdot e^{4y}\)
\(\Rightarrow \frac{dy}{e^{4y}} = e^{3x} dx \Rightarrow e^{-4y} dy = e^{3x} dx\)
Integrating both sides, we get
\(\int e^{-4y} dy = \int e^{3x} dx\)
\(\Rightarrow \frac{e^{-4y}}{-4} = \frac{e^{3x}}{3} + C_1 \Rightarrow -3e^{-4y} = 4e^{3x} + 12C_1\)
\(\Rightarrow 4e^{3x} + 3e^{-4y} = -12C_1 \Rightarrow 4e^{3x} + 3e^{-4y} = C\) … (i)
It is general solution.
Now for particular solution we put \(x = 0\) and \(y = 0\) in (i), we get
\(4 + 3 = C \Rightarrow C = 7\)
Putting \(C = 7\) in (i), we get
\(4e^{3x} + 3e^{-4y} = 7\)
It is required particular solution.

Question. Solve the following differential equation: \(2x^2 \frac{dy}{dx} - 2xy + y^2 = 0\) 
Answer: Given \(2x^2 \frac{dy}{dx} - 2xy + y^2 = 0\)
\(\Rightarrow 2x^2 \frac{dy}{dx} = 2xy - y^2 \Rightarrow \frac{dy}{dx} = \frac{2xy - y^2}{2x^2}\) ...(i)
It is homogeneous differential equation.
Let \(y = vx \Rightarrow \frac{dy}{dx} = v + x \frac{dv}{dx}\)
Equation (i) becomes
\(v + x \frac{dv}{dx} = \frac{2x \cdot vx - v^2 x^2}{2x^2}\)
\(\Rightarrow v + x \frac{dv}{dx} = \frac{2x^2 v - v^2 x^2}{2x^2} \Rightarrow v + x \frac{dv}{dx} = v - \frac{v^2}{2}\)
\(\Rightarrow x \frac{dv}{dx} = -\frac{v^2}{2} \Rightarrow \frac{dx}{x} = -\frac{2dv}{v^2}\)
Integrating both sides, we get
\(\int \frac{dx}{x} = -2 \int \frac{dv}{v^2}\)
\(\Rightarrow \log |x| + C = -2 \frac{v^{-2+1}}{-2+1} \Rightarrow \log |x| + C = 2 \cdot \frac{1}{v}\)
Putting \(v = \frac{y}{x}\), we get \(\log |x| + C = \frac{2x}{y}\)

Question. Find the particular solution of the differential equation: \(x(x^2 - 1) \frac{dy}{dx} = 1; y = 0\) when \(x = 2\) 
Answer: Given differential equation is,
\(x(x^2 - 1) \frac{dy}{dx} = 1 \Rightarrow dy = \frac{dx}{x(x^2 - 1)} \Rightarrow dy = \frac{dx}{x(x - 1)(x + 1)}\)
Integrating both sides, we get,
\(\int dy = \int \frac{dx}{x(x - 1)(x + 1)} \Rightarrow y = \int \frac{dx}{x(x - 1)(x + 1)}\) ...(i)
Let \(\frac{1}{x(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x + 1}\)
\(\Rightarrow 1 = A(x - 1)(x + 1) + Bx(x + 1) + Cx(x - 1)\)
Putting \(x = 1\), we get \(1 = 0 + B \cdot 1 \cdot 2 + 0 \Rightarrow B = \frac{1}{2}\)
Putting \(x = -1\), we get \(1 = 0 + 0 + C(-1)(-2) \Rightarrow C = \frac{1}{2}\)
Putting \(x = 0\), we get \(1 = A(-1)(1) \Rightarrow A = -1\)
Hence, \(\frac{1}{x(x - 1)(x + 1)} = \frac{-1}{x} + \frac{1}{2(x - 1)} + \frac{1}{2(x + 1)}\)
From (i) \(y = \int \left( -\frac{1}{x} + \frac{1}{2(x - 1)} + \frac{1}{2(x + 1)} \right) dx\)
\(\Rightarrow y = -\int \frac{dx}{x} + \frac{1}{2} \int \frac{dx}{x - 1} + \frac{1}{2} \int \frac{dx}{x + 1}\)
\(\Rightarrow y = -\log x + \frac{1}{2} \log |x - 1| + \frac{1}{2} \log |x + 1| + \log C_1\)
\(\Rightarrow 2y = -2 \log x + \log |x^2 - 1| + 2 \log C_1 \Rightarrow 2y = \log \left| \frac{x^2 - 1}{x^2} \right| + \log C_1^2\) ...(ii)
When \(x = 2, y = 0\)
\(\Rightarrow 0 = \log \left| \frac{4 - 1}{4} \right| + \log C_1^2 \Rightarrow \log C_1^2 = -\log \frac{3}{4}\)
Putting \(\log C_1^2 = -\log \frac{3}{4}\) in (ii), we get
\(2y = \log \left| \frac{x^2 - 1}{x^2} \right| - \log \frac{3}{4} \Rightarrow y = \frac{1}{2} \log \left| \frac{x^2 - 1}{x^2} \right| - \frac{1}{2} \log \frac{3}{4}\)

