Question Statement

Solve the differential equation:

$ydx-(y-x)dy=0$

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Background and Explanation

This is a homogeneous first-order differential equation, identifiable because it can be expressed in the form $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$. Such equations are solved using the substitution $y=ux$, which transforms the equation into a separable form that can be integrated directly.

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Solution

Step 1: Express in standard derivative form

Starting with the given equation: $ydx-(y-x)dy=0$

Rearrange to isolate $\frac{dy}{dx}$: $ydx=(y-x)dy$

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

Step 2: Verify homogeneity and substitute

Equation (1) is homogeneous since the right-hand side can be written as $\frac{(y/x)}{(y/x)-1}$, which is purely a function of $\frac{y}{x}$.

Apply the standard substitution: $y=ux$

Differentiate with respect to $x$ using the product rule: $\frac{dy}{dx}=x\frac{du}{dx}+u$

Step 3: Transform the differential equation

Substitute into equation (1): $x\frac{du}{dx}+u=\frac{ux}{ux-x}=\frac{ux}{x(u-1)}=\frac{u}{u-1}$

Isolate the derivative term: $x\frac{du}{dx}=\frac{u}{u-1}-u=\frac{u-u(u-1)}{u-1}=\frac{u-u^2+u}{u-1}=\frac{2u-u^2}{u-1}$

Factor the numerator: $x\frac{du}{dx}=\frac{-(u^2-2u)}{u-1}$

Step 4: Separate variables

Rearrange to separate $u$ and $x$ terms: $\frac{u-1}{u^2-2u}du=-\frac{dx}{x}$

Step 5: Integrate both sides

Notice that the numerator $(u-1)$ is related to the derivative of the denominator $(u^2-2u)$, since $d(u^2-2u)=(2u-2)du=2(u-1)du$.

Therefore: $\int \frac{u-1}{u^2-2u}du=\frac{1}{2}\int \frac{2u-2}{u^2-2u}du=\frac{1}{2}\ln |u^2-2u|$

Integrating both sides: $\frac{1}{2}\ln (u^2-2u)=-\ln |x|+\ln c$

(Here $\ln c$ represents the arbitrary constant of integration, and we assume $u^2-2u>0$ for the logarithm.)

Step 6: Solve for the algebraic relationship

Multiply both sides by 2: $\ln (u^2-2u)=-2\ln |x|+2\ln c=\ln \left(\frac{c^2}{x^2}\right)$

Exponentiate both sides: $u^2-2u=\frac{c^2}{x^2}$

Alternatively, working with the form given in the derivation: $\frac{1}{2}\ln (u^2-2u)=\ln \left(\frac{c}{x}\right)$

$\Rightarrow \ln \sqrt{u^2-2u}=\ln \left(\frac{c}{x}\right)$

$\Rightarrow \sqrt{u^2-2u}=\frac{c}{x}$

Step 7: Back-substitute $u=\frac{y}{x}$

Recall that $u=\frac{y}{x}$. Substitute into equation (2):

$\sqrt{{\left(\frac{y}{x}\right)}^2-2\left(\frac{y}{x}\right)}=\frac{c}{x}$

$\sqrt{\frac{y^2}{x^2}-\frac{2y}{x}}=\frac{c}{x}$

Combine terms under the radical: $\sqrt{\frac{y^2-2xy}{x^2}}=\frac{c}{x}$

$\frac{\sqrt{y^2-2xy}}{|x|}=\frac{c}{x}$

Assuming $x>0$ (or absorbing the sign into the arbitrary constant $c$): $\sqrt{y^2-2xy}=c$

This can also be written as: $\sqrt{y(y-2x)}=c$

Or, squaring both sides: y^2 - 2xy = C \quad \text{(where C = c^2 is an arbitrary positive constant)}

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Key Formulas or Methods Used

• Homogeneous DE Test: $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$
• Substitution for Homogeneous Equations: $y=ux\Rightarrow \frac{dy}{dx}=u+x\frac{du}{dx}$
• Separation of Variables: $\frac{u-1}{u^2-2u}du=-\frac{dx}{x}$
• Logarithmic Integration: $\int \frac{f^{\prime }(x)}{f(x)}dx=\ln |f(x)|+C$
• Algebraic Manipulation: Recognizing that $2(u-1)$ is the derivative of $u^2-2u$ to facilitate integration

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Summary of Steps

1. Rearrange the given equation to standard form: $\frac{dy}{dx}=\frac{y}{y-x}$
2. Identify as homogeneous and substitute $y=ux$ with $\frac{dy}{dx}=u+x\frac{du}{dx}$
3. Simplify to obtain: $x\frac{du}{dx}=\frac{2u-u^2}{u-1}$
4. Separate variables: $\frac{u-1}{u^2-2u}du=-\frac{dx}{x}$
5. Integrate both sides using the logarithmic rule to get: $\frac{1}{2}\ln (u^2-2u)=-\ln x+\ln c$
6. Exponentiate to eliminate logarithms: $\sqrt{u^2-2u}=\frac{c}{x}$
7. Substitute back $u=\frac{y}{x}$ and simplify to obtain the general solution: $\sqrt{y(y-2x)}=c$ or $y^2-2xy=C$