Question Statement

Solve the differential equation:

$\frac{dy}{dx}=\frac{y}{x+\sqrt{xy}}$

---

Background and Explanation

This is a homogeneous differential equation, where $\frac{dy}{dx}$ can be expressed as a function of $\frac{y}{x}$ alone. For such equations, we use the substitution $y=ux$ (where $u$ is a function of $x$) to transform it into a separable equation.

---

Solution

Step 1: Identify the equation type

Given: $\frac{dy}{dx}=\frac{y}{x+\sqrt{xy}}$

This is a homogeneous differential equation because both numerator and denominator are homogeneous functions of the same degree.

Step 2: Apply the substitution

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

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

Step 3: Substitute into the original equation

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

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

Simplify the denominator on the right side: $x\frac{du}{dx}+u=\frac{ux}{x+\sqrt{u{x}^2}}=\frac{ux}{x(1+\sqrt{u})}=\frac{u}{1+\sqrt{u}}$

Step 4: Separate variables

Rearranging to isolate $x\frac{du}{dx}$: $x\frac{du}{dx}=\frac{u}{1+\sqrt{u}}-u$

Combine the terms on the right side with a common denominator: $x\frac{du}{dx}=\frac{u-u(1+\sqrt{u})}{1+\sqrt{u}}=\frac{u-u-u\sqrt{u}}{1+\sqrt{u}}=\frac{-u^{3/2}}{1+\sqrt{u}}$

Separating variables: $\frac{1+\sqrt{u}}{u^{3/2}}du=-\frac{dx}{x}$

Step 5: Simplify and integrate

Split the left side: $\left(\frac{1}{u^{3/2}}+\frac{u^{1/2}}{u^{3/2}}\right)du=-\frac{dx}{x}$

Simplify exponents: $\left(u^{-3/2}+\frac{1}{u}\right)du=-\frac{dx}{x}$

Integrate both sides: $\int u^{-3/2}du+\int \frac{1}{u}du=-\int \frac{dx}{x}$

Compute the integrals: $\frac{u^{-1/2}}{-1/2}+\ln |u|=-\ln |x|+c$

Simplify: $-2u^{-1/2}+\ln |u|=-\ln |x|+c$

Rearrange: $-2\frac{1}{\sqrt{u}}+\ln |u|+\ln |x|=c$

Combine logarithms: $-2\frac{1}{\sqrt{u}}+\ln |ux|=c$

Step 6: Back-substitute

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

$-2\frac{1}{\sqrt{\frac{y}{x}}}+\ln \left|\frac{y}{x}\cdot x\right|=c$

Simplify: $-2\sqrt{\frac{x}{y}}+\ln |y|=c$

Therefore, the general solution is: $\ln y-2\sqrt{\frac{x}{y}}=c$

Or equivalently: $\ln y=2\sqrt{\frac{x}{y}}+c$

---

Key Formulas or Methods Used

• Homogeneous substitution: $y=ux\Rightarrow \frac{dy}{dx}=x\frac{du}{dx}+u$
• Power rule for integration: $\int u^{n}du=\frac{u^{n+1}}{n+1}$ for $n\ne -1$
• Logarithmic integration: $\int \frac{1}{u}du=\ln |u|$
• Logarithm properties: $\ln a+\ln b=\ln (ab)$ and $\frac{1}{\sqrt{a/b}}=\sqrt{b/a}$

---

Summary of Steps

1. Identify the equation as homogeneous and substitute $y=ux$
2. Differentiate to get $\frac{dy}{dx}=x\frac{du}{dx}+u$
3. Substitute into the original equation and simplify the algebraic expression
4. Separate variables to get $\frac{1+\sqrt{u}}{u^{3/2}}du=-\frac{dx}{x}$
5. Split the left side into $u^{-3/2}+u^{-1}$ and integrate both sides
6. Simplify the integrated result to $-2u^{-1/2}+\ln |ux|=c$
7. Back-substitute $u=\frac{y}{x}$ to obtain the final solution $\ln y-2\sqrt{\frac{x}{y}}=c$