Question. Solve the differential equation \((1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1}x}\). 
Answer: Given differential equation is
\((1 + x^2) \frac{dy}{dx} + y = e^{\tan^{-1}x} \Rightarrow \frac{dy}{dx} + \frac{1}{1 + x^2} y = \frac{e^{\tan^{-1}x}}{1 + x^2}\) …(i)
Equation (i) is of the form \(\frac{dy}{dx} + Py = Q\), where \(P = \frac{1}{1 + x^2}, Q = \frac{e^{\tan^{-1}x}}{1 + x^2}\)
\(\therefore \text{IF} = e^{\int P dx} = e^{\int \frac{1}{1 + x^2} dx} = e^{\tan^{-1}x}\)
Therefore, general solution of required differential equation is
\(y \cdot e^{\tan^{-1}x} = \int e^{\tan^{-1}x} \cdot \frac{e^{\tan^{-1}x}}{1 + x^2} dx + C \Rightarrow y \cdot e^{\tan^{-1}x} = \int \frac{e^{2\tan^{-1}x}}{1 + x^2} dx + C\) …(ii)
Let \(\tan^{-1} x = z \Rightarrow \frac{1}{1 + x^2} dx = dz\)
(ii) becomes
\(y \cdot e^{\tan^{-1}x} = \int e^{2z} dz + C \Rightarrow y \cdot e^{\tan^{-1}x} = \frac{e^{2z}}{2} + C\)
\(\Rightarrow y \cdot e^{\tan^{-1}x} = \frac{e^{2\tan^{-1}x}}{2} + C\) [Putting \(z = \tan^{-1}x\)]
\(\Rightarrow y = \frac{e^{\tan^{-1}x}}{2} + C e^{-\tan^{-1}x}\) [Dividing both sides by \(e^{\tan^{-1}x}\)]
It is the required solution.

Question. Find the particular solution of the differential equation \(e^x\sqrt{1 - y^2}dx + \frac{y}{x}dy = 0\) given that \(y = 1\) when \(x = 0\). 
Answer: We have, \(e^x\sqrt{1 - y^2}dx + \frac{y}{x}dy = 0\)
\(\Rightarrow e^x\sqrt{1 - y^2}dx = -\frac{y}{x} dy \Rightarrow xe^x dx = -\frac{y}{\sqrt{1 - y^2}} dy\)
\(\Rightarrow \int xe^x dx = -\int \frac{y}{\sqrt{1 - y^2}} dy\)
\(\Rightarrow x e^x - \int e^x dx = \frac{1}{2} \int \frac{dt}{\sqrt{t}}\), where \(t = 1 - y^2\) (Using ILATE on LHS)
\(\Rightarrow xe^x - e^x = \frac{1}{2} \left( \frac{t^{1/2}}{1/2} \right) + C \Rightarrow xe^x - e^x = \sqrt{t} + C\)
\(\Rightarrow xe^x - e^x = \sqrt{1 - y^2} + C\), is the general solution.
Putting \(y = 1\) and \(x = 0\), we get
\(0 \cdot e^0 - e^0 = \sqrt{1 - 1^2} + C \Rightarrow C = -1\)
Therefore, required particular solution is \(xe^x - e^x = \sqrt{1 - y^2} - 1\).

