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Solution of the differential equation $open parentheses fraction numerator d y over denominator d x end fraction close parentheses to the power of 2 end exponent minus fraction numerator d y over denominator d x end fraction open parentheses e to the power of x end exponent plus e to the power of negative x end exponent close parentheses plus 1 equals 0$ is given by

$y + e^{- x} = c$
$y - e^{- x} = c$
$y + e^{x} = c$
$y - e^{x} = c e^{x}$

hint Hint:

We are given a differential equation. We have to find it's solution. The first order derivative of y has power 2. We will simplify the equation before finding the solution. To find the solution, we will use integration method.

The correct answer is: $y plus e to the power of negative x end exponent equals c$

The given equation is
$open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared minus open parentheses fraction numerator d y over denominator d x end fraction close parentheses left parenthesis e to the power of x plus e to the power of negative x end exponent right parenthesis space plus space 1 space equals space 0$
We will try to simplify the equation. We will expand the the bracket in the equation. We will see if we get like terms or take anything common.
$open parentheses fraction numerator d y over denominator d x end fraction close parentheses squared minus fraction numerator d y over denominator d x end fraction left parenthesis e to the power of x plus e to the power of negative x end exponent right parenthesis space plus space 1 space equals space 0 space open parentheses fraction numerator d y over denominator d x end fraction close parentheses open parentheses fraction numerator d y over denominator d x end fraction close parentheses space minus e to the power of x fraction numerator d y over denominator d x end fraction minus e to the power of negative x end exponent space fraction numerator d y over denominator d x end fraction plus space 1 space equals space 0 space W e space c a n space t a k e space fraction numerator d y over denominator d x end fraction space c o m m o n space f r o m space f i r s t space t w o space t e r m s. B u t comma space f o r space t h i r d space a n d space f o u r t h space t e r m space w e space w i l l space h a v e space t o space m a k e space a d j u s t m e n t s space W e space c a n space w r i t e space 1 space a s space e to the power of x space end exponent e to the power of negative x end exponent. space U sin g space t h e space r u l e s space o f space i n d i c e s comma space e to the power of x space e to the power of negative x end exponent equals e to the power of x space minus x end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals e to the power of 0 space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 1$
So, the equation becomes.
$open parentheses fraction numerator d y over denominator d x end fraction close parentheses open parentheses space fraction numerator d y over denominator d x end fraction close parentheses minus e to the power of x fraction numerator d y over denominator d x end fraction minus e to the power of negative x end exponent fraction numerator d y over denominator d x end fraction plus e to the power of negative x end exponent e to the power of plus x end exponent equals space 0 fraction numerator d y over denominator d x end fraction open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses minus e to the power of x close square brackets space minus e to the power of negative x end exponent open square brackets open parentheses fraction numerator d y over denominator d x end fraction close parentheses minus e to the power of x close square brackets equals 0 open parentheses fraction numerator d y over denominator d x end fraction minus e to the power of negative x end exponent close parentheses open parentheses fraction numerator d y over denominator d x end fraction minus e to the power of x close parentheses equals 0 W e space w i l l space e q u a t e space b o t h space t h e space b r a c k e t s space w i t h space z e r o space o n e space b y space o n e space fraction numerator d y over denominator d x end fraction minus e to the power of negative x end exponent equals space 0 space fraction numerator d y over denominator d x end fraction equals e to the power of negative x end exponent d y space equals space e to the power of negative x end exponent d x I n t e g r a t i n g space b o t h space t h e space s i d e s y space equals space minus e to the power of negative x end exponent space plus space c$
y + e^-x = c ...(1)
Similarly we will equate the second bracket with 0.
$fraction numerator d y over denominator d x end fraction minus e to the power of x equals 0 fraction numerator d y over denominator d x end fraction equals e to the power of x d y space equals space e to the power of x d x I n t e g r a t i n g space b o t h space t h e space s i d e s y space equals space e to the power of x space plus space c$
y - e^x= c ...(2)
We will see which of the value matches the options. The one which matches the option is the solution.
Equation (1) matches the option. So, required solution is
y + e^-x = c

When we get multipliction of two brackets equal to zero, any of the bracket can be zero or both of the brackets can be zero. It is decided based on the conditions in the equation.

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Solution of the differential equation is given by

We are given a differential equation. We have to find it's solution. The first order derivative of y has power 2. We will simplify the equation before finding the solution. To find the solution, we will use integration method.

The correct answer is:

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The correct answer is: $y plus e to the power of negative x end exponent equals c$

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