The general solution of the differential equation $1 + x^{2} + y^{2} + x^{2} y^{2} + xy \frac{dy}{dx} = 0$ is :
(where $C$ is a constant of integration).

Question

The general solution of the differential equation 1+x^2+y^2+x^2y^2+ xy dy dx=0 is : (where C is a constant of integration).

Accepted Answer

Correct answer: A. (a) JEE Main 2020 1+x^2+y^2+x^2y^2+ xy dy dx=0 1+x^2 ∙ 1+y^2=- xy dy dx 1+x^2x dx=- y 1+y^2 dy Let I= 1+x^1x dx Let x= tan ⇒ dx= sec^2 dθ = sec^3 θdθ tan = sin^2 + cos^2 sin ∙ cos^2 = ( tan ∙ sec + cosec ) dθ = sec + ln( cosec - cot )+C = 1+x^2+ ln ( 1+x^2-1x )+C 1+x^2+ ln ( 1+x^2-1x )=- 1+y^2+C 1+y^2+ 1+x^2= ln ( x 1+x^2-1 )+C= 12 ln ( 1+x^2+1 1+x^2-1 )+C

$The general solution of the differential equation 1 + x^{2} + y^{2} + x^{2} y^{2} + xy \frac{dy}{dx} = 0 is : (where C is a constant of integration).$

Solution

More Differential Equations PYQs