Mathematics - Differential Equation Question with Solution | TestHub

Given differential equation is, $\frac{dy}{dx}+\left(2+\frac{1}{x}\right)y=e^{-2x},x>0$ $\mathrm{IF}=e^{\int \left(2+\frac{1}{x}\right)dx}=e^{2x+\ln x}=x{e}^{2x}$ Complete solution is given by $y(x)\cdot x{e}^{2x}=\int x{e}^{2x}\cdot e^{-2x}dx+c$ $=\int xdx+c$ $y(x)\cdot e^{2x}\cdot x=\frac{x^2}{2}+c$ Given, $y(1)=\frac{1}{2}{e}^{-2}$ $\therefore \frac{1}{2}{e}^{-2}\cdot e^2\cdot 1=\frac{1}{2}+c\Rightarrow c=0$ $\therefore y(x)=\frac{x^2}{2}\cdot \frac{e^{-2x}}{x}$ $y(x)=\frac{x}{2}\cdot e^{-2x}$ Differentiate both sides with respect to $x$ $y^{\prime }(x)=\frac{e^{-2x}}{2}(1-2x)<0\forall x\in \left(\frac{1}{2},1\right)$ Hence, $y(x)$ is decreasing in $\left(\frac{1}{2},1\right)$