CBSE Class 12 Mathematics Differential Equations Notes Set 04

Important Notes and Formulas

Many of the practical problems in physics and engineering can be converted into differential equations. The solution of differential equations is, therefore, of paramount importance. This chapter deals with some elementary aspects of differential equations. These are addressed through simple application of differential and integral calculus.

Definition

An equation involving independent variable x, dependent variable y and the differential coefficients \( \frac{dy}{dx}, \frac{d^2y}{dx^2}, \ldots \) is called a differential equation.
Examples: (i) \( \frac{dy}{dx} = 1 + x + y \)    (ii) \( \frac{dy}{dx} + xy = \cot x \)
(iii) \( \left(\frac{d^4y}{dx^4}\right)^3 - 4\frac{dy}{dx} + 4y = 5\cos 3x \)    (iv) \( x^2\frac{d^2y}{dx^2} + \sqrt{1 + \left(\frac{dy}{dx}\right)^2} = 0 \)

(1) Order of a differential equation: The order of a differential equation is the order of the highest derivative occurring in the differential equation. For example, the orders of the above differential equations are 1, 1, 4 and 2 respectively.
Note: The order of a differential equation is a positive integer. To determine the order of a differential equation, it is not needed to make the equation free from radicals.

(2) Degree of a differential equation: The degree of a differential equation is the degree of the highest order derivative, when differential coefficients are made free from radicals and fractions. In other words, the degree of a differential equation is the power of the highest order derivative occurring in the differential equation when it is written as a polynomial in differential coefficients.
Note: The definition of degree does not require variables x, y, t etc. to be free from radicals and fractions. The degrees of the above differential equations are 1, 1, 3 and 2 respectively.

Solution of a Differential Equation

(1) Solution of differential equations: If we have a differential equation of order n, then by solving a differential equation we mean to get a family of curves with n parameters whose differential equation is the given differential equation. Solution or integral of a differential equation is a relation between the variables, not involving the differential coefficients, such that this relation and the derivatives obtained from it satisfy the given differential equation. The solution of a differential equation is also called its primitive.
For example, \( y = e^x \) is a solution of the differential equation \( \frac{dy}{dx} = y \).
(i) General solution: The solution which contains as many arbitrary constants as the order of the differential equation is called the general solution of the differential equation. For example, \( y = A\cos x + B\sin x \) is the general solution of the differential equation \( \frac{d^2y}{dx^2} + y = 0 \). But \( y = A\cos x \) is not the general solution as it contains one arbitrary constant.
(ii) Particular solution: Solution obtained by giving particular values to the arbitrary constants in the general solution of a differential equation is called a particular solution. For example, \( y = 3\cos x + 2\sin x \) is a particular solution of the differential equation \( \frac{d^2y}{dx^2} + y = 0 \).

(2) Differential equations of first order and first degree: A differential equation of first order and first degree involves x, y and \( \frac{dy}{dx} \). So it can be put in any one of the following forms:
\( \frac{dy}{dx} = f(x, y) \) or \( f\left(x, y, \frac{dy}{dx}\right) = 0 \) or \( f(x, y)\,dx + g(x, y)\,dy = 0 \), where f(x, y) and g(x, y) are obviously the functions of x and y.

(3) Solution of first order and first degree differential equations: A first order and first degree differential equation can be written as
f(x, y) dx + g(x, y) dy = 0 or \( \frac{dy}{dx} = \frac{f(x, y)}{g(x, y)} \) or \( \frac{dy}{dx} = \phi(x, y) \)
where f(x, y) and g(x, y) are obviously the functions of x and y. It is not always possible to solve this type of equations. The solution of this type of differential equations is possible only when it falls under the category of some standard forms.

Variable Separable Type Differential Equation

Equations in variable separable form: If the differential equation is of the form
\( f_1(x)\,dx = f_2(y)\,dy \)    ...(i)
where \( f_1 \) and \( f_2 \) are functions of x and y only, then we say that the variables are separable in the differential equation.
Thus, integrating both sides of (i), we get its solution as \( \int f_1(x)\,dx = \int f_2(y)\,dy + C \), where C is an arbitrary constant.
There is no need of introducing arbitrary constants on both sides as they can be combined together to give just one.

