5.(ex + ex) dy _ (ex-C") dx = 0

Open in App

Solution

The given differential equation is,

( e x + e −x )dy−( e x − e −x )dx=0

Separate the variables.

( e x + e −x )dy=( e x − e −x )dx dy= ( e x − e −x ) ( e x + e −x ) dx

Integrating both the sides, we get,

∫ dy = ∫ ( e x − e −x ) ( e x + e −x ) dx y= ∫ ( e x − e −x ) ( e x + e −x ) dx +C (1)

Let ( e x + e −x )=t

Differentiating both the sides, we get,

d dx ( e x + e −x )= dt dx ( e x − e −x )dx=dt

Substitute the values in equation (1).

y= ∫ 1 t dt +C y=log| t |+C y=log| e x + e −x |+C

The general solution of the differential equation ( e x + e −x )dy−( e x − e −x )dx=0 is given as y=log| e x + e −x |+C.

Suggest Corrections

Similar questions

e^{x} + e^{y} = e^{x + y}

\frac{d y}{d x} + e^{y - x} = 0

e^{x} + e^{y} = e^{x + y}

\frac{d y}{d x} = e^{x - y} (\frac{e^{y} - 1}{1 - e^{x}})

\int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})}

The solution of the equation $\frac{d y}{d x} - e^{x - y} + x^{2} e^{- y}$ is
[MP PET 2004]

I = \int \frac{e^{x}}{(1 + e^{x}) (2 + e^{x})}

Join BYJU'S Learning Program