Differential Equations Reducible to Variable Separable Form

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

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