Passage: Consider the differential equation e^ x+y dy dx =e^ x-y . Question: What is the solution of the differential equation with y(0)=0 ?

Step-by-step solution

The given differential equation is e^ x+y dy dx =e^ x-y e^x e^y dy dx = e^x e^ -y e^y dy dx = e^ -y e^ 2y dy = dx Integrating both sides: e^ 2y dy = dx e^ 2y 2 = x + C Given y(0)=0 , substituting x=0 and y=0 : e^0 2 = 0 + C C = 1 2 Substituting C back into the equation: e^ 2y 2 = x + 1 2 e^ 2y = 2x + 1 2y = (2x+1)