Find the special solution of the differential equation y '= e ^ (x + y) satisfying the initial condition y (0) = 0

Find the special solution of the differential equation y '= e ^ (x + y) satisfying the initial condition y (0) = 0

y'=(e^x)(e^y)
e^(-y)dy=e^xdx
-e^(-y)=e^x+C
C = - 2
The special solution is e ^ x + e ^ (- y) = 2 or y = - ln (2-e ^ x)

Y "= e ^ 2Y y y|x = 0 = 0 y '|x = 0 = 0 to solve differential equation

For the differential equation x * (dy / DX) - 2Y = x ^ 3E ^ x under x = 1, y = 0, the answer is y = x ^ 2 (e ^ X - E),

[method 1] x * (dy / DX) - 2Y = x ^ 3 * e ^ x divided by x ^ 3 = > (x * y '- 2Y) / x ^ 3 = e ^ x left numerator and denominator multiplied by x = > (y' * x ^ 2 - y * (x ^ 2)) / x ^ 4 = (Y / x ^ 2) '= e ^ x integral on both sides = > y / x ^ 2 = e ^ x + C = > y =