CBSE Class 12 Mathematics Differential Equations Notes Set 03

Important Notes and Formulas

Definition

An equation containing an independent variable, dependent variable and differential coefficients of the dependent variable with respect to the independent variable is called a differential equation.
For example:
(i) \( \frac{dy}{dx} = \sin x \)    (ii) \( \frac{dy}{dx} + xy = \cot x \)
(iii) \( \frac{dy}{dx} = 2xy \)    (iv) \( \frac{d^2y}{dx^2} = 4x \)
(v) \( \frac{dy}{dx} = \sin x + \cos x \)    (vi) \( \frac{dy}{dx} + 2xy = x^3 \)
(vii) \( \left(\frac{d^2y}{dx^2}\right)^2 + x^2\left(\frac{dy}{dx}\right)^3 = 0 \)    (viii) \( \frac{d^2y}{dx^2} - 5\left(\frac{dy}{dx}\right) + 6y = x^2 \)

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, in the equation \( \frac{d^2y}{dx^2} + 3\frac{dy}{dx} + 2y = e^x \), the order of the highest order derivative is 2. So it is a differential equation of order 2. The equation \( \frac{d^3y}{dx^3} - 6\left(\frac{dy}{dx}\right)^2 - 4y = 0 \) is of order 3, because the order of the highest order derivative in it is 3.
Note: The order of a differential equation is a positive integer.

Degree of a Differential Equation

The degree of a differential equation is the highest power of the highest order derivative present in the equation when the involved derivatives are made free from radicals and fractions.
Thus the degree of a differential equation is the highest power of the highest order derivative present in the equation when it is written as a polynomial in the involved derivatives.

Solution of a Differential Equation

(1) General solution: The solution which contains a number of arbitrary constants equal to the order of the equation is called the general solution or complete integral or complete primitive of the differential equation.
e.g. \( y = ce^x \) is the general solution of \( \frac{dy}{dx} = y \).

(2) Particular solution: A solution obtained from the general solution by giving particular values to the constants is called a particular solution.
e.g. \( y = e^x \) is a particular solution of \( \frac{dy}{dx} = y \).

Methods of Solving a First Order First Degree Differential Equation

(1) Differential equations with variables separable
(a) Differential equation of the form \( \frac{dy}{dx} = f(x) \)
To solve this type of differential equation, we integrate both sides to obtain the general solution as discussed below:
\( \frac{dy}{dx} = f(x) \Rightarrow dy = f(x)\,dx \)
Integrating both sides, we obtain
\( \int dy = \int f(x)\,dx + C \) or \( y = \int f(x)\,dx + C \)

(b) Differential equation of the form \( \frac{dy}{dx} = f(x)\,g(y) \)
To solve this type of differential equation, we integrate both sides to obtain the general solution as discussed below:
\( \frac{dy}{dx} = f(x)\,g(y) \Rightarrow \int \frac{dy}{g(y)} = \int f(x)\,dx + C \)

(2) Homogeneous differential equations
Standard form: \( \frac{dy}{dx} = \frac{f_1(x, y)}{f_2(x, y)} \) or \( \frac{dy}{dx} = F\left(\frac{y}{x}\right) \)
where \( f_1(x, y) \) and \( f_2(x, y) \) are homogeneous functions of x and y of the same degree. On substituting y = vx or y/x = v, such an equation is reduced to the variable separable form:
\( \frac{dv}{F(v) - v} = \frac{dx}{x} \)
Now, integration will give its solution.

(3) Linear differential equations
A differential equation is said to be a linear differential equation of first order if the dependent variable and its derivative occur only in the first degree and are not multiplied together. The general form of such an equation is:
\( \frac{dy}{dx} + Py = Q \), where P and Q are constants or functions of x.
The solution of this equation is \( y(\text{I.F.}) = \int Q(\text{I.F.})\,dx + c \),
where I.F. is called the integrating factor and \( \text{I.F.} = e^{\int P\,dx} \).
So the solution of this equation is given by \( y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + c \)

Solved Examples

Illustration 1. Find the order and degree of the differential equation \( \frac{d^3y}{dx^3} = \tan x + \sec x \). [1 Mark]
Answer:
Order is 3 and degree is 1 because the highest order derivative is 3 and the power of \( \frac{d^3y}{dx^3} \) is 1. [1]

Illustration 2. Find the order and degree of the differential equation \( \left(\frac{d^4y}{dx^4}\right)^3 + \sin\left(\frac{dy}{dx}\right) + \frac{d^2y}{dx^2} = 0 \). [1 Mark]
Answer:
Since it is not a polynomial w.r.t. the derivatives, its degree is not defined, but its order is 4. [1]

Illustration 3. Find the order and degree of the differential equation \( \left(\frac{dy}{dx}\right)^{3/2} = \frac{d^2y}{dx^2} + x \). [1 Mark]
Answer:
We have \( \left(\frac{dy}{dx}\right)^{3/2} = \frac{d^2y}{dx^2} + x \)
Squaring both sides, \( \left(\frac{dy}{dx}\right)^3 = \left(\frac{d^2y}{dx^2} + x\right)^2 \)
Hence order is 2 and degree is 2.

