Question

Write the general form of $(1 + \tan y)(dx - dy) + 2xdy = 0$

Answer 1

The most general form of linear differential equations of first order is $\dfrac{{dy}}{{dx}} + Py = Q$ , where P and Q are functions of x.
To solve such an equation multiply both sides by ${e^{\smallint Pdx}}$ . Then the solution of this equation will be
$y{e^{\smallint Pdx}} = \smallint Q{e^{\smallint Pdx}}dx + c$
Another form of first order linear differential equation is $\dfrac{{dx}}{{dy}} + {P_1}x = {Q_1}$ , where $P_1$ and $Q_1$ are functions of y only. And the solution of such an equation is given by $x.{e^{\smallint {P_1}dx}} = \smallint ({Q_1} \times {e^{\smallint {P_1}dx}})dy + c$

Complete step by step answer:
Step 1: Rearranging the terms in $(1 + \tan y)(dx - dy) + 2xdy = 0$
Divide both sides of the equation by dy, we get
$(1 + \tan y)(dx - dy) + 2xdy = 0$
$(1 + \tan y)(\dfrac{{dx}}{{dy}} - 1) + 2x = 0$
Taking $\dfrac{{dx}}{{dy}}$ separately we get
$(1 + \tan y)(\dfrac{{dx}}{{dy}} - 1) + 2x = 0$
$(1 + \tan y)\dfrac{{dx}}{{dy}} - (1 + \tan y) + 2x = 0$
$(1 + \tan y)\dfrac{{dx}}{{dy}} + 2x = (1 + \tan y)$
$\dfrac{{dx}}{{dy}} + \dfrac{{2x}}{{1 + \tan y}} = 1$
The above equation is in the form of first order linear differential equation, $\dfrac{{dx}}{{dy}} + {P_1}x = {Q_1}$
Step 2: On comparing both equation $\dfrac{{dx}}{{dy}} + \dfrac{{2x}}{{1 + \tan y}} = 1$ & $\dfrac{{dx}}{{dy}} + {P_1}x = {Q_1}$ , we get
${P_1} = \dfrac{{2x}}{{1 + \tan y}}$ & ${Q_1} = 1$
Step 3: finding the integrating factor (I.F)
As we know that, integrating factor is given by
$I.F = {e^{\smallint {P_1}dy}}$
Substituting the values we get,
$I.F = {e^{\smallint \dfrac{2}{{1 + \tan y}}dy}}$
We know that $\tan y = \dfrac{{\sin y}}{{\cos y}}$ , replacing tan y with its value, we get
$\Rightarrow$ $I.F = {e^{\smallint \dfrac{{2\cos y}}{{\cos y + \sin y}}dy}}$
Now adding and subtracting siny in numerator, we get
$\Rightarrow$ $I.F = {e^{\smallint \dfrac{{\cos y + \sin y + \cos y - \sin y}}{{\cos y + \sin y}}dy}}$
$\Rightarrow$ $I.F = {e^{\smallint 1 + \dfrac{{\cos y - \sin y}}{{\cos y + \sin y}}dy}}$
$\Rightarrow$ $I.F = {e^{y + \log (\cos y + \sin y)}}$ $\\{ \int {\dfrac{{\cos y - \sin y}}{{\cos y + \sin y}}} dy = \dfrac{{d\log (\cos y + \sin y)}}{{dy}}\\}$
$\Rightarrow$ $I.F = {e^y}.(\cos y + \sin y)$
Step 4: Determining the general solution
As we know that the general solution of linear first degree differential equation is given by,
$x.{e^{\smallint {P_1}dx}} = \smallint ({Q_1} \times {e^{\smallint {P_1}dx}})dy + c$
Substituting the values, we get
$x.{e^y}.(\cos y + \sin y) = \int {1.} {e^y}(\cos y + \sin y)dy + C$
$\Rightarrow$ $x.{e^y}.(\cos y + \sin y) = \int {{e^y}(\sin y + \cos y)dy} + C$
Since, $\int {{e^y}} \left[ {f(y) + f'(y)} \right]dy = {e^y}.f(y) + C$
Therefore, we have
$x.{e^y}.(\cos y + \sin y) = {e^y}\sin y + C$
Cancelling $e^y$ from both sides, we get
$\Rightarrow$ $x.(\cos y + \sin y) = \sin y + C{e^{ - y}}$

Note: This function $g(x) = {e^{\int {Pdx} }}$ is called Integrating Factor (I.F.) of the given differential equation.
The general solution of the first order linear differential equation of the form $\dfrac{{dy}}{{dx}} + Py = Q$ is given by $y{e^{\smallint Pdx}} = \smallint Q{e^{\smallint Pdx}}dx + c$
The general solution of the first order linear differential equation of the form $\dfrac{{dx}}{{dy}} + {P_1}x = {Q_1}$ is given by $x.{e^{\smallint {P_1}dx}} = \smallint ({Q_1} \times {e^{\smallint {P_1}dx}})dy + c$