Question

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

Answer 1

Hint: 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 \[x.{e^{\smallint {P_1}dx}} = \smallint ({Q_1} \times {e^{\smallint {P_1}dx}})dy + c\]