Illustration 4. Verify that \( y = x^2 + 2x + C \) is a solution of the differential equation \( \frac{dy}{dx} - 2x - 2 = 0 \). [1 Mark]
Answer:
\( y = x^2 + 2x + C \)
⇒ \( \frac{dy}{dx} = 2x + 2 \)
or \( \frac{dy}{dx} - 2x - 2 = 0 \)
Hence verified.

Illustration 5. Verify that \( x + y = \tan^{-1} y \) is a solution of the differential equation \( y^2y' + y^2 + 1 = 0 \). [1 Mark]
Answer:
We have \( x + y = \tan^{-1} y \)
⇒ \( 1 + y' = \frac{y'}{1 + y^2} \)
⇒ \( (1 + y^2) + y'(1 + y^2) = y' \)
⇒ \( y^2y' + y^2 + 1 = 0 \)

Illustration 6. Solve the differential equation \( \frac{dy}{dx} = \sin^{-1} x \). [4 Marks]
Answer:
\( dy = \sin^{-1} x\,dx \)
⇒ \( y = \int \sin^{-1} x\,dx \) [1]
or \( y = \int \underset{\text{I}}{\sin^{-1} x} \cdot \underset{\text{II}}{(1)}\,dx \)
⇒ \( y = (\sin^{-1} x)x - \int \frac{x}{\sqrt{1 - x^2}}\,dx \) [1]
Put \( 1 - x^2 = t^2 \) ⇒ \( -x\,dx = t\,dt \)
\( y = (\sin^{-1} x)x + \int \frac{t\,dt}{t} \) [1]
\( y = x(\sin^{-1} x) + \sqrt{1 - x^2} + C \) [1]

Illustration 7. Solve the differential equation \( \frac{dy}{dx} = e^{(x + y)} \). [4 Marks]
Answer:
\( \frac{dy}{dx} = e^x \cdot e^y \) [1]
⇒ \( e^{-y}\,dy = e^x\,dx \) [2]
Integrating both sides, \( -e^{-y} = e^x + C \) [1]

Illustration 8. Find the equation of a curve passing through the point (0, −2), given that at any point (x, y) on the curve, the product of the slope of its tangent and the y-coordinate of the point is equal to the x-coordinate of the point. [4 Marks]
Answer:
Given, \( \frac{dy}{dx} \times y = x \)
⇒ \( y\,dy = x\,dx \) ...(i) [1]
or \( \frac{y^2}{2} = \frac{x^2}{2} + C \)   {Integrating both sides of (i)}
∵ The curve passes through (0, −2), [1]
⇒ \( \frac{4}{2} = 0 + C \Rightarrow C = 2 \) [1]
∴ \( y^2 = x^2 + 4 \) [1]
or \( y^2 - x^2 = 4 \)

Illustration 9. Find the equation of a curve whose tangent at any point on it, different from the origin, has slope \( y + \frac{y}{x} \). [Exemplar] [4 Marks]
Answer:
Given \( \frac{dy}{dx} = y + \frac{y}{x} \Rightarrow \frac{dy}{dx} = y\left(1 + \frac{1}{x}\right) \) [1]
or \( \frac{dy}{y} = \left(1 + \frac{1}{x}\right)dx \) [1]
Integrating both sides, we get:
\( \log y = x + \log x + C \Rightarrow \log\frac{y}{x} = x + C \)
or \( \frac{y}{x} = e^{x + C} = e^x \cdot e^C \Rightarrow \frac{y}{x} = k \cdot e^x \), where \( k = e^C \) [1]
⇒ \( y = kx \cdot e^x \) [1]

Illustration 10. In a bank, principal increases continuously at the rate of 5% per year. In how many years will Rs 1000 double itself? [4 Marks]
Answer:
Let P be the principal at any time t. According to the given problem,
\( \frac{dP}{dt} = \left(\frac{5}{100}\right) \times P \)
or \( \frac{dP}{dt} = \frac{P}{20} \) ...(1) [1]
Separating the variables in equation (1), we get
\( \frac{dP}{P} = \frac{dt}{20} \) ...(2) [½]
Integrating both sides of equation (2), we get
\( \log P = \frac{t}{20} + C_1 \) [1]
or \( P = e^{\frac{t}{20}} \cdot e^{C_1} \)
or \( P = Ce^{\frac{t}{20}} \) (where \( e^{C_1} = C \)) ...(3)
Now P = 1000, when t = 0 (given). [½]
Substituting the values of P and t in (3), we get C = 1000. Therefore, equation (3) gives:
\( P = 1000e^{\frac{t}{20}} \)
Let t years be the time required to double the principal. Then
\( 2000 = 1000e^{\frac{t}{20}} \Rightarrow t = 20\log_e 2 \) [1]

