Question

For each of the differential equation, find the general solution: $(e^{x} + e^{-x}) dy - (e^{x} - e^{-x}) dx = 0$

Answer 1

Answer:

The solution is

$\bf\:\:y=log|e^x+e^{-x}|+C$

Step-by-step explanation:

$\text{I have applied variable separable method to solve this problem}$

$(e^x+e^{-x})dy-(e^x-e^{-x})dx=0$

$(e^x+e^{-x})dy=(e^x-e^{-x})dx$

$dy=\frac{(e^x-e^{-x})dx}{e^x+e^{-x}}$

Integrating

$\int{dy}=\int{\frac{(e^x-e^{-x})dx}{e^x+e^{-x}}}$

In R.H.S integral numerator is the actual derivative of denominator

Applying, the formula

$\boxed{\bf\int{\frac{f'(x)}{f(x)}}dx=log|f(x)|+C}$

$\implies\:\bf\:\:y=log|e^x+e^{-x}|+C$

Answer 2

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{\int\limits_{}^{} \, \frac{dx}{x} = \ln |x| + c}$

Where c is integration constant

TO DETERMINE

The general Solution of the differential equation

$\sf{( {e}^{x} + {e}^{ - x} )dy -( {e}^{x} - {e}^{ - x} )dx = 0 \: \: }$

CALCULATION

$\sf{( {e}^{x} + {e}^{ - x} )dy -( {e}^{x} - {e}^{ - x} )dx = 0 \: \: }$

$\displaystyle \: \implies \: \sf{dy - \: \frac{( {e}^{x} - {e}^{ - x} )}{( {e}^{x} + {e}^{ - x} )} \: dx = 0 \: \: }$

On integration

$\displaystyle \: \implies \: \sf{ \int dy - \int \frac{( {e}^{x} - {e}^{ - x} )}{( {e}^{x} + {e}^{ - x} )} \: dx = 0 \: \: } \: \: ......(1)$

Let z =

$\sf{ z = \displaystyle \: {e}^{x} + {e}^{ - x} \: }$

Differentiating both sides with respect to x we get

$\sf{ \displaystyle \: \frac{dz}{dx} = {e}^{x} - {e}^{ - x} \: }$

$\implies \sf{ \displaystyle {dz} = ({e}^{x} - {e}^{ - x} ){dx}\: }$

So From Equation (1)

$\displaystyle \:\: \sf{ \int dy - \int \frac{dz}{z} \:= 0 \: \: } \: \:$

$\implies \sf{ y - \ln z = c\: \: }$

$\implies \sf{ y = \ln z + c\: \: }$

Putting the values of z we get

$\sf{ y = \displaystyle \ln( {e}^{x} + {e}^{ - x} ) + c \: }$

Where C is integration constant

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