tgt

## Tuesday, 5 August 2014

### CHAPTER 5 - Solving Differential Equations

In this section, we consider how to evaluate the general solution of a DE. You must appreciate the fact that evaluating the general solution of an arbitrary DE is not a simple task, in general. Over time, many methods have been developed to solve particular classes of DEs. Fortunately for us, at this level we are required to deal with only the simplest of cases.
We’ll be considering only first order and first degree DEs. Note that any such DE can be written in the general form
${M(x,y)dx + N(x,y)dy = 0}$

#### Type-1: Variable Separable Form

This is by and large the simplest type of DE that we’ll encounter. As the name suggests, in such an equation, $M$ is a function of $x$ only and $N$ is a function of $y$ only. Thus, such a DE is of the form
 $f(x)dx + g(y)dy = 0$
which can be solved by straightforward integration to obtain
 $\int {f(x)dx + } \int {g(y)dy = C}$
where $C$ is an arbitrary constant.
Observe how the “variables are separated” in this type of DE and its general solution. As a simple example, consider the DE
 $xdx + {y^2}dy = 0$
This is obviously in variable -separable form. Integrating, we obtain
 $\int {xdx + \int {{y^2}dy = C} }$ $\Rightarrow \dfrac{{{x^2}}}{2} + \dfrac{{{y^3}}}{3} = C$
This is the required general solution of the DE.
 Example: 1
 Solve the DE $y - x\dfrac{{dy}}{{dx}} = \lambda \left( {{y^2} + \dfrac{{dy}}{{dx}}} \right).$
 Solution: 1 Steps Involved: 3

#### Step-1

A little observation will show you that the variables are separable in this DE:
 $y(1 - \lambda y) = \dfrac{{dy}}{{dx}}(x + \lambda )$ $\Rightarrow \dfrac{{dy}}{{y(1 - \lambda y)}} = \dfrac{{dx}}{{x + \lambda }}$ $\Rightarrow \left( {\dfrac{1}{y} + \dfrac{\lambda }{{1 - \lambda y}}} \right)dy = \dfrac{{dx}}{{x + \lambda }}$

#### Step-2

Integrating both sides, we have
 $\ln y - \ln (1 - \lambda y) = \ln (x + \lambda ) + C$ $\Rightarrow \ln \left( {\dfrac{y}{{1 - \lambda y}}} \right) = \ln (x + \lambda ) + \ln C'$

#### Step-3

In the last step, we have written the arbitrary constant of integration $C$ as $C'$ so that the whole expression can be combined now by taking antilog on both sides. (There’s no loss of generality in doing so and it is often done to make the final expression look simpler). Thus, we now have,
 $\dfrac{y}{{1 - \lambda y}} = C'(x + \lambda )$ $\Rightarrow y = C'(x + \lambda )(1 - \lambda y)$
This is the required general solution of the DE; as expected it contains only one arbitrary constant.
 Example: 2
 Solve the DE $x{y^2}\dfrac{{dy}}{{dx}} = 1 - {x^2} + {y^2} - {x^2}{y^2}$
 Solution: 2 Steps Involved: 2

#### Step-1

Again, this DE is of the variable separable form as can be made evident by a slight rearrangement.
 $x{y^2}\dfrac{{dy}}{{dx}} = (1 - {x^2})(1 + {y^2})$ $\Rightarrow \left( {\dfrac{{{y^2}}}{{1 + {y^2}}}} \right)dy = \left( {\dfrac{{1 - {x^2}}}{x}} \right)dx$ $\Rightarrow \left( {1 - \dfrac{1}{{1 + {y^2}}}} \right)dy = \left( {\dfrac{1}{x} - x} \right)dx$

#### Step-2

Integrating both sides, we have
 $y - {\tan ^{ - 1}}y = \ln x - \dfrac{{{x^2}}}{2} + C$
This is the required general solution.