Variable Separable & Homogeneous Equations

When solving differential equations, we often need to transform complex equations into simpler forms. One powerful approach involves substitution methods.

For homogeneous equations like $x^2y' = 4x^2 + 7xy + 2y^2$, we can use the substitution y = vx (where v is a new variable). This transforms our equation by replacing y with vx and dy with vdx + xdv. Through careful algebraic manipulation, we separate the variables v and x.

After separating, we can integrate both sides of the equation. The solution often takes the form of logarithmic expressions that can be simplified using properties of logarithms. Eventually, we substitute back to find the relationship between x and y.

💡 When you see terms with similar degrees (like x²y' and x²), try the substitution y = vx. This often turns messy equations into manageable ones!

Separating Variables: Step-by-Step Examples

Separable differential equations let us move all terms with one variable to one side, and all terms with the other variable to the other side. Let's see how this works!

For equations like $3x^2cot y dx - (x^3+1)csc^2y dy = 0$, start by rearranging to get variables separated: $\frac{3x^2}{x^3+1}dx = \frac{csc^2y}{coty}dy$. Then integrate both sides and simplify using substitution techniques when helpful.

The second example, $\frac{dy}{dx} = \frac{9+y^2}{3\sqrt{4-x^2}}$, follows a similar approach. Rearranging gives $\frac{1}{9+y^2}dy = \frac{1}{3\sqrt{4-x^2}}dx$. When integrating, recognize standard forms like $\int\frac{1}{a^2+y^2}dy = \frac{1}{a}arctan\frac{y}{a}$ and $\int\frac{1}{\sqrt{a^2-x^2}}dx = arcsin\frac{x}{a}$.

⚡ Look for patterns in integrals! Expressions like $\frac{1}{a^2+x^2}$ and $\frac{1}{\sqrt{a^2-x^2}}$ appear frequently and have standard solutions.

More Separable Equation Examples

Tackling complex differential equations requires patience and strategic substitutions. Let's continue with more challenging examples.

In the equation $x(8+y^3)dx + 3y^2\sqrt{4-x^2}dy = 0$, our first step is rearranging to separate variables: $\frac{x}{\sqrt{4-x^2}}dx + \frac{3y^2}{(8+y^3)}dy = 0$. We then integrate each term separately. For integrals like $\int\frac{x}{\sqrt{4-x^2}}dx$, substituting $u = 4-x^2$ transforms it into a simpler form.

The equation $e^y(5-x)dx = xydy$ requires similar separation techniques. After dividing by $e^yx$, we get $\frac{5-x}{x}dx = \frac{y}{e^y}dy$. When integrating $\frac{y}{e^y}dy$, integration by parts helps us handle the product of y and $e^{-y}$.

After solving both sides, we combine the results and solve for the relationship between x and y, often yielding implicit solutions that can't be directly expressed as y = f(x).

🔍 When integrating rational expressions like $\frac{1}{x^2+x}$, try partial fraction decomposition to break it into simpler terms.

Solving Differential Equations with Partial Fractions

Sometimes differential equations require specific algebraic techniques to solve efficiently. Let's examine a classic case.

For the equation $\frac{dx}{x^2+x} + \frac{dy}{y^2+y} = 0$, notice both terms have similar forms. We can rewrite them using partial fraction decomposition: $\frac{1}{x^2+x} = \frac{1}{x(x+1)}$ and similarly for the y-term.

After decomposing, we integrate both sides: $\int \frac{1}{x(x+1)} dx + \int \frac{1}{y(y+1)} dy = C$. These integrate to logarithmic expressions: $-\ln(x+1) + \ln(x) - \ln(y+1) + \ln(y) = C$.

Using logarithm properties, we can simplify to: $\ln(\frac{x}{x+1} \cdot \frac{y}{y+1}) = C$. After exponentiation and algebraic manipulation, we arrive at the elegant solution $(x+1)(y+1) = -C_1$, which we can solve for y.

💫 Many differential equations yield solutions with a form of $(x+a)(y+b) = C$. This pattern appears frequently, so learn to recognize it!

Homogeneous Differential Equations

Homogeneous differential equations have a special property: when you replace x with tx and y with ty, the equation remains the same except for a power of t. The substitution y = vx $or x = vy$ is our key strategy here.

