Solution of the differential equation (d x/d y)-(x ln x/1+ ln x)=(ey/1+ ln x) if y(1)=0 is

Question

Solution of the differential equation (d x/d y)-(x ln x/1+ ln x)=(ey/1+ ln x) if y(1)=0 is (A) xx=ey ey (B) ey=xey (C) x=y ey (D) y=exy

Tardigrade Admin · Accepted Answer

(1+ ln x) (d x/d y)-x ln x=ey <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/163f13628870ba0f8b18344fc5681818-.png" /> t = ye y + e y ⋅ c x ln x = ye y + e y ⋅ c 0= c ⇒ x ln x = ye y ln x x = y e y x x = e ye y

Q. Solution of the differential equation $\frac{d x}{d y} - \frac{x l n x}{1 + l n x} = \frac{e ^{y}}{1 + l n x}$ if $y (1) = 0$ is

$x^{x} = e^{y e^{y}}$

$e^{y} = x^{e^{y}}$

$x = y e^{y}$

$y = e^{x^{y}}$

$(1 + ln x) \frac{d x}{d y} - x ln x = e^{y}$

$t = y e^{y} + e^{y} \cdot c$
$x ln x = y e^{y} + e^{y} \cdot c$
$0 = c$
$\Rightarrow x ln x = y e^{y}$
$ln x^{x} = y e^{y}$
$x^{x} = e^{y e^{y}}$

Q. Solution of the differential equation dydx​−1+lnxxlnx​=1+lnxey​ if y(1)=0 is

xx=eyey

ey=xey

x=yey

y=exy