The equation of one of the curve passing through 1-4 and satisfying the differential equation d y-d x2-x-x-y d y-d x-y A- y -8-4 xB- x2-y2-17C- 2 x2-y2-18D- 4 y -3 x -13

The correct option is B x^2+y^2=17Let dydx = t yt^2 + (x+y)t + x = 0 dydx = t = (x+y) ±√((x+y)^2 4xy)2y dydx = 1, xy dy = dx ⇒ y = x + c ydy = x ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

The equation ...

Question

The equation of one of the curve passing through $(1, 4)$ and satisfying the differential equation ${(\frac{d y}{d x})}^{2} = \frac{- x - (x + y) \frac{d y}{d x}}{y}$ is

y = 8 - 4 x

No worries! We‘ve got your back. Try BYJU‘S free classes today!

x^{2} + y^{2} = 17

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

2 x^{2} + y^{2} = 18

No worries! We‘ve got your back. Try BYJU‘S free classes today!

4 y = 3 x + 13

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $x^{2} + y^{2} = 17$
Let $\frac{d y}{d x} = t$

$y t^{2} + (x + y) t + x = 0 \frac{d y}{d x} = t = \frac{- (x + y) \pm \sqrt{(x + y)^{2} - 4 x y}}{2 y} \frac{d y}{d x} = - 1, - \frac{x}{y} d y = - d x \Rightarrow y = - x + c y d y = - x d x \Rightarrow \frac{y^{2}}{2} = - \frac{x^{2}}{2} + C$
Curve passing through $(1, 4)$
$y = - x + 5$ and $x^{2} + y^{2} = 17$

Suggest Corrections

Similar questions

Q. The equation of curve passing through (3, 4) and satisfying the differential equation

y {(\frac{d y}{d x})}^{2} + (x - y) \frac{d y}{d x} - x = 0

can be

Q. The equation of the curve passing through the origin and satisfying the differential equation

{(\frac{d y}{d x})}^{2} = (x - y)^{2}

, is

Q. The equation of the curve passing through the origin and satisfying the differential equation

{(\frac{d y}{d x})}^{2} = (x - y)^{2}

, is

The differential equation of family of curves $y^{2} = 4 a (x + a)$ is
a) $y^{2} = 4 \frac{d y}{d x} (x + \frac{d y}{d x})$
b) $2 y \frac{d y}{d x} = 4 a$
c) $y \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = 0$
d) $2 x \frac{d y}{d x} + y {(\frac{d y}{d x})}^{2} - y = 0$

Q. A curve passes thorugh the point

(x = 1, y = 0)

and satisfies the differential equation

\frac{d y}{d x} = \frac{x^{2} + y^{2}}{2 y} + \frac{y}{x}

The equation that decribes the curve is

Join BYJU'S Learning Program

The equation of one of the curve passing through (1,4) and satisfying the differential equation (dydx)2=−x−(x+y)dydxy is

The correct option is B x2+y2=17Let dydx=t yt2+(x+y)t+x=0dydx=t=−(x+y)±√(x+y)2−4xy2ydydx=−1,−xydy=−dx⇒y=−x+cydy=−xdx⇒y22=−x22+C Curve passing through (1,4) y=−x+5 and x2+y2=17

The equation of one of the curve passing through $(1, 4)$ and satisfying the differential equation ${(\frac{d y}{d x})}^{2} = \frac{- x - (x + y) \frac{d y}{d x}}{y}$ is