(i) Differential equations of the type \( \frac{dy}{dx} = f(x) \)
To solve this type of differential equations we integrate both sides to obtain the general solution as follows:
\( \frac{dy}{dx} = f(x) \Leftrightarrow dy = f(x)\,dx \)
Integrating both sides, we obtain \( \int dy = \int f(x)\,dx + C \) or \( y = \int f(x)\,dx + C \).

(ii) Differential equations of the type \( \frac{dy}{dx} = f(y) \)
To solve this type of differential equations we integrate both sides to obtain the general solution as follows:
\( \frac{dy}{dx} = f(y) \Rightarrow \frac{dx}{dy} = \frac{1}{f(y)} \Rightarrow dx = \frac{1}{f(y)}\,dy \)
Integrating both sides, we obtain \( \int dx = \int \frac{1}{f(y)}\,dy + C \) or \( x = \int \frac{1}{f(y)}\,dy + C \).

Homogeneous Differential Equation

(1) Homogeneous differential equation: A function f(x, y) is called a homogeneous function of degree n if \( f(\lambda x, \lambda y) = \lambda^n f(x, y) \).
For example, \( f(x, y) = x^2 - y^2 + 3xy \) is a homogeneous function of degree 2, because \( f(\lambda x, \lambda y) = \lambda^2x^2 - \lambda^2y^2 + 3\lambda x \cdot \lambda y = \lambda^2 f(x, y) \). A homogeneous function f(x, y) of degree n can always be written as \( f(x, y) = x^n f\left(\frac{y}{x}\right) \) or \( f(x, y) = y^n f\left(\frac{x}{y}\right) \). If a first-order first-degree differential equation is expressible in the form \( \frac{dy}{dx} = \frac{f(x, y)}{g(x, y)} \) where f(x, y) and g(x, y) are homogeneous functions of the same degree, then it is called a homogeneous differential equation. Such type of equations can be reduced to variable separable form by the substitution y = vx. The given differential equation can be written as \( \frac{dy}{dx} = \frac{x^n f(y/x)}{x^n g(y/x)} = \frac{f(y/x)}{g(y/x)} = F\left(\frac{y}{x}\right) \). If y = vx, then \( \frac{dy}{dx} = v + x\frac{dv}{dx} \). Substituting the value of \( \frac{dy}{dx} \) in \( \frac{dy}{dx} = F\left(\frac{y}{x}\right) \), we get \( v + x\frac{dv}{dx} = F(v) \Rightarrow \frac{dv}{F(v) - v} = \frac{dx}{x} \). On integration, \( \int \frac{1}{F(v) - v}\,dv = \int \frac{dx}{x} + C \), where C is an arbitrary constant of integration. After integration, v will be replaced by \( \frac{y}{x} \) in the complete solution.

(2) Algorithm for solving homogeneous differential equation
Step (i): Put the differential equation in the form \( \frac{dy}{dx} = \frac{\phi(x, y)}{\psi(x, y)} \).
Step (ii): Put y = vx and \( \frac{dy}{dx} = v + x\frac{dv}{dx} \) in the equation in step (i) and cancel out x from the right hand side. The equation reduces to the form \( v + x\frac{dv}{dx} = F(v) \).
Step (iii): Shift v on RHS and separate the variables v and x.
Step (iv): Integrate both sides to obtain the solution in terms of v and x.
Step (v): Replace v by \( \frac{y}{x} \) in the solution obtained in step (iv) to obtain the solution in terms of x and y.

