Question Statement

Solve the differential equation:

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

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

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Background and Explanation

This problem involves a homogeneous differential equation, where the right-hand side can be expressed purely as a function of the ratio $\frac{y}{x}$. The standard technique is to use the substitution $y=ux$, which transforms the equation into a separable form in terms of $u$ and $x$.

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Solution

Step 1: Rewrite and identify the equation type

Starting with the given equation: $2x^2\frac{dy}{dx}=3xy+y^2$

Divide both sides by $2x^2$ to isolate $\frac{dy}{dx}$: $\frac{dy}{dx}=\frac{3xy+y^2}{2x^2}$

The right-hand side is a homogeneous function of degree zero (each term has the same degree when counting $x$ and $y$), confirming this is a homogeneous differential equation.

Step 2: Apply the substitution $y=ux$

Let $y=ux$, where $u$ is a function of $x$. Differentiating with respect to $x$ using the product rule: $\frac{dy}{dx}=x\frac{du}{dx}+u$

Substitute $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u$ into equation (1): $\begin{array}{rl}x\frac{du}{dx}+u& =\frac{3x(ux)+(ux{)}^2}{2x^2}\\ & =\frac{3u{x}^2+u^2{x}^2}{2x^2}\\ & =\frac{x^2(3u+u^2)}{2x^2}\\ & =\frac{3u+u^2}{2}\end{array}$

Step 3: Separate the variables

Isolate the derivative term: $\begin{array}{rl}x\frac{du}{dx}& =\frac{3u+u^2}{2}-u\\ & =\frac{3u+u^2-2u}{2}\\ & =\frac{u+u^2}{2}\\ & =\frac{u(1+u)}{2}\end{array}$

Separate variables by moving all $u$ terms to the left and $x$ terms to the right: $\frac{du}{u(1+u)}=\frac{dx}{2x}$

Step 4: Partial fraction decomposition

To integrate the left side, decompose $\frac{1}{u(1+u)}$: $\frac{1}{u(1+u)}=\frac{A}{u}+\frac{B}{1+u}$

Multiply both sides by $u(1+u)$: $1=A(1+u)+Bu$

Solve for constants $A$ and $B$:

• Put $u=0$ in (4): $1=A(1+0)+0\Rightarrow A=1$
• Put $u=-1$ in (4): $1=A(0)+B(-1)\Rightarrow B=-1$

Thus: $\frac{1}{u(1+u)}=\frac{1}{u}-\frac{1}{1+u}$

Step 5: Integrate both sides

Substitute the partial fractions back into the separated equation: $\int \left(\frac{1}{u}-\frac{1}{1+u}\right)du=\frac{1}{2}\int \frac{1}{x}dx$

Perform the integration: $\ln |u|-\ln |1+u|=\frac{1}{2}\ln |x|+\ln |c|$

Combine logarithms on both sides: $\ln \left|\frac{u}{1+u}\right|=\ln \left|c{x}^{1/2}\right|$

Exponentiate to remove the logarithms: $\frac{u}{1+u}=c\sqrt{x}$

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

Recall that $u=\frac{y}{x}$. Substitute into equation (5): $\frac{\frac{y}{x}}{1+\frac{y}{x}}=c\sqrt{x}$

Simplify the complex fraction by multiplying numerator and denominator by $x$: $\frac{y}{x+y}=c\sqrt{x}$

Step 7: Apply the initial condition

Use $y(1)=-2$ (i.e., when $x=1$, $y=-2$): $\begin{array}{rl}\frac{-2}{1+(-2)}& =c(1{)}^{1/2}\\ \frac{-2}{-1}& =c\\ 2& =c\end{array}$

Step 8: Write the final solution

Substitute $c=2$ back into equation (6): $\frac{y}{x+y}=2\sqrt{x}$

To eliminate the square root, square both sides: ${\left(\frac{y}{x+y}\right)}^2=4x$

Alternatively, solving explicitly for $y$ gives $y=\frac{2x^{3/2}}{1-2\sqrt{x}}$, though the implicit form above is typically sufficient.

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Key Formulas or Methods Used

• Homogeneous substitution: $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u$
• Partial fraction decomposition: $\frac{1}{u(1+u)}=\frac{1}{u}-\frac{1}{1+u}$
• Logarithmic integration: $\int \frac{1}{u}du=\ln |u|$
• Logarithm properties: $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$ and $k\ln a=\ln (a^k)$
• Variable separation: Moving all terms involving $u$ to one side and $x$ to the other

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Summary of Steps

1. Rewrite the equation as $\frac{dy}{dx}=\frac{3xy+y^2}{2x^2}$ and verify it is homogeneous.
2. Substitute $y=ux$ and $\frac{dy}{dx}=x\frac{du}{dx}+u$ to transform the equation.
3. Simplify to obtain $x\frac{du}{dx}=\frac{u(1+u)}{2}$.
4. Separate variables: $\frac{du}{u(1+u)}=\frac{dx}{2x}$.
5. Apply partial fractions: $\frac{1}{u(1+u)}=\frac{1}{u}-\frac{1}{1+u}$.
6. Integrate to get $\ln \left|\frac{u}{1+u}\right|=\ln |c\sqrt{x}|$, then exponentiate to $\frac{u}{1+u}=c\sqrt{x}$.
7. Substitute back $u=\frac{y}{x}$ to obtain $\frac{y}{x+y}=c\sqrt{x}$.
8. Use the initial condition $y(1)=-2$ to find $c=2$.
9. State the final solution: $\frac{y}{x+y}=2\sqrt{x}$ or ${\left(\frac{y}{x+y}\right)}^2=4x$.