Linear Differential Equations

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

A differential equation of the form \( \frac{dy}{dx} + P(x)y = Q(x) \) where \( P, Q \) are functions of \( x \) or constants is called a linear differential equation of first order in \( y \).

Procedure to solve the linear differential equations:

  1. Write the differential equation in standard form \[ \frac{dy}{dx} + P(x)y = Q(x) \]
  2. Find the integrating factor \[ I.F = e^{\int P(x)dx} \]
  3. Write down the general solution as \[ y \cdot (I.F) = \int Q(x) (I.F) dx + c \]

    4. or

    \[ y e^{\int P(x)dx} = \int e^{\int P(x)dx} Q(x) dx + c \]

Note: A differential equation of the form \( \frac{dx}{dy} + P(y)x = Q(y) \) where \( P, Q \) are functions of \( y \) or constants is called a linear differential equation of first order in \( x \) and its general solution is given by

\[ x e^{\int P(y) dy} = \int Q(y) e^{\int P(y) dy} dy + c , where I.F = e^{\int P(y) dy} \]

1. Solve \( x \log x \frac{dy}{dx} + y = \log x^2 \)

Solution: Given differential equation can be written as \( \frac{dy}{dx} + \frac{y}{x \log x} = \frac{2}{x} \)

Which is in the form of \( \frac{dy}{dx} + P(x)y = Q(x) \) and is linear in \( y \), where \( P(x) = \frac{1}{x \log x} \), \( Q(x) = \frac{2}{x} \)

\[ I.F =e^{\int P(x) \, dx}= e^{\int \frac{1}{x \log x} dx} = e^{\log \log x} = \log x \]

The general solution is \[ y \cdot (I.F) = \int Q(x) (I.F) dx + c \]

\[ y \log x = \int \frac{2 \log x}{x} dx + c \]

\[ y \log x = (log x)^2 + c \]

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

Solution: The given differential equation can be written as:

\[ \frac{dx}{dy} + \left( -\frac{1}{y} \right) x = 2y^2 \]

Which is in the form of \( \frac{dx}{dy} + P(y)x = Q(y) \) and is linear in \( x \), where \( P(y) = -\frac{1}{y} \), \( Q(y) = 2y^2 \).

Therefore,

\[ \text{I.F.} = e^{\int P(y) \, dy} = e^{\int \left( -\frac{1}{y} \right) \, dy} = e^{-\log y} = e^{\log y^{-1}} = y^{-1} = \frac{1}{y} \]

The general solution is:

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

\[ \frac{x}{y} = \int 2y \, dy + c \]

\[ \frac{x}{y} = y^2 + c \quad \text{or} \quad x = y^3 + cy \]

Model Problems

Solve the following differential equations:

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

Answers

  1. \( \frac{y}{x+1} = e^{\frac{x}{3}} \)
  2. \( \frac{y}{1-x} = x \)
  3. \( \frac{y}{\sqrt{1-x^2}} = c \)
  4. \( y e^x = 2 \sqrt{x} + c \)
  5. \( y e^x = \sin x + c \)
  6. \( \frac{x^2}{2} + (1+\sin x) = c \)
  7. \( y e^x = 2x + c \)
  8. \( x e^{\tan^{-1} y} = e^{\tan^{-1} y}(\tan^{-1} y -1) + c \)
  9. \( x e^{\sin^{-1} y} = e^{\sin^{-1} y}(\sin^{-1} y -1) + c \)
  10. \( y \sec^2 x = \sec x + c \)

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