An equation of the curve satisfying xdy - ydx - x 2- y 2 dx and y 1-0 isA- y-2 x2x-1B- y-x2log-sin xC- y - x sin log - x -D- y2-xx-12

The correct option is C y = x sin (log|x|)The equation can be written as x^2 xdy ydx/x^2 = x √(1 (y/x)^2)dx ⇒d(y/x)/√(1 (y/x)^2) = dx/x⇒ sin^ 1y/x = log ...

You visited us 2 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Variable Separable Method

An equation o...

Question

An equation of the curve satisfying $x d y - y d x = \sqrt{x^{2} - y^{2}} d x$ and y(1) = 0 is

y = x^{2} l o g | s i n x |

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

y = x s i n (l o g | x |)

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

y^{2} = x (x - 1)^{2}

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

y = 2 x^{2} (x - 1)

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

Open in App

Solution

The correct option is B $y = x s i n (l o g | x |)$
The equation can be written as $x^{2} \frac{x d y - y d x}{x^{2}} = x \sqrt{1 - (\frac{y}{x})^{2}} d x$
$\Rightarrow \frac{d (\frac{y}{x})}{\sqrt{1 - (\frac{y}{x})^{2}}} = \frac{d x}{x} \Rightarrow s i n^{- 1} \frac{y}{x} = l o g | x | +$ const
Since y(1) = 0 so const =0. Hence y = x~ sin(log|x|)

Suggest Corrections

Similar questions

Q. An equation of the curve satisfying

x d y - y d x = \sqrt{x^{2} - y^{2}}

dx and

y (1) = 0

Q. An equation of the curve satisfying

x d y - y d x = \sqrt{x^{2} - y^{2}} d x

and

y (1) = 0

Suppose y = f(x) satisfies the differential equation $y d x + y^{2} d x = x d y$ . If y(x) > 0 for all x $ϵ$ R and y(1) = 1, then y (-3) =

Q. In the following, show that the given differential equation is homogeneous and solve each of them.
1.

(x^{2} + x y) d y = (x^{2} + y^{2}) d x

y^{'} = \frac{x + y}{x}

(x y) d y (x + y) d x = 0

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

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

x d y . y d x = \sqrt{x^{2} + y^{2}} d x

{x cos (\frac{y}{x}) + y sin (\frac{y}{x})} y d x = {y sin (\frac{y}{x}) - x cos (\frac{y}{x})} x d y

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

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

Q. Solve each of the following initial value problems:
(i) (x² + y²) dx = 2xy dy, y (1) = 0

(ii)

x e^{y / x} - y + x \frac{d y}{d x} = 0, y (e) = 0

(iii)

\frac{d y}{d x} - \frac{y}{x} + cosec \frac{y}{x} = 0, y (1) = 0

(iv) (xy − y²) dx − x² dy = 0, y(1) = 1

(v)

\frac{d y}{d x} = \frac{y (x + 2 y)}{x (2 x + y)}, y (1) = 2

(vi) (y⁴ − 2x³ y) dx + (x⁴ − 2xy³) dy = 0, y (1) = 1

(vii) x (x² + 3y²) dx + y (y² + 3x²) dy = 0, y (1) = 1

(viii)

\{x \sin^{2} (\frac{y}{x}) - y\} d x + x d y = 0, y (1) = \frac{π}{4}

(ix)

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

Join BYJU'S Learning Program

An equation of the curve satisfying xdy−ydx=√x2−y2dx and y(1) = 0 is

The correct option is B y=x sin(log|x|)The equation can be written as x2xdy−ydxx2=x√1−(yx)2dx ⇒d(yx)√1−(yx)2=dxx⇒sin−1yx=log|x|+const Since y(1) = 0 so const =0. Hence y = x~ sin(log|x|)

An equation of the curve satisfying $x d y - y d x = \sqrt{x^{2} - y^{2}} d x$ and y(1) = 0 is