If `y_1` and `y_2` are two solutions to the differential equation `(dy)/(dx)+P(x)y=Q(x)` . Then prove that `y=y_1+c(y_1-y_2)` is the general solution to the equation where `c` is any constant.

Explanation

`y_(1), y_(2)` are the solutions of the differential equation
`(dy)/(dx) + P(x)y=Q(x)` .............(1)
Then `(dy_(1))/(dt) + P(x)y_(1)=Q(x)`............(2) and `(dy_(2))/(dx) + P(x)y_(2)=Q(x)`.............(3)
From equations (1) and (2), we get
`(d(y-y_(1)))/(dx) + P(x)(y-y_(1))=0`..............(4)
and from equation (1) and (2), we get
`(d(y_(1)-y_(2)))/(dx) + P(x)(y_(1)-y_(2))=0`.............(5)
Now, from equation (4) and (5), we get
`(d/(dx)(y-y_(1)))/(d/(dx)(y_(1)-y_(2))) = (y-y_(1))/(y_(1)-y_(2))`
or `int(d(y-y_(1))/(y-y_(1))) = int(d(y_(1)-y_(2))/(y_(1)-y_(2)))`
or `"ln"(y-y_(1))="ln"(y-y_(2))+"ln"c`
or `y-y_(1)=c(y_(1)-y_(2))`
or `y=y_(1)+c(y_(1)-y_(2))`
or `y-y_(1)=c(y_(1)-y_(2)`
or `y=y_(1)+c(y_(1)-y_(2))`