Differential Equations in Variable Separable Form

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.

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