If k is an arbitrary constant, then what is the general solution of the equation (x + y)^2 dy dx = k^2 ?

Step-by-step solution

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 )