If `y_1` and `y_2` are the solution of the differential equation `(dy)/(dx)+P y=Q` , where `P` and `Q` are functions of `x` alone and `y_2=y_1z` , then prove that `z=1+cdote^(-fQ/(y_1)dx),` where `c` is an arbitrary constant.

Explanation

Given, `(dy_(1))/(dx)+Py_(1)=Q`...............(1)
`(dy_(2))/(dx) + Py_(2)=Q`..............(2)
Clearly, we have to eliminate P and `y_(2)=zy_(1)`
or `(d(zy_(1)))/(dx) +Pzy_(1)=Q`
or `y_(1)(dz)/(dx) + z(dy_(1))/(dx) + Pzy_(1)=Q`
From equation (1), put the value of `Py_(1)`.
We have `y_(1)(dz)(dx) +z(dy_(1)/(dx))=Q`
or `y_(1)(dz)/(dx) = Q(1-z)`
or `int(dz)/(z-1) = -intQ/y_(1)dx`
or `log(z-1)=-intQ/y_(1)dx+logc`
or `log(z-1)/(c) = -intQ/y_(1)dx`
or `z=1+ce^(int(Q/y_(1))dx` or `z=1+ce^(int(Q/y_(1))dx`