(2𝑥𝑦 𝑑𝑥+ 𝑑𝑦)𝑒𝑥2=0 𝑦(0)=2

Question

solve this exact equation

Answer 1

(2xydx+dy)e^xx =0

Couldn't fit x² in this notation so I wrote it xx but you know its derivative is 2x. The expanded version gives you M = 2xye^xx and N= e^xxdy. Note M_y = 2xe^xx= N_x so the equation is exact

2xye^xx dx + e^xx dy = 0

∫2xye^xxdx = ye^xx + h(y)

∫e^xxdy = ye^xx + g(x)

Since no extra terms show up then the integral gives

ye^xx + c =0 or ye^xx = C where c ==C is a constant.

the information y(0) = 2 makes y and explicit function of x but we don't need to rewrite the equation explicitly, just remember the information says y =2 when x = 0. Substituting we have

2 e⁰ =2 so ye^xx =2 is the solution to the equation.