If the equation of the curve which passes thr | vedclass

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(D) Given the differential equation: $\frac{dy}{dx} = \frac{2x-5y+3}{5x+2y-3}$ Rearranging the terms: $(5x+2y-3)dy = (2x-5y+3)dx$ $(5x+2y-3)dy - (2x-5

Vedclass · Accepted Answer

$x^2-5xy-y^2+3x+3y-1=0$

Vedclass · Answer

(D) Given the differential equation: $\frac{dy}{dx} = \frac{2x-5y+3}{5x+2y-3}$ Rearranging the terms: $(5x+2y-3)dy = (2x-5y+3)dx$ $(5x+2y-3)dy - (2x-5y+3)dx = 0$ $5x dy + 2y dy - 3 dy - 2x dx + 5y dx - 3 dx = 0$ Group the terms: $5(x dy + y dx) + (2y dy - 2x dx) - (3 dy + 3 dx) = 0$ This can be written as: $5 d(xy) + d(y^2) - d(x^2) - 3 d(x+y) = 0$ Integrating both sides: $5xy + y^2 - x^2 - 3(x+y) = C$ Since the curve passes through $(1,1)$,substitute $x=1$ and $y=1$: $5(1)(1) + (1)^2 - (1)^2 - 3(1+1) = C$ $5 + 1 - 1 - 6 = C \Rightarrow C = -1$ Substituting $C$ back into the equation: $5xy + y^2 - x^2 - 3x - 3y = -1$ Rearranging: $x^2 - 5xy - y^2 + 3x + 3y - 1 = 0$ Thus,the correct option is $D$.

If the equation of the curve which passes through the point $(1,1)$ satisfies the differential equation $\frac{dy}{dx} = \frac{2x-5y+3}{5x+2y-3}$,then the equation of that curve is:

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