If y = y(x) is the solution of the differential equation [5 + e^x 2 + y.dy dx + e^x = 0] s…

Question

If y = y(x) is the solution of the differential

equation $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$  satisfying
y(0) = 1, then a value of y(loge13) is :

Accepted Answer

$${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} + {e^x} = 0$$

$$ \Rightarrow $$ $${{5 + {e^x}} \over {2 + y}}.{{dy} \over {dx}} = -{e^x} $$

Integrating both sides,

$$ \Rightarrow $$ $$\int {{{dy} \over {2 + y}}}  = \int {{{ - {e^x}} \over {{e^x} + 5}}} dx$$

$$ \Rightarrow $$ ln (y + 2) = -- ln(ex
 + 5) + k

$$ \Rightarrow $$ (y + 2) (ex
 + 5) = C

$$ \because $$ y(0) = 1

$$ \Rightarrow $$ C = 18

$$ \therefore $$ y + 2 = $${{{18} \over {{e^x} + 5}}}$$

At x = loge13

y + 2 = $${{{18} \over {13 + 5}}}$$ = 1

$$ \Rightarrow $$ y = -1

Solutions · 1