Linear Differential Equation

(1) Linear and non-linear differential equations: A differential equation is a linear differential equation if it is expressible in the form \( P_0\frac{d^ny}{dx^n} + P_1\frac{d^{n-1}y}{dx^{n-1}} + P_2\frac{d^{n-2}y}{dx^{n-2}} + \ldots + P_{n-1}\frac{dy}{dx} + P_ny = Q \) where \( P_0, P_1, P_2, \ldots, P_{n-1}, P_n \) and Q are either constants or functions of the independent variable x.
Thus, if a differential equation when expressed in the form of a polynomial involves the derivatives and dependent variable in the first power and there are no products of these, and also the coefficients of the various terms are either constants or functions of the independent variable, then it is said to be a linear differential equation. Otherwise, it is a non-linear differential equation.
It follows from the above definition that a differential equation will be a non-linear differential equation if (i) its degree is more than one (ii) any of the differential coefficients has exponent more than one (iii) exponent of the dependent variable is more than one (iv) products containing the dependent variable and its differential coefficients are present.

(2) Linear differential equation of first order: The general form of a linear differential equation of first order is
\( \frac{dy}{dx} + Py = Q \)    ...(i)
where P and Q are functions of x (or constants).
For example, \( \frac{dy}{dx} + xy = x^3 \), \( x\frac{dy}{dx} + 2y = x^3 \), \( \frac{dy}{dx} + 2y = \sin x \) etc. are linear differential equations. This type of differential equations are solved when they are multiplied by a factor, which is called the integrating factor, because by multiplication of this factor the left hand side of the differential equation (i) becomes the exact differential of some function.
Multiplying both sides of (i) by \( e^{\int P\,dx} \), we get \( e^{\int P\,dx}\left(\frac{dy}{dx} + Py\right) = Qe^{\int P\,dx} \Rightarrow \frac{d}{dx}\left\{ye^{\int P\,dx}\right\} = Qe^{\int P\,dx} \)
On integrating both sides w.r.t. x, we get \( ye^{\int P\,dx} = \int Qe^{\int P\,dx}\,dx + C \)    ...(ii)
which is the required solution, where C is the constant of integration. \( e^{\int P\,dx} \) is called the integrating factor. The solution (ii) in short may also be written as \( y \cdot (\text{I.F.}) = \int Q \cdot (\text{I.F.})\,dx + C \)

(3) Algorithm for solving a linear differential equation:
Step (i): Write the differential equation in the form \( \frac{dy}{dx} + Py = Q \) and obtain P and Q.
Step (ii): Find the integrating factor (I.F.) given by \( \text{I.F.} = e^{\int P\,dx} \).
Step (iii): Multiply both sides of the equation in step (i) by I.F.
Step (iv): Integrate both sides of the equation obtained in step (iii) w.r.t. x to obtain \( y(\text{I.F.}) = \int Q(\text{I.F.})\,dx + C \)
This gives the required solution.

(4) Linear differential equations of the form \( \frac{dx}{dy} + Rx = S \): Sometimes a linear differential equation can be put in the form \( \frac{dx}{dy} + Rx = S \) where R and S are functions of y or constants. Note that y is the independent variable and x is the dependent variable.

(5) Algorithm for solving linear differential equations of the form \( \frac{dx}{dy} + Rx = S \):
Step (i): Write the differential equation in the form \( \frac{dx}{dy} + Rx = S \) and obtain R and S.
Step (ii): Find I.F. by using \( \text{I.F.} = e^{\int R\,dy} \).
Step (iii): Multiply both sides of the differential equation in step (i) by I.F.
Step (iv): Integrate both sides of the equation obtained in step (iii) w.r.t. y to obtain the solution given by \( x(\text{I.F.}) = \int S(\text{I.F.})\,dy + C \), where C is the constant of integration.

Application of Differential Equation

Differential equations are applied in various practical fields of life. They are used to define various physical laws and quantities and are widely used in physics, chemistry, engineering etc.
Some important fields of application are: (i) Rate of change (ii) Geometrical problems etc.
Differential equations are used for finding the family of curves for which some conditions involving the derivatives are given.

Particular Solution Type Problems

To solve such a problem, we proceed according to the type of the problem (i.e. variable separable, homogeneous, linear etc.) and then we apply the given conditions to find the particular values of the arbitrary constants.

