A curve passing through (2,3) and satisfying the differential equation ∫ limits0x t y(t) d t=x2 y(x),(x0) is -

Question

A curve passing through (2,3) and satisfying the differential equation ∫ limits0x t y(t) d t=x2 y(x),(x>0) is - (A) x2+y2=13 (B) y2=(9/2) x (C) (x2/8)+(y2/18)=1 (D) x y=6

Tardigrade Admin · Accepted Answer

∫ limits0x t y(t) d t=x2 y(x) Differentiating, we get x y=2 x y+x2 (d y/d x) ⇒ x2 (d y/d x)+x y=0 x (d y/d x)+y=0 ⇒ x d y+y d x=0 d(x y)=0 ⇒ x y=c ∴ text since it passes through (2,3) ∴ text c =6 text Hence x y=6 .

Q. A curve passing through $(2, 3)$ and satisfying the differential equation $0 \int x t y (t) d t = x^{2} y (x), (x > 0)$ is -

$x^{2} + y^{2} = 13$

$y^{2} = \frac{9}{2} x$

$\frac{x ^{2}}{8} + \frac{y ^{2}}{18} = 1$

$x y = 6$

$0 \int x t y (t) d t = x^{2} y (x)$
Differentiating, we get
$x y = 2 x y + x^{2} \frac{d y}{d x} \Rightarrow x^{2} \frac{d y}{d x} + x y = 0$
$x \frac{d y}{d x} + y = 0 \Rightarrow x d y + y d x = 0$
$d (x y) = 0 \Rightarrow x y = c$
$∴ since it passes through (2, 3)$
$∴ c = 6$
$Hence x y = 6.$

Q. A curve passing through (2,3) and satisfying the differential equation 0∫x​ty(t)dt=x2y(x),(x>0) is -

x2+y2=13

y2=29​x

8x2​+18y2​=1

xy=6

0∫x​ty(t)dt=x2y(x) Differentiating, we get xy=2xy+x2dxdy​⇒x2dxdy​+xy=0 xdxdy​+y=0⇒xdy+ydx=0 d(xy)=0⇒xy=c ∴ since it passes through (2,3) ∴ c =6 Hence xy=6.

Q. A curve passing through $(2, 3)$ and satisfying the differential equation $0 \int x t y (t) d t = x^{2} y (x), (x > 0)$ is -

$x^{2} + y^{2} = 13$

$y^{2} = \frac{9}{2} x$

$\frac{x ^{2}}{8} + \frac{y ^{2}}{18} = 1$

$x y = 6$