solve-the-differential-equation-y-2e-x-y-2e-x-y-

Question Number 108406 by pticantor last updated on 16/Aug/20

s o l v e t h e d i f f e r e n t i a l

e q u a t i o n

y^{^{'}} - 2 e^{x} y = 2 e^{x} \sqrt{y}

Answered by Dwaipayan Shikari last updated on 16/Aug/20

\frac{d y}{d x} = 2 e^{x} (y + \sqrt{y})

\int \frac{d y}{y + \sqrt{y}} = \int 2 e^{x} d x

\int \frac{d y}{\sqrt{y} (1 + \sqrt{y})} d y = 2 e^{x} + C

\int \frac{2 t}{t (1 + t)} d t = 2 e^{x} + C y = t^{2}, 1 = 2 t \frac{d t}{d y}

2 l o g (1 + t) = 2 e^{x} + C

l o g (1 + t) = e^{x} + \frac{C}{2}

l o g (1 + \sqrt{y}) = e^{x} + \frac{C}{2}