Solved Examples

Example 1. The order and degree of the differential equation \( y = x\frac{dy}{dx} + \sqrt{a^2\left(\frac{dy}{dx}\right)^2 + b^2} \) are
(a) 1, 2
(b) 2, 1
(c) 1, 1
(d) 2, 2
Answer:
Ans. (a)
Clearly, the highest order derivative involved is \( \frac{dy}{dx} \), having order 1.
Expressing the above differential equation as a polynomial in derivative, we have \( \left(y - x\frac{dy}{dx}\right)^2 = a^2\left(\frac{dy}{dx}\right)^2 + b^2 \)
i.e., \( (x^2 - a^2)\left(\frac{dy}{dx}\right)^2 - 2xy\frac{dy}{dx} + y^2 - b^2 = 0 \)
In this equation, the power of the highest order derivative is 2. So its degree is 2.

Example 2. The order and degree of the differential equation \( \frac{d^2y}{dx^2} + \left(\frac{dy}{dx}\right)^{\frac{1}{3}} + x^{\frac{1}{4}} = 0 \) are respectively
(a) 2, 3
(b) 3, 3
(c) 2, 6
(d) 2, 4
Answer:
Ans. (a)
The highest order derivative involved is \( \frac{d^2y}{dx^2} \) which is the 2nd order derivative. Hence the order of the differential equation is 2. Making the above equation free from radicals, as far as the derivatives are concerned, we have
\( \left(\frac{d^2y}{dx^2} + x^{\frac{1}{4}}\right)^3 = -\frac{dy}{dx} \) i.e. \( \left(\frac{d^2y}{dx^2} + x^{\frac{1}{4}}\right)^3 + \frac{dy}{dx} = 0 \)
The exponent of the highest order derivative \( \frac{d^2y}{dx^2} \) will be 3. Hence the degree of the differential equation is 3.

Example 3. The degree of the differential equation \( \frac{d^2y}{dx^2} + 3\left(\frac{dy}{dx}\right)^2 = x^2\log\left(\frac{d^2y}{dx^2}\right) \) is
(a) 1
(b) 2
(c) 3
(d) None of these
Answer:
Ans. (d)
The above equation cannot be written as a polynomial in derivatives due to the term \( x^2\log\left(\frac{d^2y}{dx^2}\right) \).
Hence the degree of the differential equation is 'not defined'.

Example 4. The order of the differential equation whose solution is \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
Ans. (c)
To eliminate the arbitrary constants g, f and c, we need 3 more equations, that is by differentiating the equation 3 times. Hence the highest order derivative will be \( \frac{d^3y}{dx^3} \). Hence the order of the differential equation will be 3.

Example 5. The order of the differential equation whose general solution is \( y = c_1e^x + c_2e^{2x} + c_3e^{3x} + c_4e^{x + c_5} \), where \( c_1, c_2, c_3, c_4, c_5 \) are arbitrary constants, is
(a) 5
(b) 4
(c) 3
(d) None of these
Answer:
Ans. (c)
Rewriting the given general solution, we have
\( y = c_1e^x + c_2e^{2x} + c_3e^{3x} + c_4e^x \cdot e^{c_5} \)
\( = (c_1 + c_4e^{c_5})e^x + c_2e^{2x} + c_3e^{3x} = c_1'e^x + c_2e^{2x} + c_3e^{3x} \)
where \( c_1' = c_1 + c_4e^{c_5} \). So there are 3 arbitrary constants associated with different terms. Hence the order of the differential equation formed will be 3.

Example 6. The order and degree of \( y = 1 + \frac{dy}{dx} + \frac{1}{2!}\left(\frac{dy}{dx}\right)^2 + \frac{1}{3!}\left(\frac{dy}{dx}\right)^3 + \ldots \) is
(a) 1, 2
(b) 1, 1
(c) Order 1, degree not defined
(d) None of these
Answer:
Ans. (b)
The given differential equation can be re-written as \( y = e^{\frac{dy}{dx}} \Rightarrow \ln y = \frac{dy}{dx} \)
This is a polynomial in derivative. Hence order is 1 and degree 1.

