Here is the question :
If $y_1$ and $y_2$ be the solutions of the differential equation $\frac{dy}{dx} +Py = Q$, where P & Q are functions of $x$ alone, and $y_2=y_1z $, then prove that $$z=1+ae^{-\int \frac{Q}{y_1}\,dx} $$, 'a' being an arbitrary constant.
I have been at this for hours now, what I tried is:
- I tried to differentiate the relation $y_2 = y_1 z$ and relate the individual equations of $y_1$ and $y_2$ .
2.Dividing equations of $y_1$ and $y_2$ to get some relation between the constants of integration in both case
But I haven't been able to reach anything conclusive.
Any help would be appreciated. Thanks
We are given that $y_1$ and $y_2$ are solutions of
$$\tag 1 \dfrac{dy}{dx} +Py = Q$$
where $P$ and $Q$ are functions of $x$ alone, and $y_2=y_1z$, and asked to prove that
$$z=1+ae^{-\int \frac{Q}{y_1}\,dx} $$
where $a$ is an arbitrary constant.
Given $y_2 = y_1 z$, taking the derivative $\tag 2 y_2' = y_1' z + y_1 z'$
Since $y_2$ is a solution to $(1)$, we have
$$y_2' + P y_2 = Q$$
Substituting
$$y_1' z + y_1 z' + P y_1 z = Q$$
Simplifying
$$z(y_1' + Py _1) + y_1 z' = Q$$
Using $(1)$ for the expression in parenthesis, this reduces to
$$z Q + y_1 z' = Q$$
This is a Separable Equation
$$\displaystyle \int \dfrac{1}{z-1}~dz = \int -\dfrac{Q}{y_1}~dx$$
Integrating
$$\ln(z-1) = c + \int -\dfrac{Q}{y_1}~dx$$
Solving for $z$
$$z = 1 + a e^{-\int \frac{Q}{y_1}~dx}$$