For $ydx + (2x-y)dy = 0$, we substitute x = vy, which means dx = vdy + ydv. After substitution and rearranging, we get $\frac{1}{3v-1}dv + \frac{1}{y}dy = 0$.

Integrating both sides: $\frac{\ln(3v-1)}{3} + \ln(y) = C$. We simplify using logarithm properties, eventually getting $(3v-1)(y^3) = C_1$. Substituting back v = x/y gives us the final solution $3xy^2-y^3 = C_1$.

The beauty of this method is how it transforms a complex equation into a separable one, making it much easier to solve step by step.

🌟 To check if an equation is homogeneous, see if all terms have the same degree when counting x and y together. For example, xy and x² + y² are both degree 2.

More Homogeneous Equations

Homogeneous equations can look intimidating at first, but our substitution method makes them manageable. Let's tackle a few more examples.

For $\frac{dy}{dx} = \frac{y}{x} + e^{y/x}$, we use the substitution y = vx, which gives dy = vdx + xdv. After substituting and simplifying, we get $\frac{dv}{e^v} = \frac{dx}{x}$. Integrating both sides yields $-e^{-v} = \ln(x) + c$. Substituting back v = y/x gives us $e^{-e^{-y/x}} = x + C_1$.

For $y^2 dx = (xy - x^2) dy$, we use x = vy, leading to $\frac{dv}{v^2} = \frac{-dy}{y}$. After integration and substitution, our solution becomes $\frac{y}{x} = \ln(y) + c$.

These solutions may look complex, but they follow from the systematic application of our substitution method and careful integration. The pattern becomes clearer with practice.

🔮 When solving homogeneous equations, after substituting y = vx, watch for terms that can be factored out, making the separation of variables possible.

Tackling Complex Homogeneous Equations

Even the most complicated-looking homogeneous equations yield to our substitution approach. Let's see how to handle one with square roots.

For $ydx - (x - \sqrt{x^2 + y^2})dy = 0$, we substitute x = vy, transforming it to $y^2dv + y\sqrt{v^2 + 1}dy = 0$. This rearranges to $\frac{dv}{\sqrt{v^2 + 1}} + \frac{dy}{y} = 0$.

The integral $\int \frac{dv}{\sqrt{v^2 + 1}}$ requires a trigonometric substitution. Setting v = tan(u) transforms it into $\int sec(u)du$, which equals $\ln(\sqrt{v^2 + 1} + v)$. Combined with $\int \frac{dy}{y} = \ln(y)$, we get $\ln((\sqrt{v^2 + 1} + v)(y)) = C$.

After substituting back v = x/y and simplifying, our final solution is $y\sqrt{\frac{x^2}{y^2} + 1} + x = C_1$, which can be further simplified to $\sqrt{x^2 + y^2} + x = C_1$.

📐 Trigonometric substitutions are powerful for integrals containing $\sqrt{a^2 + x^2}$, $\sqrt{a^2 - x^2}$, or $\sqrt{x^2 - a^2}$. Remember that v = tan(u) works well for $\sqrt{v^2 + 1}$.

Final Example: Homogeneous Equation with Quadratics

Let's conclude with a differential equation that combines several techniques we've learned.

For $(4x^2-y^2)dx - xydy = 0$, we use the substitution y = vx, giving dy = vdx + xdv. After substituting and collecting like terms, we get $2x^2(2-v^2)dx - x^3vdv = 0$.

Rearranging to separate variables: $\frac{2}{x}dx - \frac{v}{(2-v^2)}dv = 0$. We integrate $\int\frac{2}{x}dx = 2\ln(x)$ and for the other term, we use substitution u = 2-v² to get $-\ln|2-v^2| = -\ln(\frac{2-v^2}{2})$.

Combining results and using logarithm properties leads to $(x^4)(\frac{y^2}{x^2}-2) = C_1$, which simplifies to $x^2y^2 - 2x^4 = C_1$.

Solving for y gives us the explicit solution $y = \sqrt{\frac{C_1}{x^2}+2}$, showing how our homogeneous equation approach leads to a clear final answer.

🧩 When the final form involves y², you can often get an explicit solution by taking the square root of both sides. Just remember to consider both positive and negative solutions if needed.