Let y(x) be the solution of the differential equation 2x2 dy + (ey - 2x)dx = 0, x 0.

Question

Let y(x) be the solution of the differential equation

2x2 dy + (ey - 2x)dx = 0, x > 0. If y(e) = 1, then y(1) is equal to :

Accepted Answer

$2{x^2}dy + ({e^y} - 2x)dx = 0$

$${{dy} \over {dx}} + {{{e^y} - 2x} \over {2{x^2}}} = 0 \Rightarrow {{dy} \over {dx}} + {{{e^y}} \over {2{x^2}}} - {1 \over x} = 0$$

$${e^{ - y}}{{dy} \over {dx}} - {{{e^{ - y}}} \over x} =  - {1 \over {2{x^2}}} \Rightarrow $$ Put ${e^{ - y}} = z$

$${{ - dz} \over {dx}} - {z \over x} =  - {1 \over {2{x^2}}} \Rightarrow xdz + zdx = {{dx} \over {2x}}$$

$$d(xz) = {{dx} \over {2x}} \Rightarrow xz = {1 \over 2}{\log _e}x + c$$

$$x{e^{ - y}} = {1 \over 2}{\log _e}x + c$$, passes through (e, 1)

$$ \Rightarrow C = {1 \over 2}$$

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

$${e^{ - y}} = {1 \over 2} \Rightarrow y = {\log _e}2$$

Solutions · 1