Question

Find the general solution of the differential equation $(1+tany)(dx-dy)+2xdy=0.$

OR Solve the differential equation $(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0.$

Answer 1

Here $(1+tany)(dx-dy)+2xdy=0\Rightarrow (1+tany)dx-(1+tany)dy+2xdy=0$

$\Rightarrow \frac{dx}{dy}-\frac{1+tany-2x}{1+tany}=0\Rightarrow \frac{dx}{dy}+\left(\frac{2}{1+tany}\right)x=1$

Clearly it is in the form of $\frac{dx}{dy}+P\left(y\right)x=Q\left(y\right)$ where $P\left(y\right)=\frac{2}{1+tany},Q\left(y\right)=1$

I.F.$=e^{\int P\left(y\right)dy}=e^{\int \frac{2}{1+tany}dy}=e^{\int \frac{(cosy+siny)+(-siny+cosy)}{cosy+siny}dy}=e^{\int (1+\frac{-siny+cosy}{cosy+siny})}dy$

$\Rightarrow =e^{y+log(cosy+siny)}=e^y\times e^{log(cosy+siny)}=e^y(cosy+siny)$

Therefore, the solution is given by $x(I.F.)=\int (I.F.)\times Q\left(y\right)dy+C$

$\therefore x{e}^y(cosy+siny)=\int e^y(cosy+siny)\times 1dy+C$

$\Rightarrow x{e}^y(cosy+siny)=e^{y}siny+C.$

OR We have$(1+e^{\frac{x}{y}})dx+e^{\frac{x}{y}}(1-\frac{x}{y})dy=0\Rightarrow (1+e^{\frac{x}{y}})dx=e^{\frac{x}{y}}(\frac{x}{y}-1)dy$

$\Rightarrow \frac{dx}{dy}=\frac{e^{\frac{x}{y}(\frac{x}{y}-1)}}{(1+e^{\frac{x}{y}})}=f\left(\frac{x}{y}\right)$ say

Put $x=\lambda x,y=\lambda y$ then, $f\left(\frac{\lambda x}{\lambda y}\right)=\frac{e^{\frac{\lambda x}{\lambda y}(\frac{\lambda x}{\lambda y}-1)}}{(1+e^{\frac{\lambda x}{\lambda y}})}=\frac{e^{\frac{x}{y}(\frac{x}{y}-1)}}{(1+e^{\frac{x}{y}})}=f\left(\frac{x}{y}\right)$

Hence the diff. eq. is homogeneous.

To solve, let $x=vy\Rightarrow \frac{dx}{dy}=v+y\frac{dv}{dy}$

$\therefore v+y\frac{dv}{dy}=\frac{e^v(v-1)}{(1+e^v)}\Rightarrow y\frac{dv}{dy}=\frac{v{e}^v-e^v-v-v{e}^v}{(1+e^v)}=\frac{-e^v-v}{(1+e^v)}$

$\Rightarrow \frac{1+e^v}{v+e^v}dv=-\frac{dy}{y}\Rightarrow log|v+e^v|=-log|y|+logC$

$\Rightarrow log\left|\right(v+e^v\left)y\right|=logC\Rightarrow (v+e^v)y=\pm C$

$\therefore (\frac{x}{y}+e^{x/y})y=A,$ where $A=\pm C.$