Solution of `(dy)/(dx)+2x y=y` is (a) `( b ) (c) y=c (d) e^( e ) (f) x-( g ) x^((( h )2( i ))( j ) (k))( l ) (m)` (n) (b) `( o ) (p) y=c( q ) e^( r ) (s) (t) x^((( u )2( v ))( w )-x (x))( y ) (z)` (aa) (c) `( d ) (e) y=c( f ) e^(( g ) x (h))( i ) (j)` (k) (d) `( l ) (m) y=c (n) e^( o ) (p)-( q ) x^((( r )2( s ))( t ) (u))( v ) (w)` (x)

Explanation

Give that, `(dy)/(dx)+2xy=y` ltbr `Rightarrow(dy)/(dx)+2xy-y=0`
`Rightarrow(dy)/(dx)+(2x-1)y=0`
Which is linear differential equation.
On comparing it with `(dy)/(dx)+Py=Q`, we get
`P=(2x-1),Q=0`
`IF=e^(intPdx)=e^(int(2x-1))dx`
`=e((2x^(2))/(2)-x)=e^(x^(2)-x)`
The complete solution is
`y.e^(x^(2)-x)=intQ.e^(x^(2)-x)dx+C`
`Rightarrow y.e^(x^(2-x))=0+C`
`Rightarrow y=C e^(x-x^(2))`