Here you will learn how to find the solution of the differential equations in variable separable form with examples.
Let’s begin –
Differential Equations in Variable Separable Form
If the differential equation can be put in the form f(x) dx = g(y) dy, we say that the variables are seperable and such equations can be solved by integrating on both sides. The solution is given by
\(\int\) f(x) dx = \(\int\) g(y) dy + C, where C is an arbitrary constant
Note : There is no need of introducing arbitrary constants of integration on both sides as they can be combined together to give just one arbitrary constant.
Example : Solve the differential equation : (x + 1)\(dy\over dx\) = 2xy
Solution : We have,
(x + 1)\(dy\over dx\) = 2xy
\(\implies\) (x + 1)dy = 2xy dx
\(\implies\) \(dy\over y\) = \(2x\over x + 1\) dx
Now, integrating on both sides,
\(\int\) \(1\over y\) dy = 2 \(\int\) \(x\over x + 1\) dx
\(\implies\) \(1\over y\) dy = 2 \(\int\) \(x + 1 – 1\over x + 1\) dx
\(\implies\) \(1\over y\) dy = 2 \(\int\) \(1 – {1\over x + 1}\) dx
log y = 2{x – log| x + 1 |} + C, which is the solution of the given differential equation.
Example : Solve the differential equation : cos x(1 + cos y) dx – sin y(1 + sin x)dy = 0
Solution : We have,
cos x(1 + cos y) dx – sin y(1 + sin x)dy = 0
\(\implies\) \(cos x\over 1 + sin x\) dx – \(sin y\over 1 + cos y\) dy = 0
\(\implies\) \(cos x\over 1 + sin x\) dx + \(-sin y\over 1 + cos y\) dy = 0
Now, integrating on both sides,
\(\int\) \(cos x\over 1 + sin x\) dx + \(\int\) \(-sin y\over 1 + cos y\) dy = 0
\(\implies\) log|1 + sin x| + log|1 + cos y| = log C
\(\implies\) log{|1 + sin x|.|1 + cos y|} = log C
|1 + sin x|.|1 + cos y| = C
Hence, (1 + sin x)(1 + cos y) = C is the solution of the given differential equation.