Example 7. The order and degree of \( \frac{d^2y}{dx^2} = \sin\left(\frac{dy}{dx}\right) + x \) is
(a) 2, 1
(b) Order 2, degree not defined
(c) 2, 0
(d) None of these
Answer:
Ans. (b)
As the highest order derivative involved is \( \frac{d^2y}{dx^2} \), the order is 2.
The given differential equation cannot be written as a polynomial in derivatives, so the degree is not defined.

Example 8. \( y = \frac{x}{x + 1} \) is a solution of the differential equation
(a) \( y^2\frac{dy}{dx} = x^2 \)
(b) \( x^2\frac{dy}{dx} = y^2 \)
(c) \( y\frac{dy}{dx} = x \)
(d) \( x\frac{dy}{dx} = y \)
Answer:
Ans. (b)
We have \( y = \frac{x}{x + 1} \Rightarrow \frac{1}{y} = \frac{x + 1}{x} = 1 + \frac{1}{x} \)
Differentiating w.r.t. x,
\( -\frac{1}{y^2}\frac{dy}{dx} = 0 - \frac{1}{x^2} \)
∴ \( x^2\frac{dy}{dx} = y^2 \)

Example 9. The solution of the differential equation \( (1 + x^2)\frac{dy}{dx} = x(1 + y^2) \) is [AISSE 1983]
(a) \( 2\tan^{-1}y = \log(1 + x^2) + c \)
(b) \( \tan^{-1}y = \log(1 + x^2) + c \)
(c) \( 2\tan^{-1}y + \log(1 + x^2) + c = 0 \)
(d) None of these
Answer:
Ans. (a)
Separating the variables, we can re-write the given differential equation as
\( \frac{x\,dx}{1 + x^2} = \frac{dy}{1 + y^2} \Rightarrow \int \frac{2x\,dx}{1 + x^2} = 2\int \frac{dy}{1 + y^2} \Rightarrow 2\tan^{-1}y = \log_e(1 + x^2) + c \)

Example 10. The solution of the differential equation \( \frac{dy}{dx} = x^2 + \sin 3x \) is [DSSE 1981]
(a) \( y = \frac{x^3}{3} + \frac{\cos 3x}{3} + c \)
(b) \( y = \frac{x^3}{3} - \frac{\cos 3x}{3} + c \)
(c) \( y = \frac{x^3}{3} + \sin 3x + c \)
(d) None of these
Answer:
Ans. (b)
We have \( dy = (x^2 + \sin 3x)\,dx \Rightarrow \int dy = \int (x^2 + \sin 3x)\,dx \Rightarrow y = \frac{x^3}{3} - \frac{\cos 3x}{3} + c \)

Example 11. The solution of \( \frac{dy}{dx} = \frac{1}{y^2 + \sin y} \) is
(a) \( x = \frac{y^3}{3} - \cos y + c \)
(b) y + cos y = x + c
(c) \( x = \frac{y^3}{3} + \cos y + c \)
(d) None of these
Answer:
Ans. (a)
The given equation may be re-written as \( dx = (y^2 + \sin y)\,dy \)
Integrating, \( \int dx = \int (y^2 + \sin y)\,dy \)
∴ \( x = \frac{y^3}{3} - \cos y + c \)

Example 12. The solution of the differential equation \( x\frac{dy}{dx} = y(\log y - \log x + 1) \) is [IIT 1986]
(a) \( y = xe^{cx} \)
(b) \( y + xe^{cx} = 0 \)
(c) \( y + e^x = 0 \)
(d) None of these
Answer:
Ans. (a)
The given equation may be expressed as \( \frac{dy}{dx} = \frac{y}{x}\left[\log\left(\frac{y}{x}\right) + 1\right] \)    ...(i)
Let \( \frac{y}{x} = v \Rightarrow y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \)
∴ From (i), \( v + x\frac{dv}{dx} = v(\log v + 1) \Rightarrow x\frac{dv}{dx} = v\log v \Rightarrow \frac{dv}{v\log v} = \frac{dx}{x} \Rightarrow \int \frac{1}{\log v}\,d(\log v) = \int \frac{dx}{x} \)
∴ log (log v) = log x + log c ⇒ log (log v) = log (cx) ⇒ log v = cx ⇒ \( v = e^{cx} \Rightarrow \frac{y}{x} = e^{cx} \), ∴ \( y = xe^{cx} \)