Question. Solve the differential equation: \((\tan^{-1} y - x) dy = (1 + y^2) dx\) 
Answer: The given differential equation can be written as
\(\frac{dx}{dy} + \frac{x}{1 + y^2} = \frac{\tan^{-1} y}{1 + y^2}\) ...(i)
Now, (i) is of the form \(\frac{dx}{dy} + Px = Q\), where \(P = \frac{1}{1 + y^2}\) and \(Q = \frac{\tan^{-1} y}{1 + y^2}\)
Therefore, \(\text{IF} = e^{\int \frac{1}{1 + y^2} dy} = e^{\tan^{-1} y}\)
Thus, the solution of the given differential equation is
\(x e^{\tan^{-1} y} = \int \left( \frac{\tan^{-1} y}{1 + y^2} \right) e^{\tan^{-1} y} dy + C\) ...(ii)
Let \(I = \int \left( \frac{\tan^{-1} y}{1 + y^2} \right) e^{\tan^{-1} y} dy\)
Substituting \(\tan^{-1} y = t\) so that \(\left( \frac{1}{1 + y^2} \right) dy = dt\), we get
\(I = \int t e^t dt = t e^t - \int 1 \cdot e^t dt = t e^t - e^t = e^t(t - 1)\)
or \(I = e^{\tan^{-1} y}(\tan^{-1} y - 1)\)
Substituting the value of \(I\) in equation (ii), we get
\(x \cdot e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y - 1) + C\)
or \(x = (\tan^{-1} y - 1) + C e^{-\tan^{-1} y}\) is the required solution.

Question. Solve the following differential equation: \(\left[ \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1, x \neq 0\) 
Answer: Given \(\left[ \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \right] \frac{dx}{dy} = 1, x \neq 0\)
\(\Rightarrow \frac{dy}{dx} = \frac{e^{-2\sqrt{x}}}{\sqrt{x}} - \frac{y}{\sqrt{x}} \Rightarrow \frac{dy}{dx} + \frac{1}{\sqrt{x}} \cdot y = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)
It is in the form \(\frac{dy}{dx} + Py = Q\), where \(P = \frac{1}{\sqrt{x}}, Q = \frac{e^{-2\sqrt{x}}}{\sqrt{x}}\)
\(\therefore \text{IF} = e^{\int P dx} = e^{\int \frac{1}{\sqrt{x}} dx} = e^{\int x^{-1/2} dx} = e^{\frac{x^{1/2}}{1/2}} = e^{2\sqrt{x}}\)
Therefore general solution is
\(y \cdot e^{2\sqrt{x}} = \int Q \times \text{IF} dx + C \Rightarrow y \cdot e^{2\sqrt{x}} = \int \frac{e^{-2\sqrt{x}}}{\sqrt{x}} \cdot e^{2\sqrt{x}} dx + C\)
\(\Rightarrow y \cdot e^{2\sqrt{x}} = \int \frac{dx}{\sqrt{x}} + C \Rightarrow y \cdot e^{2\sqrt{x}} = \frac{x^{-1/2+1}}{-1/2+1} + C\)
\(\Rightarrow y \cdot e^{2\sqrt{x}} = 2\sqrt{x} + C\)

