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If $y_1$ and $y_2$ are the solutions of the differential equation $\frac{dy}{dx}+Py=Q$,where $P$ and $Q$ are functions of $x$ alone and $y_2=y_1\cdot z$,then prove that $$z=1+c\cdot\exp\left(-\int\frac{Q}{y_2}dx\right),$$ where $c$ is an arbitrary constant.

My Attempt— The integrating factor of the given differential equation is $\exp(\int \ Pdx)$ and so the general solution of the differential equation is $y\exp(\int Pdx)=\int (Q\exp(\int Pdx))dx$. Putting this in the given equation, I don't get the value of $z$ as it is required in the question. Please help me out. Thanks in advance.

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  • $\begingroup$ Your math is very hard to read, please use proper MathJax formatting. $\endgroup$ – Bobson Dugnutt Sep 4 '16 at 20:12
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Starting with $y_2=y_1z$ we get $y_2' = y_1'z+y_1z'$

So

$$\dfrac{dy_2}{dx} + Py_2 = y_1'z+y_1z' + Py_1z = z(y_1'+Py_1)+y_1z'$$

We know $y_1' + Py_1=Q$ so we have

$$\dfrac{dy_2}{dx} + Py_2 = Q \iff \dfrac{z'}{z-1}=-\dfrac{Q}{y_1} $$

Integrating both sides gives

$$\ln|z-1| = \int\left(-\dfrac{Q}{y_1}\right) dx + C$$

And this should be easy to finish off

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