Non-linear Differential Equation of First Order

Contents

Introduction

Exact Differential Equations

Equations reducible to exact form

Linear Differential Equations

Non-Linear Differential Equations

Applications

Newton’s law of cooling

Law of natural growth and decay

Orthogonal Trajectories

1M Questions

Bernoulli’s Equation:

A differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x)y^n \), \( (n \neq 1) \) where \( P, Q \) are functions of \( x \) is called a Bernoulli’s equation in \( y \).

Procedure to Solve Bernoulli’s Equation:

Consider the Bernoulli’s equation \[ \frac{dy}{dx} + P(x)y = Q(x)y^n \quad ...(1)\]

Divide equation (1) with \( y^n \), we get \[ y^{-n} \frac{dy}{dx} + P(x)y^{1-n} = Q(x) \quad ...(2)\]

Put \( y^{1-n} = t \Rightarrow (1-n)y^{-n} \frac{dy}{dx} = \frac{dt}{dx} \Rightarrow y^{-n} \frac{dy}{dx} = \frac{1}{1-n}\frac{dt}{dx} \)

Substituting these in (2), we get \[ \frac{1}{1-n} \frac{dt}{dx} + P(x)t = Q(x) \Rightarrow \frac{dt}{dx} + (1-n)P(x)t = (1-n)Q(x) \]

Which is linear differential equation in \( t \).

Note: A differential equation of the form \( \frac{dx}{dy} + P(y)x = Q(y)x^n \), \( (n \neq 1) \) where \( P, Q \) are functions of \( y \) is called a Bernoulli’s equation in \( x \).

1. Solve \( x\frac{dy}{dx} + y = x^3 y^6 \).

Solution: Given differential equation can be written as \( \frac{dy}{dx} + \frac{y}{x} = x^2 y^6 \), which is Bernoulli’s equation in \( y \).

Dividing with \( y^6 \) on both sides, we get \[ y^{-6} \frac{dy}{dx} + \frac{y^{-5}}{x} = x^2 \]

Put \( y^{-5} = t \), then \( -5y^{-6} \frac{dy}{dx} = \frac{dt}{dx} \implies y^{-6} \frac{dy}{dx} = \frac{-1}{5} \frac{dt}{dx} \)

Substituting these in the (1), we get:

\[ \frac{-1}{5} \frac{dt}{dx} + \frac{1}{x} t = x^2 \implies \frac{dt}{dx} - \frac{5}{x} t = -5x^2 \quad \text{… (2)} \]

Which is linear in \( t \), where \( P(x) = -\frac{5}{x} \), \( Q(x) = -5x^2 \).

\[ \therefore \text{I.F.} = e^{\int P(x) \, dx} = e^{\int \left( -\frac{5}{x} \right) \, dx} = e^{-5 \log x} = e^{\log x^{-5}} = x^{-5} = \frac{1}{x^5} \]

The general solution of (2) is

\[ t (\text{I.F.}) = \int Q(x) (\text{I.F.}) \, dx + c \implies t \left( \frac{1}{x^5} \right) = \int (-5x^2) \left( \frac{1}{x^5} \right) \, dx + c \]

\[ \frac{t}{x^5} = \int (-5x^{-3}) \, dx + c \]

\[ \frac{t}{x^5} = \frac{-5}{-2} x^{-2} + c = \frac{5}{2x^2} + c \]

\[ \frac{1}{x^5y^5} = \frac{5}{2x^2} + c \quad (\because t = y^{-5} = \frac{1}{y^5}) \]

Note: A differential equation of the form \[ f'(y) \frac{dy}{dx} + P(x) f(y) = Q(x) \] can be reduced to linear differential equation in \( t \) by using th substitution \( f(y) = t \) and \( f'(y) \frac{dy}{dx} = \frac{dt}{dx} \).

2. Solve \( 2y \cos y^2 \frac{dy}{dx} - \frac{2}{1+x} \sin y^2 = (1+x)^3 \).

Solution: Given differential equation is in the form \[ f'(y) \frac{dy}{dx} + P(x) f(y) = Q(x) \]

Put \( \sin y^2 = t \Rightarrow 2y \cos y^2 \frac{dy}{dx} = \frac{dt}{dx} \)

Substituting these in given D.E., we get

\[ \frac{dt}{dx} - \frac{2}{1+x} t = (1+x)^3 \]

Which is a linear equation in \( t \), where

\[ P(x) = \frac{-2}{1+x}, \quad Q(x) = (1+x)^3 \] \[ \therefore I.F = e^{\int P(x)dx} = e^{\int \frac{-2}{1+x}dx} = e^{-2 \log(1+x)} = (1+x)^{-2} = \frac{1}{(1+x)^2} \]

The general solution is

\[ t (I.F) = \int Q(x)(I.F)dx + c \] \[ \Rightarrow t \frac{1}{(1+x)^2}= \int (1+x)^3 \frac{1}{(1+x)^2} dx + c \] \[ \Rightarrow \frac{t}{(1+x)^2} = \int (1+x) dx + c \] \[ \Rightarrow \frac{t}{(1+x)^2} = x+\frac{x^2}{2} + c \]

Since \( t = \sin y^2 \), we get

\[ \frac{\sin y^2}{(1+x)^2} = \frac{x^2}{2} + x + c \]

Model Problems

Solve the following differential equations:

    1. \( (x^3 y^2 + xy) dx = dy \)
    2. \( (1 - x^2) \frac{dy}{dx} + xy = y^3 \sin^{-1} x \)
    3. \( y' + y = xy^{5/3} \)
    4. \( \frac{dy}{dx} + y \cos x = y^3 \sin 2x \)
    5. \( \frac{dy}{dx} + y \tan x = y^2 \sec x \)
    6. \( 2xy' = 10x^3 y^5 + y \)
    7. \( xy(1 + xy^2) \frac{dy}{dx} = 1 \)
    8. \( \frac{dy}{dx} + x \sin 2y = x^3 \cos^2 y \)
    9. \( \frac{dy}{dx} - \frac{\tan y}{1+x} = (1+x)e^x \sec y \)

    Answers

    1. \( \frac{1}{y} e^{x^{2}/2} = -2e^{x^{2}/2} \left( \frac{x^2}{2} - 1 \right) + c \)
    2. \( \frac{1 - x^2}{y^2} = -2 \left( x \sin^{-1} x + \sqrt{1 - x^2} \right) + c \)
    3. \( y^{-2/3} e^{-2x/3} = \frac{3}{2} e^{-2x/3} \left( \frac{-2x}{3} - 1 \right) + c \)
    4. \( \frac{1}{y^2} e^{-2\sin x} = e^{-2\sin x} \left(1 + 2\sin x\right) + c \)
    5. \( \frac{1}{y} \cos x = -x + c \)
    6. \( \frac{x^2}{y^4} = -4x^5 + c \)/li>
    7. \( \frac{1}{x} e^{y^{2}/2} = -2e^{y^{2}/2} \left( \frac{y^2}{2} - 1 \right) + c \)
    8. \( e^{x^2} \tan y = \frac{1}{2} e^{x^2} \left( x^2 - 1 \right) + c \)
    9. \( \frac{\sin y}{1 + x} = e^x + c \)

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