Example 13. The solution of the differential equation \( yy' = x\left(\frac{y^2}{x^2} + \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)}\right) \) is
(a) \( \phi(y^2/x^2) = cx^2 \)
(b) \( x^2\phi(y^2/x^2) = c^2y^2 \)
(c) \( x^2\phi(y^2/x^2) = c \)
(d) \( \phi(y^2/x^2) = \frac{cy}{x} \)
Answer:
Ans. (a)
The given equation may be re-written as \( \frac{y}{x} \cdot \frac{dy}{dx} = \left(\frac{y}{x}\right)^2 + \frac{\phi((y/x)^2)}{\phi'((y/x)^2)} \)    ...(i)
Let \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \) and \( \frac{y}{x} = v \)
∴ From (i), \( v\left(v + x\frac{dv}{dx}\right) = v^2 + \frac{\phi(v^2)}{\phi'(v^2)} \Rightarrow vx\frac{dv}{dx} = \frac{\phi(v^2)}{\phi'(v^2)} \Rightarrow \frac{\phi'(v^2)(2v\,dv)}{\phi(v^2)} = 2\frac{dx}{x} \)
Integrating, \( \ln(\phi(v^2)) = 2\ln x + \ln c \Rightarrow \phi(v^2) = cx^2 \)
∴ \( \phi(y^2/x^2) = cx^2 \)

Example 14. The solution of \( \frac{dy}{dx} = \frac{y^3 + 2x^2y}{x^3 + 2xy^2} \) is
(a) \( (x^2 - y^2)^3 = Bx^2y^2 \)
(b) \( (x^2 + y^2)^3 = Bx^2y^2 \)
(c) \( (x^2 - y^2)^3 = x^2y^2 \)
(d) None of these
Answer:
Ans. (a)
The given equation is homogeneous. Let y = vx  ∴ \( \frac{dy}{dx} = v + x\frac{dv}{dx} \)
⇒ \( \frac{y^3 + 2x^2y}{x^3 + 2xy^2} = v + x\frac{dv}{dx} \Rightarrow \frac{(y/x)^3 + 2(y/x)}{1 + 2(y/x)^2} = v + x\frac{dv}{dx} \Rightarrow \frac{v^3 + 2v}{1 + 2v^2} = v + x\frac{dv}{dx} \Rightarrow x\frac{dv}{dx} = v\left\{\frac{v^2 + 2}{1 + 2v^2} - 1\right\} = v\left\{\frac{1 - v^2}{1 + 2v^2}\right\} \)
⇒ \( \frac{1 + 2v^2}{v(1 - v^2)}\,dv = \frac{dx}{x} \Rightarrow \frac{1 + 2v^2}{v(1 - v)(1 + v)}\,dv = \frac{dx}{x} \Rightarrow \left(\frac{A}{v} + \frac{B}{1 - v} + \frac{D}{1 + v}\right)dv = \frac{dx}{x} \)
where \( A(1 - v)(1 + v) + Bv(1 + v) + Dv(1 - v) = 1 + 2v^2 \)
Putting v = 0, A = 1; v = 1, \( B = \frac{3}{2} \); v = −1, \( D = -\frac{3}{2} \)
∴ \( \left(\frac{1}{v} + \frac{3}{2}\cdot\frac{1}{1 - v} - \frac{3}{2}\cdot\frac{1}{1 + v}\right)dv = \frac{dx}{x} \)
Integrating both sides, we get \( \ln v + \frac{3}{2}\cdot\frac{\ln(1 - v)}{-1} - \frac{3}{2}\ln(1 + v) = \ln x + \ln c \Rightarrow \ln v - \frac{3}{2}\ln(1 - v) - \frac{3}{2}\ln(1 + v) = \ln cx \)
⇒ \( \frac{v}{\{(1 - v)(1 + v)\}^{3/2}} = cx \Rightarrow \left(\frac{v}{cx}\right)^2 = (1 - v^2)^3 \Rightarrow \left(\frac{y}{cx^2}\right)^2 = \left(1 - \frac{y^2}{x^2}\right)^3 \Rightarrow (x^2 - y^2)^3 = \frac{x^2y^2}{c^2} \)
∴ \( (x^2 - y^2)^3 = Bx^2y^2 \), \( \left(\because \frac{1}{c^2} = B\right) \)

