Question

In Problems 11–14 solve the given initial-value problem.

$\left(x+y{e}^{\raisebox{1ex}{$y$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}\right)\mathit{d}\mathit{x}\mathbf{-}\mathit{x}{\mathit{e}}^{\raisebox{1ex}{$\mathbf{y}$}\!\left/ \!\raisebox{-1ex}{$\mathbf{x}$}\right.}\mathit{d}\mathit{y}\mathbf{=}\mathbf{0}\mathbf{,}\mathit{y}\left(1\right)\mathbf{=}\mathbf{0}$

Short answer

The solution of the given differential equation is e^y/x= Inx + 1.

Step-by-step solution

Define substitution method for differentiation.

Often, the initial step in solving a differential equation is to use a substitution to change it into another differential equation.

The substitutions that can be employed to solve a homogeneous differential equation are suggested by their properties. A homogeneous equation can be reduced to a separable first-order differential equation by substituting y = ux or x = vy, where u and v are new dependent variables.

Substitution of variables by equations using property.

The given differential equation is a homogeneous function. So, let y = ux. Then,$\frac{dy}{dx}=u+x\frac{du}{dx}$.

Substitute them into the differential equation.

$\begin{array}{rcl}\left(x+y{e}^{y/x}\right)dx-x{e}^{y/x}dy& =& 0\\ \left(x+\left[ux\right]e^{\left[ux\right]/x}\right)dx-x{e}^{\left[ux\right]/x}\left[udx+xdu\right]& =& 0\\ \left(x+ux{e}^u\right)dx-x{e}^u\left[udx+xdu\right]& =& 0\\ xdx+ux{e}^{u}dx-ux{e}^{u}dx-x^2{e}^{u}du& =& 0\\ xdx-x^2{e}^{u}du& =& 0\\ dx-x{e}^{u}du& =& 0\end{array}$

Integrate the function.

Let separate the variables be,

$e^{u}du=\frac{1}{x}dx$

Integrate on both sides.

$\begin{array}{rcl}\int e^{u}du& =& \int \frac{1}{x}dx\\ e^u& =& Inx+C\end{array}$

Substitute the equation$u=\frac{y}{x}$into the above equation.

$e^{y/x}=Inx+C$

To find the value of constant, substitute y (1) = 0.

1 = 0 + C

C = 1

Hence, the solution is, e^y/x = Inx + 1.