Question Statement

Solve the differential equation:

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

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Background and Explanation

This is a homogeneous differential equation because both the numerator and denominator are homogeneous functions of the same degree (degree 1). Such equations can be solved using the substitution $y=ux$, which transforms the equation into a separable form.

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Solution

Step 1: Substitution for Homogeneous Equation

Given:

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

Since this is a homogeneous differential equation, we use the substitution:

$y=ux$

Differentiating both sides with respect to $x$ using the product rule:

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

Step 2: Transform and Simplify

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

$x\frac{du}{dx}+u=\frac{x+3ux}{3x+ux}$

Factor out $x$ from the right-hand side:

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

Isolate $x\frac{du}{dx}$:

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

Simplify the numerator:

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

Step 3: Separate Variables

Rearranging to separate variables $u$ and $x$:

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

Factor the denominator $1-u^2=(1+u)(1-u)$:

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

Step 4: Partial Fraction Decomposition

We decompose the left side:

$\frac{3+u}{(1+u)(1-u)}=\frac{A}{1+u}+\frac{B}{1-u}$

Multiplying both sides by $(1+u)(1-u)$:

$3+u=A(1-u)+B(1+u)$

Finding $A$: Set $u=-1$ (making $1+u=0$): $3+(-1)=A(1-(-1))+B(0)$ $2=2A\Rightarrow A=1$

Finding $B$: Set $u=1$ (making $1-u=0$): $3+1=A(0)+B(1+1)$ $4=2B\Rightarrow B=2$

Substituting back into equation (3):

$\frac{3+u}{(1+u)(1-u)}=\frac{1}{1+u}+\frac{2}{1-u}$

Step 5: Integration

Equation (2) becomes:

$\int \left(\frac{1}{1+u}+\frac{2}{1-u}\right)du=\int \frac{1}{x}dx$

Integrating term by term (note: $\int \frac{2}{1-u}du=-2\ln |1-u|$):

$\ln |1+u|-2\ln |1-u|=\ln |x|+\ln |c|$

Combine logarithms on both sides:

$\ln \left|\frac{1+u}{(1-u{)}^2}\right|=\ln |cx|$

Removing logarithms:

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

Step 6: Back-Substitution

Substitute $u=\frac{y}{x}$ back into the equation:

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

Simplify the complex fraction by multiplying numerator and denominator by $x$ and $x^2$ respectively:

$\frac{\frac{x+y}{x}}{{\left(\frac{x-y}{x}\right)}^2}=cx$

$\frac{x+y}{x}\cdot \frac{x^2}{(x-y{)}^2}=cx$

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

Divide both sides by $x$ (assuming $x\ne 0$):

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

Or equivalently:

$(x+y)=c(x-y{)}^2$

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Key Formulas or Methods Used

• Homogeneous Equation Substitution: $y=ux⟹\frac{dy}{dx}=u+x\frac{du}{dx}$
• Partial Fraction Decomposition: $\frac{3+u}{(1+u)(1-u)}=\frac{1}{1+u}+\frac{2}{1-u}$
• Logarithmic Integration: $\int \frac{1}{ax+b}dx=\frac{1}{a}\ln |ax+b|+C$
• Logarithm Properties: $\ln a-\ln b=\ln \left(\frac{a}{b}\right)$ and $k\ln a=\ln (a^k)$

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Summary of Steps

1. Identify the equation as homogeneous (degree 1 in numerator and denominator).
2. Substitute $y=ux$ and $\frac{dy}{dx}=u+x\frac{du}{dx}$.
3. Simplify to obtain $x\frac{du}{dx}=\frac{1-u^2}{3+u}$.
4. Separate variables: $\frac{3+u}{1-u^2}du=\frac{1}{x}dx$.
5. Decompose using partial fractions: $\frac{1}{1+u}+\frac{2}{1-u}$.
6. Integrate both sides to get $\ln |1+u|-2\ln |1-u|=\ln |cx|$.
7. Simplify logarithmically to $\frac{1+u}{(1-u{)}^2}=cx$.
8. Substitute back $u=\frac{y}{x}$ and simplify to final form $\frac{x+y}{(x-y{)}^2}=c$.