Example 15. Which of the following is a linear differential equation?
(a) \( \left(\frac{d^2y}{dx^2}\right)^2 + x^2\left(\frac{dy}{dx}\right)^2 = 0 \)
(b) \( y = \frac{dy}{dx} + \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \)
(c) \( \frac{dy}{dx} + \frac{y}{x} = \log x \)
(d) \( y\frac{dy}{dx} - 4 = x \)
Answer:
Ans. (c)
(a), (b), (d) do not fulfil the criteria of a linear differential equation but (c) does.
\( \frac{dy}{dx} + \frac{y}{x} = \log x \) is a linear differential equation.

Example 16. Find the integrating factor of the equation \( (x^2 + 1)\frac{dy}{dx} + 2xy = x^2 - 1 \) [UPSEAT 2002]
(a) \( x^2 + 1 \)
(b) \( \frac{2x}{x^2 + 1} \)
(c) \( \frac{x^2 - 1}{x^2 + 1} \)
(d) None of these
Answer:
Ans. (a)
The given equation may be written as \( \frac{dy}{dx} + \frac{2x}{x^2 + 1}y = \frac{x^2 - 1}{x^2 + 1} \)
Comparing with \( \frac{dy}{dx} + Py = Q \), \( P = \frac{2x}{x^2 + 1} \)
\( \text{I.F.} = e^{\int P\,dx} = e^{\int \frac{2x\,dx}{1 + x^2}} = e^{\ln(1 + x^2)} = 1 + x^2 \)

Example 17. The solution of \( \frac{dy}{dx} + 2y\tan x = \sin x \) is
(a) \( y\sec^3 x = \sec^2 x + c \)
(b) \( y\sec^2 x = \sec x + c \)
(c) \( y\sin x = \tan x + c \)
(d) None of these
Answer:
Ans. (b)
Comparing with \( \frac{dy}{dx} + Py = Q \), P = 2 tan x, Q = sin x
\( \text{I.F.} = e^{\int 2\tan x\,dx} = e^{2\ln\sec x} = e^{\ln\sec^2 x} = \sec^2 x \)
Multiplying the given equation by I.F. and integrating, \( y\sec^2 x = \int \sin x \cdot \sec^2 x\,dx = \int \sec x\tan x\,dx \)
∴ \( y\sec^2 x = \sec x + c \)

Example 18. The solution of the differential equation \( (1 + y^2) + (x - e^{\tan^{-1}y})\frac{dy}{dx} = 0 \) is [AIEEE 2003]
(a) \( (x - 2) = ke^{\tan^{-1}y} \)
(b) \( 2xe^{\tan^{-1}y} = e^{2\tan^{-1}y} + k \)
(c) \( xe^{\tan^{-1}y} = \tan^{-1}y + k \)
(d) \( xe^{2\tan^{-1}y} = e^{\tan^{-1}y} + k \)
Answer:
Ans. (b)
We have \( (x - e^{\tan^{-1}y})\frac{dy}{dx} = -(1 + y^2) \Rightarrow \frac{dx}{dy} = -\left(\frac{x - e^{\tan^{-1}y}}{1 + y^2}\right) \Rightarrow \frac{dx}{dy} + \frac{1}{1 + y^2}x = \frac{e^{\tan^{-1}y}}{1 + y^2} \)    ...(i)
This is a linear differential equation of the form \( \frac{dx}{dy} + R(y) \cdot x = S(y) \)
\( R = \frac{1}{1 + y^2} \), \( S = \frac{e^{\tan^{-1}y}}{1 + y^2} \)
Integrating factor \( = e^{\int R\,dy} = e^{\int \frac{dy}{1 + y^2}} = e^{\tan^{-1}y} \)
Multiplying (i) by I.F. and integrating, \( xe^{\tan^{-1}y} = \int \frac{e^{\tan^{-1}y}}{1 + y^2} \cdot e^{\tan^{-1}y}\,dy = \int \frac{(e^{\tan^{-1}y})^2\,dy}{1 + y^2} = \frac{(e^{\tan^{-1}y})^2}{2} + \frac{k}{2} \)
∴ \( 2xe^{\tan^{-1}y} = e^{2\tan^{-1}y} + k \)

