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 $${\displaystyle \frac{dy}{dx}}+Py=Q$$ , where P and Q are functions of x.
To solve such an equation multiply both sides by $$e^{\int Pdx}$$ . Then the solution of this equation will be
 $$y{e}^{\int Pdx}=\int Q{e}^{\int Pdx}dx+c$$
Another form of first order linear differential equation is $${\displaystyle \frac{dx}{dy}}+P_{1}x=Q_1$$ , where $P_1$ and $Q_1$ are functions of y only. And the solution of such an equation is given by $$x.e^{\int P_{1}dx}=\int (Q_1\times e^{\int P_{1}dx})dy+c$$

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