Question Number 123417 by sdfg last updated on 25/Nov/20
Answered by TANMAY PANACEA last updated on 25/Nov/20
![(dy/dx)−2y=3e^x e^(−2x) (dy/dx)−2ye^(−2x) =3e^(−x) [intregating factor=e^(∫pdx) =e^(∫−2dx) when eqn is (dy/dx)+py=q] (d/dx)(ye^(−2x) )=3e^(−x) ∫d(ye^(−2x) )=∫3e^(−x) ye^(−2x) =((3e^(−x) )/(−1))+c 0×e^0 =((3e^(−0) )/(−1))+c→c=3 ye^(−2x) +3e^(−x) =3](https://www.tinkutara.com/question/Q123432.png)
$$\frac{{dy}}{{dx}}−\mathrm{2}{y}=\mathrm{3}{e}^{{x}} \\ $$$${e}^{−\mathrm{2}{x}} \frac{{dy}}{{dx}}−\mathrm{2}{ye}^{−\mathrm{2}{x}} =\mathrm{3}{e}^{−{x}} \:\:\left[{intregating}\:{factor}={e}^{\int{pdx}} ={e}^{\int−\mathrm{2}{dx}} \right. \\ $$$$\left.{when}\:{eqn}\:{is}\:\frac{{dy}}{{dx}}+{py}={q}\right] \\ $$$$\frac{{d}}{{dx}}\left({ye}^{−\mathrm{2}{x}} \right)=\mathrm{3}{e}^{−{x}} \\ $$$$\int{d}\left({ye}^{−\mathrm{2}{x}} \right)=\int\mathrm{3}{e}^{−{x}} \\ $$$${ye}^{−\mathrm{2}{x}} =\frac{\mathrm{3}{e}^{−{x}} }{−\mathrm{1}}+{c} \\ $$$$\mathrm{0}×{e}^{\mathrm{0}} =\frac{\mathrm{3}{e}^{−\mathrm{0}} }{−\mathrm{1}}+{c}\rightarrow{c}=\mathrm{3} \\ $$$${ye}^{−\mathrm{2}{x}} +\mathrm{3}{e}^{−{x}} =\mathrm{3} \\ $$