The correct option is A y=C1e3x+C2e4x
The given equation can be written as (ddx−3)(dydx−4y)=0
If dydx−4y=u then (I) reduces to dudx−3u=0
⇒duu=3dx⇒u=C1e3x. Therefore, we have dydx−4y=C1e3x which is a linear equation whose I.F. is
e4x.Soddx(ye−4x)=C1e−x⇒ye−4x=−C1e−x+C2⇒y=C3x1+C2e4x