How do you solve the separable differential equation $\frac{d y}{d x} = (cos x) e^{y - sin x}$ $dy/dx=(cosx)e^(y-sinx)$ ?

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Answer 1 · 2017-03-10T18:30:22

The answer is $y = - ln (e^{- sin x} - C)$ $y=-ln(e^-sinx-C)$

We need

$e^{a - b} = \frac{e^{a}}{e^{b}}$ $e^(a-b)=e^(a)/e^b$

Let's rewrite and simplify the equation

$\frac{d y}{d x} = cos x e^{y - sin x}$ $dy/dx=cosx e^(y-sinx)$

$\frac{d y}{d x} = cos x \frac{e^{y}}{e^{sin x}}$ $dy/dx=cosx e^y/e^sinx$

$\frac{d y}{e^{y}} = \frac{cos x}{e^{sin x}} d x$ $dy/e^y=cosx/e^sinxdx$

Integrating both sides

$\int \frac{d y}{e^{y}} = \int \frac{cos x}{e^{sin x}} d x$ $intdy/e^y=intcosx/e^sinxdx$

$\int \frac{d y}{e^{y}} = - e^{- y}$ $intdy/e^y=-e^-y$

For the $R H S$ $RHS$ , we perform a substitution

Let $u = sin x$ $u=sinx$ , $\Rightarrow$ $=>$ , $d u = cos x d x$ $du=cosx dx$

$\int \frac{cos x}{e^{sin x}} d x = \int \frac{d u}{e^{u}}$ $intcosx/e^sinxdx=int(du)/e^(u)$

$= - e^{- u}$ $=-e^(-u)$

$= - e^{- sin x}$ $=-e^(-sinx)$

Therefore,

$- e^{- y} = - e^{- sin x} + C$ $-e^-y=-e^(-sinx)+C$

$e^{- y} = e^{- sin x} - C$ $e^-y=e^(-sinx)-C$

$- y = ln (e^{- sin x} - C)$ $-y=ln(e^-sinx-C)$

$y = - ln (e^{- sin x} - C)$ $y=-ln(e^-sinx-C)$

How do you solve the separable differential equation dydx=(cosx)ey−sinxdy/dx=(cosx)e^(y-sinx)?