Find the solution of ex-e-xdydx-ex-e-x

Given, (e^x+e^ x)dydx=(e^x e^ x) dydx=e^x e^ xe^x+e^ x integrating on both sides, we get, ∫dydx=∫e^x e^ xe^x+e^ x substitute u=e^x+e^ x→ du=(e ...

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Standard XII

Mathematics

Variable Separable Method

find the solu...

Question

find the solution of
$(e^{x} + e^{- x}) \frac{d y}{d x} = e^{x} - e^{- x}$

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Solution

Given,

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

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

integrating on both sides, we get,

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

substitute $u = e^{x} + e^{- x} \to d u = (e^{x} - e^{- x}) d x$

$y + c = log (e^{x} + e^{- x}) + c$

$∴ y = log (e^{x} + e^{- x})$

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

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(e^{x} + e^{- x}) d y - (e^{x} - e^{- x}) d x = 0

The general solution of the differential equation $\frac{d y}{d x} = e^{x + y}$ is
(a) $e^{x} + e^{- y} = C$
(b) $e^{x} + e^{y} = C$
(c) $e^{- x} + e^{y} = C$
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Q. The general solution of the differential equation

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

, is
(a) e^x+ e^−y = C
(b) e^x + e^y = C
(c) e⁻^x + e^y = C
(d) e^−x + e^−y = C

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find the solution of (ex+e−x)dydx=ex−e−x

Given,(ex+e−x)dydx=(ex−e−x)dydx=ex−e−xex+e−xintegrating on both sides, we get,∫dydx=∫ex−e−xex+e−xsubstitute u=ex+e−x→du=(ex−e−x)dxy+c=log(ex+e−x)+c∴y=log(ex+e−x)

find the solution of
$(e^{x} + e^{- x}) \frac{d y}{d x} = e^{x} - e^{- x}$