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# Solve $\bigg( 1 + e^{\Large\frac{x}{y}} \bigg) dx + e^{\Large\frac{x}{y}} \bigg( 1 -\large \frac{x}{y} \bigg) $$dy= 0 \begin{array}{1 1} (A)\;x+ye^{\Large\frac{x}{y}}=0 \\ (B)\;x-e^{\Large\frac{x}{y}}=0 \\ (C)\;x+e^{\Large\frac{2x}{y}}=0\\ (D)\;x+e^{\Large\frac{x}{2y}}=0 \end{array} Can you answer this question? ## 1 Answer 0 votes Toolbox: • If the homogenous differential equation is in the form \large\frac{dx}{dy} =$$ F(x,y)$, where F(x,y) is homogenous function of degree zero, then we make substitution $x = vy$, hence $\large\frac{dx}{dy}$$= v + y\large\frac{dv}{dy}. Step 1: To check whether the given problem is a homogenous function of degree zero we rewrite the given equation as \large\frac{dx}{dy} = -\large\frac{ e^{\Large\frac{x}{y}}(1 - \large\frac{x}{y})}{1 + e^{\Large\frac{x}{y}}} Let F(x,y) = -\large\frac{ e^{\Large\frac{x}{y}}(1 - \large\frac{x}{y})}{ 1+ e^{\Large\frac{x}{y}}} F(kx,ky) =\large\frac{ e^{\Large\frac{-kx}{ky}}(1-\large\frac{kx}{ky})}{1 + e^{\Large kx/ky} }= -\large\frac{ e^x/y(1 - x/y)}{1 + e^x/y}$$= k^0.F(x,y)$
Hence this is a homogenous differential equation.
Step 2:
To solve this put $x = vy$ and $\large\frac{dx}{dy}$$= v + y\large\frac{dv}{dy} So v + y\large\frac{dv}{dy }= -e^{\Large\frac{vy}{y}}$$(1- vy/y)/1+e^{\Large\frac{ vy}{y}}$
Cancelling the common factor y we get
$v + v \large\frac{dv}{dy} = - e^v\large\frac{(1 - v)}{1+e^v}$
$v\large\frac{dv}{dy}= -e^v\large\frac{(1 - v)}{1+e^v}$$-v v\large\frac{dv}{dy} = v+\large\frac{ e^v}{1+e^v} Step 3: Bringing v and dv to the LHS and x and dx to the RHS we get, [\large\frac{1+e^v}{v+e^v} ] = \frac{- dy}{y} Integrating on both sides we get, \log(v+e^y) = - \log y + \log C Step 4: Substituting back for v we get \large\frac{x}{y} + e^{\Large\frac{x}{y }}= \frac{C}{y} On cancelling y throughout,we get the solution as x + ye^{\Large\frac{x}{y}}$$= C$