Illustration 11. Show that the differential equation \( x\cos\left(\frac{y}{x}\right)\frac{dy}{dx} = y\cos\left(\frac{y}{x}\right) + x \) is homogeneous and solve it. [4 Marks]
Answer:
The given differential equation can be written as \( \frac{dy}{dx} = \frac{y\cos\left(\frac{y}{x}\right) + x}{x\cos\left(\frac{y}{x}\right)} \) ...(i)
It is a differential equation of the form \( \frac{dy}{dx} = F(x, y) \).
Here, \( F(x, y) = \frac{y\cos\left(\frac{y}{x}\right) + x}{x\cos\left(\frac{y}{x}\right)} \) [1]
Replacing x by λx and y by λy, we get \( F(\lambda x, \lambda y) = \frac{\lambda\left[y\cos\left(\frac{y}{x}\right) + x\right]}{\lambda\left(x\cos\left(\frac{y}{x}\right)\right)} = \lambda^0[F(x, y)] \)
Thus, F(x, y) is a homogeneous function of degree zero.
Therefore, the given differential equation is a homogeneous differential equation.
To solve it, we make the substitution
y = vx ...(ii)
Differentiating equation (ii) with respect to x, we get
\( \frac{dy}{dx} = v + x\frac{dv}{dx} \) ...(iii) [1]
Substituting the values of y and \( \frac{dy}{dx} \) in equation (i), we get:
\( v + x\frac{dv}{dx} = \frac{v\cos v + 1}{\cos v} \)
⇒ \( x\frac{dv}{dx} = \frac{v\cos v + 1}{\cos v} - v \) [1]
⇒ \( x\frac{dv}{dx} = \frac{1}{\cos v} \)
⇒ \( \cos v\,dv = \frac{dx}{x} \)
Therefore, \( \int \cos v\,dv = \int \frac{1}{x}\,dx \)
⇒ sin v = log|x| + log|C|
⇒ sin v = log|Cx|
Replacing v by \( \frac{y}{x} \), we get \( \sin\left(\frac{y}{x}\right) = \log|Cx| \) [1]
which is the general solution of the differential equation (i).

Illustration 12. Find the equation of a curve passing through \( \left(1, \frac{\pi}{4}\right) \) if the slope of the tangent to the curve at any point P(x, y) is \( \frac{y}{x} - \cos^2\frac{y}{x} \). [Exemplar] [4 Marks]
Answer:
According to the given condition,
\( \frac{dy}{dx} = \frac{y}{x} - \cos^2\frac{y}{x} \) [1]
This is a homogeneous differential equation. Substituting y = vx, we get
\( v + x\frac{dv}{dx} = v - \cos^2 v \Rightarrow x\frac{dv}{dx} = -\cos^2 v \) [1]
⇒ \( \int \sec^2 v\,dv = -\int \frac{dx}{x} \Rightarrow \tan v = -\log x + C \)
⇒ \( \tan\frac{y}{x} + \log x = C \) [1]
Substituting x = 1, \( y = \frac{\pi}{4} \), we get C = 1. Thus, we get
\( \tan\frac{y}{x} + \log x = 1 \), which is the required equation. [1]

Illustration 13. Solve the differential equation \( \frac{dy}{dx} - y\tan x = 2\sin x \). [4 Marks]
Answer:
\( \frac{dy}{dx} + (-\tan x)y = 2\sin x \)
It is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \),
where P = −tan x and Q = 2 sin x.
Therefore, \( \text{I.F.} = e^{\int P\,dx} = e^{-\int \tan x\,dx} \) [1]
Hence, the solution of the given differential equation is
\( y\,e^{-\int \tan x\,dx} = \int 2\sin x\,e^{-\int \tan x\,dx}\,dx + C \)
\( \left(\because e^{-\int \tan x\,dx} = e^{-\ln\sec x} = e^{\ln\cos x} = \cos x\right) \) [1]
\( y\cos x = 2\int \sin x\cos x\,dx + C \) [1]
\( y\cos x = \int \sin 2x\,dx + C \)
\( y\cos x = -\frac{\cos 2x}{2} + C \) [1]

