If y(x) is the solution of the differential equation [dy dx + ( 2x + 1 x )y = e^ - 2x,,,x …

Question

If y(x) is the solution of the differential equation $${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}},\,\,x > 0,\,$$ where $$y\left( 1 \right) = {1 \over 2}{e^{ - 2}},$$ then

Accepted Answer

$${{dy} \over {dx}} + \left( {{{2x + 1} \over x}} \right)y = {e^{ - 2x}}$$

I.F. $$ = {e^{\int {\left( {{{2x + 1} \over x}} \right)dx} }} = {e^{\int {\left( {2 + {1 \over x}} \right)dx} }} = {e^{2x + \ell nx}} = {e^{2x}}.x$$

So,  $$y\left( {x{e^{2x}}} \right) = \int {{e^{ - 2x}}.x{e^{2x}} + C} $$

$$ \Rightarrow xy{e^{2x}} = \int {xdx + C} $$

$$ \Rightarrow 2xy{e^{2x}} = {x^2} + 2C$$

It passes through $$\left( {1,{1 \over 2}{e^{ - 2}}} \right)$$ we get C $=$ 0

$$y = {{x{e^{ - 2x}}} \over 2}$$

$$ \Rightarrow {{dy} \over {dx}} = {1 \over 2}{e^{ - 2x}}\left( { - 2x + 1} \right)$$

$$ \Rightarrow f(x)$$ is decreasing in $$\left( {{1 \over 2},1} \right)$$

$$y\left( {{{\log }_e}2} \right) = {{\left( {{{\log }_e}2} \right){e^{ - 2({{\log }_e}2)}}} \over 2}$$

$$ = {1 \over 8}{\log _e}2$$

Solutions · 1