Q.

The solution of  xdydx+y=ex is:

1 y=exx+kx  
2 y=xex+Cx  
3 y=xex+k  
4 x=eyy+ky  

Ans.

(1)

Ans.        y=exx+kx

Explanation:

Given that,

                xdydx+y=ex

⇒        dydx+yx=exx

which is a linear differential equation.

∴     IF=e∫1x dx=elogx=x

The general solution is

⇒      y·x=∫(exx·x)dx

⇒      y·x=∫ex dx

⇒      y·x=ex+k

⇒           y=exx+kx