Here you will learn how to find the solution of the differential equations reducible to variable separable form with examples.
Let’s begin –
Differential Equations Reducible to Variable Separable Form
Differential Equations of the form \(dy\over dx\) = f(ax + by + c) can be reduce to variable separable form by the substitution ax + by + c = 0 which can be cleared by the examples given below.
Example : Solve the differential equation : \(sin^{-1}\) \(dy\over dx\) = x + y
Solution : We are given that,
\(sin^{-1}\) \(dy\over dx\) = x + y
\(\implies\) \(dy\over dx\) = sin(x + y)
Let x + y = v. Then
1 + \(dy\over dx\) = \(dv\over dx\)
\(\implies\) \(dy\over dx\) = \(dv\over dx\) – 1
Putting x + y = v and \(dy\over dx\) = \(dv\over dx\) – 1 in the given differential equation, we get
\(dv\over dx\) – 1 = sin v
\(dv\over dx\) = 1 + sin v
\(1\over 1 + sin v\) dv = dx
Now, Integrating on both sides,
\(\int\) dx = \(\int\) \(1\over 1 + sin v\) dv
\(\int\) dx = \(\int\) \(1 – sin v\over 1 – sin^2 v\) dv
\(\int\) dx = \(\int\) \(1 – sin v\over cos^2 v\) dv
\(\int\) dx = \(\int\) \((sec^2 v – tan v sec v)\) dv
x = tan v – sec v + C
\(\implies\) x = tan (x + y) – sec (x + y) + C, which is the required solution.
Example : Solve the differential equation : \(dy\over dx\) = cos(x + y)
Solution : We are given that,
\(dy\over dx\) = cos(x + y)
Let x + y = v. Then
1 + \(dy\over dx\) = \(dv\over dx\)
\(\implies\) \(dy\over dx\) = \(dv\over dx\) – 1
Putting x + y = v and \(dy\over dx\) = \(dv\over dx\) – 1 in the given differential equation, we get
\(dv\over dx\) – 1 = cos v
\(dv\over dx\) = 1 + cos v
\(1\over 1 + cos v\) dv = dx
Now, Integrating on both sides,
\(\int\) dx = \(\int\) \(1\over 1 + cos v\) dv
\(\int\) dx = \(\int\) \(1\over 2\) \(sec^2 v/2\) dv
x = \(tan v/2\) + C
\(\implies\) x = x = \(tan (x + y)/2\) + C, which is the required solution
