Find the solution of $(2 x + 3 y - 5) d y + (3 x + 2 y - 5) d x = 0$

4 x y + 3 (x^{2} + y^{2}) - 10 (x + y) = k .

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4 x y - 3 (x^{2} + y^{2}) - 10 (x + y) = k .

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2 x y + 3 (x^{2} + y^{2}) - 10 (x + y) = k .

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

2 x y - 3 (x^{2} + y^{2}) - 10 (x + y) = k .

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

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Solution

The correct option is A $4 x y + 3 (x^{2} + y^{2}) - 10 (x + y) = k .$
Given, $(2 x + 3 y - 5) d y + (3 x + 2 y - 5) d x = 0$
$2 (x d y + y d x) + (3 y - 5) d y + (3 x - 5) d x = 0.$
$2 d x y + (3 y - 5) d y + (3 x - 5) d x = 0.$
Integrating we get,
$2 (x y) + 3 \frac{y^{2}}{2} - 5 y + 3 \frac{x^{2}}{2} - 5 x = k$
$4 x y + 3 (x^{2} + y^{2}) - 10 (x + y) = k .$

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Similar questions

Q. If the circle

3 x^{2} + 3 y^{2} + 10 x + y - 27 = 0

bisects the circumference of the circle

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

then

k^{2} - 1 =

Q. Find the real points

(x, y)

satisfying

3 x^{2} + 3 y^{2} - 4 x y + 10 x - 10 y + 10 = 0

Q. Simplify:

\frac{3 y (x - y) - 2 x (y - 2 x)}{7 x (x - y) - 3 (x^{2} {- y}^{2})}

Q. Fnd the real points

(x, y)

satisfying

3 x^{2} + 3 y^{2} - 4 x y + 10 x + 10 y + 10 = 0.

Match the column

$\begin{matrix} Equation & Name of the curve 1) x^{2} - 2 x - y - 3 = 0 & P) Circle 2) x^{2} + 3 x y + 2 y^{2} - x - 4 y - 6 = 0 & Q) Parabola 3) x^{2} + y^{2} - 20 = 0 & R) Ellipse 4) 7 x^{2} + 7 y^{2} + 2 x y + 10 x - 10 y + 7 = 0 & S) Hyperbola 5) 6 x^{2} - x y - y^{2} - 23 x + 4 y + 15 = 0 & T) Pair of straight lines \end{matrix}$

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Find the solution of (2x+3y−5)dy+(3x+2y−5)dx=0

The correct option is A 4xy+3(x2+y2)−10(x+y)=k.Given, (2x+3y−5)dy+(3x+2y−5)dx=02(xdy+ydx)+(3y−5)dy+(3x−5)dx=0. 2dxy+(3y−5)dy+(3x−5)dx=0. Integrating we get, 2(xy)+3y22−5y+3x22−5x=k 4xy+3(x2+y2)−10(x+y)=k.

Find the solution of $(2 x + 3 y - 5) d y + (3 x + 2 y - 5) d x = 0$