Let y = y(x) be the solution of the differential equation [dy dx = 1 + xe^y - x, - 2

Question

Let y = y(x) be the solution of the differential equation $${{dy} \over {dx}} = 1 + x{e^{y - x}}, - \sqrt 2  < x < \sqrt 2 ,y(0) = 0$$

then, the minimum value of $$y(x),x \in \left( { - \sqrt 2 ,\sqrt 2 } \right)$$ is equal to :

Accepted Answer

$${{dy - dx} \over {{e^{y - x}}}} = xdx$$

$$ \Rightarrow {{dy - dx} \over {{e^{y - x}}}} = xdx$$

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

At x = 0, y = 0 $$\Rightarrow$$ c = -1

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

$$ \Rightarrow y = x - \ln \left( {{{2 - {x^2}} \over 2}} \right)$$

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

So minimum value occurs at $$x = 1 - \sqrt 3 $$

$$y\left( {1 - \sqrt 3 } \right) = \left( {1 - \sqrt 3 } \right) - \ln \left( {{{2 - \left( {4 - 2\sqrt 3 } \right)} \over 2}} \right)$$

$$ = \left( {1 - \sqrt 3 } \right) - \ln \left( {\sqrt 3  - 1} \right)$$

Solutions · 1