Question Statement

Solve the differential equation:

$\left(x+y{e}^{\frac{y}{x}}\right)dx-x{e}^{\frac{y}{x}}dy=0$

subject to the initial condition $y(1)=0$.

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Background and Explanation

This is a first-order homogeneous differential equation, identifiable by the fact that all terms have the same degree (degree 1). Such equations are solved using the substitution $y=ux$ (where $u=\frac{y}{x}$), which transforms them into separable equations in terms of $u$ and $x$.

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Solution

Rearranging to Standard Form

Starting with the given equation:

$\left(x+y{e}^{y/x}\right)\mathrm{dx}-x{e}^{y/x}dy=0$

Rearrange to isolate the differential terms:

$\left(x+y{e}^{y/x}\right)dx=x{e}^{y/x}dy$

Divide both sides by $dx$ and rearrange to express $\frac{dy}{dx}$:

$\Rightarrow \frac{dy}{dx}=\frac{x+y{e}^{y/x}}{x{e}^{y/x}}$

Substitution for Homogeneous Equation

Equation (1) is homogeneous. Apply the standard substitution:

$y=ux\Rightarrow \frac{y}{x}=u$

Differentiate $y=ux$ with respect to $x$ using the product rule:

$\frac{dy}{dx}=x\frac{du}{dx}+u$

Transforming the Equation

Substitute into equation (1):

$x\frac{du}{dx}+u=\frac{x+ux\cdot e^u}{x{e}^u}$

Simplify the right-hand side by factoring out $x$:

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

Subtract $u$ from both sides to isolate the derivative term:

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

Separation of Variables

Rearrange to separate variables $u$ and $x$:

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

Integration

Integrate both sides:

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

$e^u=\ln |x|+c$

Substitute back $u=\frac{y}{x}$:

$e^{\frac{y}{x}}=\ln x+c$

(Note: Since the initial condition is given at $x=1>0$, we write $\ln x$ for the domain $x>0$)

Applying the Initial Condition

Given $y(1)=0$, substitute $x=1$ and $y=0$ into equation (2):

$e^{\frac{0}{1}}=\ln 1+c$

$e^0=0+c$

$1=c$

Final Solution

Substitute $c=1$ back into equation (2):

$e^{\frac{y}{x}}=\ln x+1$

This can also be expressed explicitly for $y$ as:

$y=x\ln (\ln x+1)$

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Key Formulas or Methods Used

• Homogeneous equation test: $f(tx,ty)=t^{n}f(x,y)$ for some degree $n$
• Substitution method: $y=ux$ where $u=\frac{y}{x}$
• Product rule differentiation: $\frac{d}{dx}(ux)=x\frac{du}{dx}+u$
• Variable separation: Rearranging to form $f(u)du=g(x)dx$
• Standard integrals: $\int e^{u}du=e^u$ and $\int \frac{1}{x}dx=\ln |x|$

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Summary of Steps

1. Rearrange the equation to express $\frac{dy}{dx}$ explicitly as $\frac{x+y{e}^{y/x}}{x{e}^{y/x}}$
2. Identify the equation as homogeneous and substitute $y=ux$
3. Differentiate to get $\frac{dy}{dx}=x\frac{du}{dx}+u$
4. Substitute and simplify to obtain $x\frac{du}{dx}=e^{-u}$
5. Separate variables: $e^{u}du=\frac{dx}{x}$
6. Integrate both sides to get $e^u=\ln x+c$
7. Back-substitute $u=\frac{y}{x}$ to get $e^{y/x}=\ln x+c$
8. Apply initial condition $y(1)=0$ to determine $c=1$
9. State final solution: $e^{y/x}=\ln x+1$ or $y=x\ln (\ln x+1)$