Solve the differential equation? 2dy/dx-y=e^x/2 .

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Answer 1 · 2018-06-10T18:06:26

$y = \frac{1}{2} x e^{\frac{x}{2}} + C e^{\frac{x}{2}}$ $y=1/2xe^(x/2)+Ce^(x/2)$

$2 \frac{d y}{d x} - y = e^{\frac{x}{2}}$ $2(dy)/(dx)-y=e^(x/2)$

$\Rightarrow \frac{d y}{d x} + (- \frac{1}{2}) y = \frac{1}{2} e^{\frac{x}{2}} - - (1)$ $=>(dy)/(dx)+(-1/2)y=1/2e^(x/2)--(1)$

Integrating factor

$e^{- (\frac{1}{2}) \int d x} = e^{- \frac{x}{2}}$ $e^(-(1/2)intdx)=e^(-x/2)$

multiply $(1)$ $(1)$ by $e^{- \frac{x}{2}}$ $e^(-x/2)$

$e^{- \frac{x}{2}} \frac{d y}{d x} - \frac{1}{2} e^{- \frac{x}{2}} y = \frac{1}{2}$ $e^(-x/2)(dy)/(dx)-1/2e^(-x/2)y=1/2$

$\Rightarrow \frac{d}{d x} (y e^{- \frac{1}{2} x}) = \frac{1}{2}$ $=>d/(dx)(ye^(-1/2x))=1/2$

integrating

$y e^{- \frac{1}{2} x} = \frac{1}{2} x + C$ $ye^(-1/2x)=1/2x+C$

$':.y=1/2xe^(x/2)+Ce^(x/2)$