For the given differential equation, identify whether it is separable, homogeneous, or exact - justify your answer. Also, find the general solution to the given DE.

Question

a. (1+x²+y²+x²y²)dy = y²dx

b. dy/dx = (y/x)+(x²/y²)+1

c. dN/dt +N = Nte¹⁺²

Answer 1

a. Is equation in separate variables becouse we can factor right side: dy/dx = (1 + x²)(1+ y²);

dy/(1 + y²) = (1 + x²)dx; ∫dy/(1 + y²) = ∫(1 + x²)dx; tan^-1y = x + x³/3 + C, y = tan(x + x³/3 + C).

b. This equation is homogenius, becouse right side is homogenius function

to solve we take new function v: y = xv, y' = v + xv' and we get v + xv' = v + 1/v² + 1, or xv' = 1 + 1/v²;

v²/(1 + v²)dv = dx/x; ∫ (1 - 1/(1 + v²))dv = ∫dx/x; v - tan^-1v = lnx + C; y/x - tan^-1(y/x) - lnx = C

c. This equation is separable: dN/dt = N(te^2t - 1); ; dN/N = ∫(te^2t - 1)dt; lnN = 1/2te^2t - 1/4e^2t - t + C