Solution of differential equation $\frac{dy}{dx}+Py=Q$

Question

If $\ y_1$ and $\ y_2$ be the solution of the differential equation $\frac{dy}{dx}+Py=Q$ where P and Q are the function of x alone and $\ y_2=z\ y_1$, then prove that $z=1+ae^{{-\int\frac{Qdx}{y}}}$, where a is being an arbitrary constant.

My approach is as follow Step 1: $\ y_2-\ y_1=z(\ y_1-1)$

Step 2: $\frac{dy_1}{dx}+Py_1=Q$ &$\frac{dy_2}{dx}+Py_2=Q$

Step 3: $\frac{dy_2}{dx}-\frac{dy_1}{dx}=z\frac{dy_1}{dx}$

Step 4: $\frac{dy_2}{dx}-\frac{dy_1}{dx}+P(\ y_2-\ y_1)=0$ or $\frac{dy_2}{dx}-\frac{dy_1}{dx}+Pz(\ y_1-1)=0$

Step 5: $z\frac{dy_1}{dx}+Pz(\ y_1-1)=0$

AFter this step I am confused

Answer 1

A rather terse presentation:

$y_2 = zy_1; \; y_i' + Py_i = Q, \; i = 1, 2; \tag 0$

$y_2' = z'y_1 + zy_1'; \tag 1$

$y_2' + Py_2 = z'y_1 + zy_1' + Pzy_1 = z'y_1 + z(y_1' + Py_1); \tag 2$

$Q = z'y_1 + zQ; \tag 3$

$z' + \dfrac{Q}{y_1} z = \dfrac{Q}{y_1}; \tag 4$

$(e^{\int \frac{Q}{y_1} dx}) z' + (e^{\int \frac{Q}{y_1} dx}) \dfrac{Q}{y_1} z = (e^{\int \frac{Q}{y_1} dx}) \dfrac{Q}{y_1}; \tag{4.5}$

$(e^{\int \frac{Q}{y_1} dx} z)' = (e^{\int \frac{Q}{y_1} dx}) \dfrac{Q}{y_1} = (e^{\int \frac{Q}{y_1} dx})'; \tag 5$

$e^{\int \frac{Q}{y_1} dx} z + C_1 = e^{\int \frac{Q}{y_1} dx} + C_2; \tag 6$

$e^{\int \frac{Q}{y_1} dx} z = e^{\int \frac{Q}{y_1} dx} + C_2 - C_1; \tag 7$

$a = C_2 - C_1; \tag 8$

$e^{\int \frac{Q}{y_1} dx} z = e^{\int \frac{Q}{y_1} dx} + a; \tag 9$

$z = 1 + ae^{-\int \frac{Q}{y_1} dx}. \tag{10}$

Note Added in Edit, Saturday 20 October 2018 12:15 PM PST: We may in fact discover more details about the function $z(x)$ and the constant $a$ by intercepting the stream of the preceding argument at (4.5) and replacing the indefinite integrals with more concrete calculations, performing definite integrations over the interval $[x_0, x]$:

$\left (\exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) z'(x) + \left (\exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) \dfrac{Q(x)}{y_1(x)} z(x)$ $= \left ( \exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) \dfrac{Q(x)}{y_1(x)}; \tag{11}$

$\left ( \left (\exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) z(x) \right )' = \left ( \exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right) \right )'; \tag{12}$

$\left (\exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) z(x) - z(x_0)$ $= \displaystyle \int_{x_0}^x \left ( \left (\exp \left (\displaystyle \int_{x_0}^u \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) z(u) \right )' \; du = \int_{x_0}^x \left ( \exp \left ( \int_{x_0}^u \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right )' \; du$ $= \exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) - 1; \tag{13}$

$\left (\exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) z(x)= \exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) + z(x_0) - 1, \tag{14}$

or, with

$a = z(x_0) - 1, \tag{15}$

$\left (\exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) \right ) z(x)= \exp \left (\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ) + a, \tag{16}$

and therefore,

$z(x) = 1 + a \exp \left (-\displaystyle \int_{x_0}^x \dfrac{Q(s)}{y_1(s)} \; ds \right ), \tag{17}$

the more specific form of (10). End of Note.

Answer 2

Hint: Here's another way. Suppose $y_1 = zy_2$, then \begin{align} y_1'+Py_1 = Q \ \ \implies& \ \ z'y_2+zy_2'+zPy_2=(zy_2)'+P(zy_2) = Q\\ \implies&\ \ z'y_2+z(y_2'+Py_2) = Q\\ \implies&\ \ z'+qz = q \end{align} where $q = Q/y_2$.