Question Statement

Solve the differential equation:

$\frac{dy}{dx}=\frac{3x^3+y^3}{x{y}^2}$

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Background and Explanation

This is a homogeneous differential equation, where the function $f(x,y)$ satisfies $f(tx,ty)=f(x,y)$. Such equations are solved using the substitution $y=ux$ (or $y=vx$), which transforms the equation into a separable form in terms of $u$ and $x$.

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Solution

Step 1: Identify the equation type

Given: $\frac{dy}{dx}=\frac{3x^3+y^3}{x{y}^2}$

This is a homogeneous differential equation because every term in the numerator and denominator has the same degree (degree 3).

Step 2: Apply the substitution

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

Differentiating with respect to $x$ using the product rule: $\frac{dy}{dx}=\frac{du}{dx}\cdot x+u\cdot 1=x\frac{du}{dx}+u$

Step 3: Transform the equation

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

$x\frac{du}{dx}+u=\frac{3x^3+(ux{)}^3}{x\cdot (ux{)}^2}$

Simplify the right-hand side: $x\frac{du}{dx}+u=\frac{3x^3+u^3{x}^3}{x\cdot u^2{x}^2}=\frac{x^3(3+u^3)}{u^2{x}^3}$

Cancel $x^3$ from numerator and denominator: $x\frac{du}{dx}+u=\frac{3+u^3}{u^2}$

Step 4: Separate variables

Isolate $x\frac{du}{dx}$: $x\frac{du}{dx}=\frac{3+u^3}{u^2}-u$

Combine the terms on the right-hand side with common denominator $u^2$: $x\frac{du}{dx}=\frac{3+u^3-u^3}{u^2}=\frac{3}{u^2}$

Rearrange to separate variables: $u^{2}du=\frac{3dx}{x}$

Step 5: Integrate both sides

$\int u^{2}du=3\int \frac{1}{x}dx$

Perform the integration: $\frac{u^3}{3}=3\ln |x|+c_1$

Multiply both sides by 3: $u^3=9\ln |x|+3c_1$

Let $c=3c_1$ (combining constants): $u^3=9\ln |x|+c$

Step 6: Back-substitute to original variables

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

${\left(\frac{y}{x}\right)}^3=9\ln |x|+c$

Therefore, the general solution is: $\frac{y^3}{x^3}=9\ln |x|+c$

Or equivalently: $y^3=x^3(9\ln |x|+c)$

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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: $y=ux⟹\frac{dy}{dx}=x\frac{du}{dx}+u$
• Power rule for integration: $\int u^{n}du=\frac{u^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Logarithmic integration: $\int \frac{1}{x}dx=\ln |x|+C$
• Separation of variables: Rearranging to form $f(u)du=g(x)dx$

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Summary of Steps

1. Verify homogeneity: Confirm that $\frac{dy}{dx}=\frac{3x^3+y^3}{x{y}^2}$ is homogeneous (all terms degree 3).
2. Substitute: Let $y=ux$, so $\frac{dy}{dx}=x\frac{du}{dx}+u$.
3. Simplify: Substitute into the original equation and cancel powers of $x$ to get $x\frac{du}{dx}+u=\frac{3+u^3}{u^2}$.
4. Isolate: Rearrange to $x\frac{du}{dx}=\frac{3}{u^2}$.
5. Separate variables: Write as $u^{2}du=\frac{3}{x}dx$.
6. Integrate: Obtain $\frac{u^3}{3}=3\ln |x|+c_1$, which simplifies to $u^3=9\ln |x|+c$.
7. Back-substitute: Replace $u=\frac{y}{x}$ to get the final solution $\frac{y^3}{x^3}=9\ln |x|+c$.