We #160;have,dydx=ex+1 #160;y #8658;1ydy=ex+1 #160;dxIntegrating #160;both #160;sides, #160;we #160;get #160; #8747;1ydy= #8747;ex+1 #16 ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Springs in Parallel and Series

dyd x=e x+1 y

Question

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

Open in App

Solution

$We have, \frac{d y}{d x} = (e^{x} + 1) y \Rightarrow \frac{1}{y} d y = (e^{x} + 1) d x Integrating both sides, we get \int \frac{1}{y} d y = \int (e^{x} + 1) d x \Rightarrow \log |y| = e^{x} + x + C Hence, \log |y| = e^{x} + x + C is the required solution .$

Suggest Corrections

Similar questions

y = e^{x}

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

If e^{x} + e^{y} = e^{x + y}, prove that \frac{d y}{d x} = - \frac{e^{x} (e^{y} - 1)}{e^{y} (e^{x} - 1)} or \frac{d y}{d x} + e^{y - x} = 0

\frac{d y}{d x} - y \tan x = e^{x}

\frac{d y}{d x} + 4 x = e^{x}

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

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

Join BYJU'S Learning Program