Solution of Different Types of Differential Equations

4 Marks Questions

1. Find the particular solution of the differential equation <math>\frac{dy}{dx} = 1 + x + y + xy</math>, given that <math>y = 0</math>

when <math>x = 1</math>. All India 2014

Given differential equation is

<math display="block">\frac{dy}{dx} = 1 + x + y + xy</math><math display="block">\frac{dy}{dx} = 1(1+x) + y(1+x)</math>

<math>\frac{dy}{dx} = (1+x)(1+y)</math> ...(i) (1)

On separating variables, we get

<math display="block">\frac{1}{(1+y)}dy = (1+x)dx</math>

...(ii)

On integrating both sides of Eq. (ii), we get

<math display="block">\int \frac{1}{1+y} \, dy = \int (1+x) \, dx</math>

<math>\Rightarrow</math> <math>\log |1+y| = x + \frac{x^2}{2} + C</math> ...(iii) (1)

Also, given that <math>y = 0</math>, when <math>x = 1</math>.

On substituting <math>x = 1</math>, <math>y = 0</math> in Eq. (iii), we get <math>\log |1+0| = 1 + \frac{1}{2} + C \Rightarrow C = -\frac{3}{2} [\because \log 1 = 0]</math> (1) Now, on substituting the value of C in Eq. (iii), we get

<math>\log |1+y| = x + \frac{x^2}{2} - \frac{3}{2}</math>

which is the required particular solution of given differential equation. (1)

1. Find the particular solution of the differential equation <math>x \frac{dy}{dx} - y + x \csc\left(\frac{y}{x}\right) = 0</math> or <math>\frac{dy}{dx} - \frac{y}{x} + \csc\left(\frac{y}{x}\right) = 0</math>, given that <math>y = 0</math>, when <math>x = 1</math>. All India 2014C, 2011; Delhi 2009