−y=2x Now, F=e−∫1dx=e−x y.e−x=∫2xe−xdx+k,k be the constant of integration =2[x‌∫e−xdx−∫1.(−e−x)dx]+k [using integration by parts] ⇒y.e−x=−2xe−x−2e−x+k As curve (i) passes through (0,0) ∴0=0−2+k ⇒k=2 Thus, the curve is ye−x=−2xe−x−2e−x+2 ∴y=2(ex−x−1) Hence, option (b) is correct.