Question
If k is an arbitrary constant, then what is the general solution of the equation (x + y)^2 dy dx = k^2?
If k is an arbitrary constant, then what is the general solution of the equation (x + y)^2 dy dx = k^2?
B. x + y = k ( y - c k )
Let x + y = v Differentiating with respect to x, we get 1 + dy dx = dv dx dy dx = dv dx - 1 Substituting in the given differential equation: v^2 ( dv dx - 1 ) = k^2 v^2 dv dx = v^2 + k^2 v^2 v^2 + k^2 dv = dx ( 1 - k^2 v^2 + k^2 ) dv = dx Integrating both sides: ( 1 - k^2 v^2 + k^2 ) dv = dx v - k ^ -1 ( v k ) = x + c Substituting v = x + y: x + y - k ^ -1 ( x + y k ) = x + c y - c = k ^ -1 ( x + y k ) y - c k = ^ -1 ( x + y k ) x + y = k ( y - c k ) Answer: x + y = k ( y - c k )
Related: Mathematics — Differential Equations · All PYQ Banks