Question Statement

Solve the differential equation:

$\frac{dy}{dx}=\frac{y-x}{x+y}$

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Background and Explanation

This is a homogeneous differential equation because the right-hand side can be expressed as a function of $\frac{y}{x}$ alone. Such equations are solved using the substitution $y=ux$, which transforms the equation into a separable form.

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Solution

Step 1: Identify the Type of Equation

The given differential equation is:

$\frac{dy}{dx}=\frac{y-x}{x+y}$

This is a homogeneous differential equation since both numerator and denominator are homogeneous functions of degree 1.

Step 2: Apply the Substitution

Put $y=ux$, where $u$ is a function of $x$.

Differentiating with respect to $x$ using the product rule:

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

Step 3: Transform the Equation

Substituting $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u$ into equation (1):

$x\frac{du}{dx}+u=\frac{ux-x}{x+ux}$

Factor out $x$ from numerator and denominator on the right-hand side:

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

Step 4: Separate the Variables

Rearranging to isolate $\frac{du}{dx}$:

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

Combine the terms on the right-hand side:

$x\frac{du}{dx}=\frac{u-1-u(1+u)}{1+u}=\frac{u-1-u-u^2}{1+u}=\frac{-(1+u^2)}{1+u}$

Separating the variables:

$\frac{1+u}{1+u^2}du=-\frac{dx}{x}$

Step 5: Integrate Both Sides

$\int \frac{1+u}{1+u^2}du=-\int \frac{dx}{x}$

Split the left-hand side integral:

$\int \left(\frac{1}{1+u^2}+\frac{u}{1+u^2}\right)du=-\int \frac{dx}{x}$

$\int \frac{1}{1+u^2}du+\int \frac{u}{1+u^2}du=-\int \frac{dx}{x}$

Evaluating each integral:

• $\int \frac{1}{1+u^2}du={\tan }^{-1}u$
• $\int \frac{u}{1+u^2}du=\frac{1}{2}\ln (1+u^2)$ (using the form $\frac{1}{2}\int \frac{2u}{1+u^2}du$)
• $\int \frac{dx}{x}=\ln |x|$

Therefore:

${\tan }^{-1}u+\frac{1}{2}\ln (1+u^2)=-\ln x+c$

Rearranging:

${\tan }^{-1}u+\frac{1}{2}\ln (1+u^2)+\ln x=c$

Step 6: Back-Substitute $u=\frac{y}{x}$

Since $y=ux$, we have $u=\frac{y}{x}$. Substituting into equation (2):

${\tan }^{-1}\left(\frac{y}{x}\right)+\frac{1}{2}\ln \left(1+\frac{y^2}{x^2}\right)+\ln x=c$

Simplifying the logarithmic term:

${\tan }^{-1}\left(\frac{y}{x}\right)+\frac{1}{2}\ln \left(\frac{x^2+y^2}{x^2}\right)+\ln x=c$

Note: This can be further simplified to ${\tan }^{-1}\left(\frac{y}{x}\right)+\frac{1}{2}\ln (x^2+y^2)=c$ since $\frac{1}{2}\ln \left(\frac{x^2+y^2}{x^2}\right)+\ln x=\frac{1}{2}\ln (x^2+y^2)-\ln x+\ln x$.

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Key Formulas or Methods Used

• Homogeneous substitution: $y=ux⟹\frac{dy}{dx}=u+x\frac{du}{dx}$
• Separation of variables: Rearranging to form $f(u)du=g(x)dx$
• Standard integral: $\int \frac{1}{1+u^2}du={\tan }^{-1}u$
• Logarithmic integral: $\int \frac{f^{\prime }(u)}{f(u)}du=\ln |f(u)|$ (applied to $\int \frac{u}{1+u^2}du$)

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Summary of Steps

1. Identify the equation as homogeneous since $\frac{y-x}{x+y}$ can be written as a function of $\frac{y}{x}$
2. Substitute $y=ux$ and $\frac{dy}{dx}=u+x\frac{du}{dx}$ into the original equation
3. Simplify to obtain $x\frac{du}{dx}=\frac{-(1+u^2)}{1+u}$
4. Separate variables to get $\frac{1+u}{1+u^2}du=-\frac{dx}{x}$
5. Integrate both sides to obtain ${\tan }^{-1}u+\frac{1}{2}\ln (1+u^2)+\ln x=c$
6. Substitute back $u=\frac{y}{x}$ to get the final solution in terms of $x$ and $y$