Example 19. A particle moves in a straight line with a velocity given by \( \frac{dx}{dt} = x + 1 \) (x is the distance described). The time taken by the particle to traverse a distance of 99 metres is
(a) \( \log_{10} e \)
(b) \( 2\log_e 10 \)
(c) \( 2\log_{10} e \)
(d) \( \frac{1}{2}\log_{10} e \)
Answer:
Ans. (b)
We have \( \frac{dx}{x + 1} = dt \)
Integrating, \( \int_0^{99} \frac{dx}{x + 1} = \int_0^t dt \Rightarrow [\ln(x + 1)]_0^{99} = t \)
∴ \( t = \ln 100 = \log_e(10)^2 = 2\log_e 10 \)

Example 20. The slope of the tangent at (x, y) to a curve passing through \( \left(1, \frac{\pi}{4}\right) \) is given by \( \frac{y}{x} - \cos^2\left(\frac{y}{x}\right) \), then the equation of the curve is [Kurukshetra CEE 2002]
(a) \( y = \tan^{-1}\left[\log\left(\frac{e}{x}\right)\right] \)
(b) \( y = x\tan^{-1}\left[\log\left(\frac{x}{e}\right)\right] \)
(c) \( y = x\tan^{-1}\left[\log\left(\frac{e}{x}\right)\right] \)
(d) None of these
Answer:
Ans. (c)
We have \( \frac{dy}{dx} = \frac{y}{x} - \cos^2\left(\frac{y}{x}\right) \)
Let \( y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx} \Rightarrow v + x\frac{dv}{dx} = v - \cos^2 v \Rightarrow x\frac{dv}{dx} = -\cos^2 v \Rightarrow \sec^2 v\,dv = -\frac{dx}{x} \Rightarrow \tan v = -\ln x + c \)
⇒ tan (y/x) = − ln x + c
For x = 1, y = π/4
⇒ \( \tan\frac{\pi}{4} = -\ln 1 + c \Rightarrow 1 = 0 + c \)  ∴ c = 1
∴ tan (y/x) = 1 − ln x
⇒ \( \frac{y}{x} = \tan^{-1}(1 - \ln x) = \tan^{-1}(\ln e - \ln x) = \tan^{-1}\left[\ln\left(\frac{e}{x}\right)\right] \)
∴ \( y = x\tan^{-1}\left[\ln\left(\frac{e}{x}\right)\right] \)

Example 21. A continuously differentiable function φ(x) in (0, π) satisfying \( y' = 1 + y^2 \), y(0) = 0 = y(π) is
(a) tan x
(b) x(x − π)
(c) \( (x - \pi)(1 - e^x) \)
(d) Not possible
Answer:
Ans. (d)
For φ(x) = y, \( y' = 1 + y^2 \Rightarrow \frac{dy}{dx} = 1 + y^2 \Rightarrow \int \frac{dy}{1 + y^2} = \int dx \Rightarrow \tan^{-1}y = x + c \)
∴ y = tan (x + c)
i.e., φ(x) = tan (x + c)
As y(0) = 0, 0 = tan c and y(π) = 0 ⇒ 0 = tan (π + c) = tan c
∴ c = 0
∴ φ(x) = y = tan x.
But tan x is not continuous in (0, π), since \( \tan\frac{\pi}{2} \) is not defined.
Hence there exists no function satisfying the given condition.