JEE Main 2019 — Mathematics Calculus

Given differential equation is, $\frac{d y}{d x}+\left(2+\frac{1}{x}\right) y=e^{-2 x}, x>0$dxdy​+(2+x1​)y=e−2x,x>0 $\displaystyle \mathrm{IF}=e^{\int\left(2+\frac{1}{x}\right) d x}=e^{2 x+\ln x}=x e^{2 x}$IF=e∫(2+x1​)dx=e2x+lnx=xe2x Complete solution is given by $\displaystyle y(x) \cdot x e^{2 x}=\int x e^{2 x} \cdot e^{-2 x} d x+c$y(x)⋅xe2x=∫xe2x⋅e−2xdx+c $\displaystyle =\int x d x+c$=∫xdx+c $y(x) \cdot e^{2 x} \cdot x=\frac{x^{2}}{2}+c$y(x)⋅e2x⋅x=2x2​+c Given, $y(1)=\frac{1}{2} e^{-2}$y(1)=21​e−2 $\therefore \quad \frac{1}{2} e^{-2} \cdot e^{2} \cdot 1=\frac{1}{2}+c \Rightarrow c=0$∴21​e−2⋅e2⋅1=21​+c⇒c=0 $\therefore \quad y(x)=\frac{x^{2}}{2} \cdot \frac{e^{-2 x}}{x}$∴y(x)=2x2​⋅xe−2x​ $y(x)=\frac{x}{2} \cdot e^{-2 x}$y(x)=2x​⋅e−2x Differentiate both sides with respect to $x$x $y^{\prime}(x)=\frac{e^{-2 x}}{2}(1-2 x) < 0 \forall x \in\left(\frac{1}{2}, 1\right)$y′(x)=2e−2x​(1−2x)<0∀x∈(21​,1) Hence, $y(x)$y(x) is decreasing in $\left(\frac{1}{2}, 1\right)$(21​,1)