Question. Solve the differential equation \((x^2 - 1) \frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1}\), where \(x \in (-\infty, -1) \cup (1, \infty)\).
Answer: The given differential equation is \((x^2 - 1) \frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1}\)
\(\Rightarrow \frac{dy}{dx} + \frac{2x}{x^2 - 1} y = \frac{2}{(x^2 - 1)^2}\) ...(i)
This is a linear differential equation of the form \(\frac{dy}{dx} + Py = Q\), where \(P = \frac{2x}{x^2 - 1}\) and \(Q = \frac{2}{(x^2 - 1)^2}\)
\(\therefore \text{IF} = e^{\int P dx} = e^{\int 2x/(x^2-1) dx} = e^{\log |x^2-1|} = x^2 - 1\)
Multiplying both sides of (i) by \(\text{IF} = x^2 - 1\), we get \((x^2 - 1) \frac{dy}{dx} + 2xy = \frac{2}{x^2 - 1}\)
Integrating both sides, we get
\(y(x^2 - 1) = \int \frac{2}{x^2 - 1} dx + C\)
\(\Rightarrow y(x^2 - 1) = 2 \cdot \frac{1}{2} \log \left| \frac{x - 1}{x + 1} \right| + C \Rightarrow y(x^2 - 1) = \log \left| \frac{x - 1}{x + 1} \right| + C\)
This is the required solution.

Question. Find the particular solution of the differential equation \(\frac{dy}{dx} = 1 + x + y + xy\) given that \(y = 0\) when \(x = 1\). 
Answer: Given differential equation is \(\frac{dy}{dx} = 1 + x + y + xy\)
\(\Rightarrow \frac{dy}{dx} = (1 + x) + y(1 + x) \Rightarrow \frac{dy}{dx} = (1 + x)(1 + y) \Rightarrow \frac{dy}{1 + y} = (1 + x)dx\)
Integrating both sides, we get \(\log |1 + y| = \int (1 + x)dx\)
\(\Rightarrow \log |1 + y| = x + \frac{x^2}{2} + C\) is the general solution.
Putting \(x = 1, y = 0\), we get
\(\log |1 + 0| = 1 + \frac{1}{2} + C \Rightarrow 0 = \frac{3}{2} + C \Rightarrow C = -\frac{3}{2}\)
Hence, particular solution is \(\log |1 + y| = x + \frac{x^2}{2} - \frac{3}{2}\).

Question. Solve the differential equation \(x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x\). 
Answer: Given differential equation is \(x \log x \frac{dy}{dx} + y = \frac{2}{x} \log x\)
\(\Rightarrow \frac{dy}{dx} + \left( \frac{1}{x \log x} \right) y = \frac{2}{x^2}\) (Divide each term by \(x \log x\))
It is in the form \(\frac{dy}{dx} + Py = Q\), where \(P = \frac{1}{x \log x}\) and \(Q = \frac{2}{x^2}\).
\(\therefore \text{IF} = e^{\int P dx} = e^{\int \frac{dx}{x \log x}}\)
Put \(\log x = z \Rightarrow \frac{dx}{x} = dz \Rightarrow e^{\int \frac{1}{z} dz} = e^{\log z} = z = \log x\)
\(\therefore\) General solution is
\(y \cdot \log x = \int \log x \cdot \frac{2}{x^2} dx + C \Rightarrow y \log x = 2 \int \frac{\log x}{x^2} dx + C\)
Let \(\log x = z \Rightarrow \frac{1}{x} dx = dz\) also \(\log x = z \Rightarrow x = e^z\)
\(\therefore y \log x = 2 \int \frac{z}{e^{2z}} \frac{e^z dz}{e^z} + C \Rightarrow y \log x = 2 \int z \cdot e^{-z} dz + C\)
\(\Rightarrow y \log x = 2 \left[ z \cdot \frac{e^{-z}}{-1} - \int \frac{e^{-z}}{-1} dz \right] + C \Rightarrow y \log x = 2 \left[ -z e^{-z} + \int e^{-z} dz \right] + C\)
\(\Rightarrow y \log x = -2z e^{-z} - 2 e^{-z} + C \Rightarrow y \log x = -2 \log x \cdot e^{-\log x} - 2 e^{-\log x} + C\)
\(\Rightarrow y \log x = -2 \log x \cdot \frac{1}{x} - 2 \cdot \frac{1}{x} + C\) [\(\because e^{-\log x} = e^{\log \frac{1}{x}} = \frac{1}{x}\)]
\(\Rightarrow y \log x = -\frac{2}{x} (1 + \log x) + C\)