Illustration 14. Solve the differential equation \( \frac{dy}{dx} - 3y\cot x = \sin 2x \). [4 Marks]
Answer:
\( \frac{dy}{dx} + (-3\cot x)y = \sin 2x \)
It is a linear differential equation of the form \( \frac{dy}{dx} + Py = Q \), [1]
where P = −3 cot x and Q = sin 2x.
Therefore, \( \text{I.F.} = e^{\int P\,dx} = e^{-\int 3\cot x\,dx} \)
Hence, the solution of the given differential equation is
\( y\,e^{-\int 3\cot x\,dx} = \int \sin 2x\,e^{-\int 3\cot x\,dx}\,dx + C \) [1]
\( \left(\because e^{-\int 3\cot x\,dx} = e^{-3\ln\sin x} = \frac{1}{\sin^3 x}\right) \)
⇒ \( \frac{y}{\sin^3 x} = 2\int \frac{\sin x\cos x}{\sin^3 x}\,dx + C \) [1]
or \( y\,\text{cosec}^3 x = 2\int \text{cosec}\,x\cot x\,dx + C \)
⇒ \( y\,\text{cosec}^3 x = -2\,\text{cosec}\,x + C \) [1]

Illustration 15. Solve the differential equation \( \frac{dy}{dx} - \frac{y}{x} = \frac{\log x}{x} \). [4 Marks]
Answer:
This differential equation is of the form \( \frac{dy}{dx} + Py = Q \),
where \( P = -\frac{1}{x} \) and \( Q = \frac{\log x}{x} \).
Its solution is \( y\,e^{\int P\,dx} = \int Q\,e^{\int P\,dx}\,dx + C \) [1]
⇒ \( y\,e^{-\int \frac{dx}{x}} = \int \frac{\log x}{x}\,e^{-\int \frac{dx}{x}}\,dx + C \) [1]
or \( y\,e^{-\ln x} = \int \frac{\log x}{x} \times e^{-\ln x}\,dx + C \) [1]
⇒ \( \frac{y}{x} = \int \underset{\text{I}}{\log x} \cdot \underset{\text{II}}{\frac{1}{x^2}}\,dx + C \) [1]
∴ \( \frac{y}{x} = \frac{\log x\,(x^{-1})}{(-1)} - \int \frac{1}{x} \cdot \frac{x^{-1}}{(-1)}\,dx + C \)
⇒ \( \frac{y}{x} = -\frac{\log x}{x} - \frac{1}{x} + C \), i.e. \( y = Cx - \log x - 1 \) [1]

Do Yourself

Do Yourself 1
Find the order and degree of the following differential equations:
(i) \( \left(\frac{dy}{dx}\right)^3 + 4x\frac{d^2y}{dx^2} = 0 \)    (ii) \( (y''')^2 + 2y'' + y' + \sin x = 0 \)

Do Yourself 2
(i) Verify that \( y = e^{-x} + ax + b \) is a solution of the differential equation \( \frac{d^2y}{dx^2} = e^{-x} \).
(ii) Verify that \( y = 4\sin 3x \) is a solution of the differential equation \( \frac{d^2y}{dx^2} + 9y = 0 \).

Do Yourself 3
(i) Solve the differential equation \( \frac{dy}{dx} = 1 + x + y + xy \).
(ii) Solve the differential equation (1 + cos x) dy = (1 − cos x) dx.

Do Yourself 4
(i) Solve the differential equation \( y' = \frac{x + y}{x} \).
(ii) Solve the differential equation \( (x^2 - y^2)\,dx + 2xy\,dy = 0 \), given that at x = 1, y = 1.

Do Yourself 5
(i) Solve: \( y + \frac{d}{dx}(xy) = x(\sin x + \log x) \)
(ii) Solve the differential equation \( y' + y = -2xe^{-x} \).

Answers for Do Yourself

Do Yourself 1
(i) Order 2 and degree 1    (ii) Order 3 and degree 2

Do Yourself 2
(i) \( y' = -e^{-x} + a \), \( y'' = e^{-x} \). Hence verified.
(ii) \( y' = 12\cos 3x \), \( y'' = -36\sin 3x = -9y \), so \( y'' + 9y = 0 \). Hence verified.

Do Yourself 3
(i) \( \ln|1 + y| = x + \frac{x^2}{2} + C \)    (ii) \( y = 2\tan\frac{x}{2} - x + C \)

Do Yourself 4
(i) \( y = x\ln|x| + Cx \)    (ii) \( x^2 + y^2 = 2x \)

Do Yourself 5
(i) \( x^2y = -x^2\cos x + 2x\sin x + 2\cos x + \frac{x^3}{3}\log x - \frac{x^3}{9} + C \)
(ii) \( y \cdot e^x + x^2 = C \)