Question Statement

Solve the differential equation:

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

subject to the initial condition $y(1)=2$.

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Background and Explanation

This is a homogeneous differential equation, where $\frac{dy}{dx}$ can be expressed as a function of the ratio $\frac{y}{x}$. Such equations are typically solved using the substitution $y=ux$, which transforms the equation into a separable form.

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Solution

First, rewrite the differential equation in standard form by dividing both sides by $x{y}^2$:

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

Notice that the right-hand side can be written as a function of $\frac{y}{x}$ alone, confirming this is a homogeneous equation.

Step 1: Substitution

Let $y=ux$, where $u$ is a function of $x$. Differentiating with respect to $x$ using the product rule:

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

Step 2: Transform the equation

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

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

Simplify the right-hand side by factoring out $x^3$ in the numerator and calculating the denominator ($x\cdot u^2{x}^2=u^2{x}^3$):

$x\frac{du}{dx}+u=\frac{x^3(u^3-1)}{x^3{u}^2}=\frac{u^3-1}{u^2}$

Step 3: Separate variables

Subtract $u$ from both sides and combine terms:

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

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

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

Step 4: Integrate both sides

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

$\frac{u^3}{3}=-\ln |x|+c$

Rearranging to group the constant:

$\frac{u^3}{3}+\ln |x|=c$

Step 5: Back-substitute $u=\frac{y}{x}$

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

$\frac{1}{3}{\left(\frac{y}{x}\right)}^3+\ln |x|=c$

$\frac{y^3}{3x^3}+\ln |x|=c$

Step 6: Apply the initial condition

Given $y(1)=2$, substitute $x=1$ and $y=2$ into equation (3):

$\frac{2^3}{3(1{)}^3}+\ln (1)=c$

$\frac{8}{3}+0=c$

$c=\frac{8}{3}$

Step 7: Final solution

Substitute $c=\frac{8}{3}$ back into equation (3):

$\frac{y^3}{3x^3}+\ln |x|=\frac{8}{3}$

Multiply through by 3 to clear the fraction:

$\frac{y^3}{x^3}+3\ln |x|=8$

This can also be written as:

$y^3=x^3(8-3\ln |x|)$

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Key Formulas or Methods Used

• Homogeneous equation identification: $\frac{dy}{dx}=f\left(\frac{y}{x}\right)$
• Substitution for homogeneous equations: $y=ux⟹\frac{dy}{dx}=u+x\frac{du}{dx}$
• Variable separation: Rearranging to form $f(u)du=g(x)dx$
• Power rule integration: $\int u^{n}du=\frac{u^{n+1}}{n+1}$ for $n\ne -1$
• Logarithmic integration: $\int \frac{1}{x}dx=\ln |x|+C$

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Summary of Steps

1. Rewrite as $\frac{dy}{dx}=\frac{y^3-x^3}{x{y}^2}$ and verify homogeneity
2. Substitute $y=ux$ and $\frac{dy}{dx}=u+x\frac{du}{dx}$
3. Simplify to obtain $x\frac{du}{dx}=-\frac{1}{u^2}$
4. Separate variables: $u^{2}du=-\frac{dx}{x}$
5. Integrate to get $\frac{u^3}{3}+\ln |x|=c$
6. Substitute back $u=\frac{y}{x}$ yielding $\frac{y^3}{3x^3}+\ln |x|=c$
7. Apply $y(1)=2$ to find $c=\frac{8}{3}$
8. Present final answer: $\frac{y^3}{x^3}+3\ln |x|=8$ or $y^3=x^3(8-3\ln |x|)$