BITSAT 2018 Slot 2 Solved Paper

We have,

dy

dx

=y+2x
⇒

dy

dx

−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.