Homogeneous Differential Equations: JEE Questions & Solutions

Let the solution curve of the differential equation x dy - y dx = x^2+y^2 dx, x0, with y(1)=0, be y=y(x). Then y(3) is equal to

Let x = x(y) be the solution of the differential equation: y = ( x - y dxdy) ( xy), y 0 and x(1) = 2. Then (x(2)) is equal to:

If ( yx) = |x| + 2 is the solution of the differential equation x ( yx) dydx = y ( yx) + x, and y(1) = 3, then ^2 is equal to:

The solution of the differential equation dydx = x^2 + 3y^23x^2 + y^2, y(1) = 0 is:

Let y = y(x) be the solution of the differential equation (3y^2 - 5x^2) y dx + 2x(x^2 - y^2) dy = 0, such that y(1) = 1. Then ( y(2))^3 -…

The slope of the tangent at any point (x, y) on a curve y = y(x) is x^2 + y^22xy, x 0. If y(2) = 0, then a value of y(8) is: