Question Statement

Solve the differential equation:

$\frac{dy}{dx}=\frac{x^2{y}^2}{1+x}$

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Background and Explanation

This is a separable differential equation, where we can rearrange the equation to isolate all $y$ terms on one side and all $x$ terms on the other. The key challenge here is integrating the rational function $\frac{x^2}{1+x}$, which requires polynomial long division or algebraic manipulation to simplify before integration.

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Solution

Step 1: Separate the variables

We begin by dividing both sides by $y^2$ and multiplying by $dx$ to separate the variables:

$\frac{dy}{y^2}=\frac{x^2}{1+x}dx$

Step 2: Simplify the rational function

To integrate $\frac{x^2}{1+x}$, we perform polynomial division. Notice that:

$x^2=(x-1)(x+1)+1=x^2-1+1$

Therefore:

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

Step 3: Integrate both sides

Now we integrate both sides of the separated equation:

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

Breaking this into individual integrals:

$\int y^{-2}dy=\int xdx-\int 1dx+\int \frac{1}{1+x}dx$

Step 4: Evaluate the integrals

Left side: $\int y^{-2}dy=\frac{y^{-2+1}}{-2+1}=\frac{y^{-1}}{-1}=-\frac{1}{y}$

Right side: $\int xdx-\int 1dx+\int \frac{1}{1+x}dx=\frac{x^2}{2}-x+\ln (1+x)+c_1$

Note: We assume $x>-1$ so that $\ln (1+x)$ is defined. If $x<-1$, we would write $\ln |1+x|$.

Step 5: Combine and solve for $y$

Equating both sides:

$-\frac{1}{y}=\frac{x^2}{2}-x+\ln (1+x)+c_1$

Multiply both sides by $-1$:

$\frac{1}{y}=-\frac{x^2}{2}+x-\ln (1+x)-c_1$

Redefining the constant $c=-c_1$ (since an arbitrary constant remains arbitrary regardless of sign):

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

Or equivalently:

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

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Key Formulas or Methods Used

• Separation of variables: $\frac{dy}{dx}=f(x)g(y)⟹\frac{dy}{g(y)}=f(x)dx$
• Polynomial long division: $\frac{x^2}{x+1}=x-1+\frac{1}{x+1}$
• Power rule for integration: $\int x^{n}dx=\frac{x^{n+1}}{n+1}+C$ (for $n\ne -1$)
• Logarithmic integration: $\int \frac{1}{u}du=\ln |u|+C$
• Constant absorption: $-c_1$ can be rewritten as $+c$ where $c$ is an arbitrary constant

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Summary of Steps

1. Separate variables: Rearrange to get $\frac{dy}{y^2}=\frac{x^2}{1+x}dx$
2. Simplify: Use polynomial division to write $\frac{x^2}{1+x}=x-1+\frac{1}{1+x}$
3. Integrate: Compute $\int y^{-2}dy=\int (x-1+\frac{1}{1+x})dx$
4. Evaluate: Obtain $-\frac{1}{y}=\frac{x^2}{2}-x+\ln (1+x)+c_1$
5. Solve: Multiply by $-1$ and redefine constants to get $\frac{1}{y}=-\frac{x^2}{2}+x-\ln (1+x)+c$