Let y = y(x) be the solution of the differential equation [e^x 1 - y^2 dx + ( y x )dy = 0]…

Question

Let y = y(x) be the solution of the differential equation $${e^x}\sqrt {1 - {y^2}} dx + \left( {{y \over x}} \right)dy = 0$$, y(1) = -1. Then the value of (y(3))2 is equal to :

Accepted Answer

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

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

$$ \Rightarrow \int {{{ - y} \over {\sqrt {1 - {y^2}} }}} dy = \int_{II}^{{e^x}} {_1^xdx} $$

$$ \Rightarrow \sqrt {1 - {y^2}}  = {e^x}(x - 1) + c$$

Given : At x = 1, y = -1

$$\Rightarrow$$ 0 = 0 + c $$\Rightarrow$$ c = 0

$$\therefore$$ $$\sqrt {1 - {y^2}}  = {e^x}(x - 1)$$

At x = 3

$$1 - {y^2} = {({e^3}2)^2} \Rightarrow {y^2} = 1 - 4{e^6}$$

Solutions · 1