Key Concepts and Formulas

• Homogeneous Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(tx, ty) = f(x,y)$ for any scalar $t$. Such equations can be solved using the substitution $y=vx$.
• Separable Differential Equations: A differential equation of the form $g(y)dy = h(x)dx$ can be solved by direct integration: $\int g(y) dy = \int h(x) dx + C$.

Step-by-Step Solution

Step 1: Rewriting the Differential Equation

The given differential equation is: $ydx + (x + x^2y)dy = 0$

Let's rewrite it to isolate $\frac{dy}{dx}$: $ydx = -(x + x^2y)dy$ $\frac{dy}{dx} = -\frac{y}{x + x^2y} = -\frac{y}{x(1+xy)}$ $\frac{dy}{dx} = -\frac{y}{x} \cdot \frac{1}{1+xy}$

Step 2: Recognizing a Non-Standard Form and Attempting Simplification

This form doesn't immediately fit a standard separable, linear, or Bernoulli equation. Let's take the reciprocal to analyze $\frac{dx}{dy}$: $\frac{dx}{dy} = -\frac{x + x^2y}{y} = -\frac{x}{y} - x^2$ $\frac{dx}{dy} + \frac{x}{y} = -x^2$

Step 3: Recognizing and Applying the Bernoulli Transformation

The equation $\frac{dx}{dy} + \frac{x}{y} = -x^2$ is a Bernoulli equation in the form $\frac{dx}{dy} + P(y)x = Q(y)x^n$, where $P(y) = \frac{1}{y}$, $Q(y) = -1$, and $n=2$. We use the substitution $v = x^{1-n} = x^{1-2} = x^{-1} = \frac{1}{x}$.

Then, $\frac{dv}{dy} = -x^{-2} \frac{dx}{dy}$. Multiplying the Bernoulli equation by $x^{-2}$, we get: $x^{-2} \frac{dx}{dy} + x^{-2} \frac{x}{y} = -x^{-2}x^2$ $x^{-2} \frac{dx}{dy} + \frac{1}{x y} = -1$ Since $v = \frac{1}{x}$, then $\frac{1}{xy} = \frac{v}{y}$. Also, $x^{-2} \frac{dx}{dy} = -\frac{dv}{dy}$. Substituting: $-\frac{dv}{dy} + \frac{v}{y} = -1$ $\frac{dv}{dy} - \frac{v}{y} = 1$

Step 4: Solving the Linear Differential Equation

This is a linear first-order differential equation. The integrating factor (I.F.) is: $I.F. = e^{\int -\frac{1}{y} dy} = e^{-\ln|y|} = e^{\ln|y^{-1}|} = \frac{1}{y}$

Multiplying the differential equation by the integrating factor: $\frac{1}{y} \frac{dv}{dy} - \frac{1}{y^2} v = \frac{1}{y}$ $\frac{d}{dy} \left( \frac{v}{y} \right) = \frac{1}{y}$

Integrating both sides with respect to $y$: $\int \frac{d}{dy} \left( \frac{v}{y} \right) dy = \int \frac{1}{y} dy$ $\frac{v}{y} = \ln|y| + C$

Step 5: Substituting Back and Solving for the Constant

Substitute $v = \frac{1}{x}$: $\frac{1}{xy} = \ln|y| + C$ $\ln|y| = \frac{1}{xy} - C$ $\ln|y| = \frac{1 - Cxy}{xy}$

This doesn't match the option (A) $\log y = Cx$. Let's go back to the original equation and consider a homogeneous equation substitution. The original equation is $ydx + (x + x^2y)dy = 0$ which can be written as $\frac{dy}{dx} = -\frac{y}{x+x^2y} = -\frac{y}{x(1+xy)}$.

Let's try rewriting the original equation as: $ydx + xdy + x^2ydy = 0$ $d(xy) + x^2ydy = 0$ Divide by $x^2y$: $\frac{d(xy)}{x^2y} + dy = 0$ Let $z=xy$, then $dz = d(xy)$. Thus, the equation becomes $\frac{dz}{z^2} + dy = 0$ Integrate: $\int \frac{dz}{z^2} + \int dy = 0$ $-\frac{1}{z} + y = C$ $-\frac{1}{xy} + y = C$

Still doesn't match. However, the correct answer is $\log y = Cx$, which means $y = e^{Cx}$. If $y = e^{Cx}$, then $\frac{dy}{dx} = Ce^{Cx} = Cy$. Substituting this into the original equation, we have $ydx + xdy + x^2ydy = 0$ $ydx + x(Cy dx) + x^2y (Cy dx) = 0$ $y + Cx + Cx^2y = 0$ $e^{Cx} + Cxe^{Cx} + Cx^2e^{2Cx} = 0$ $1 + Cx + Cx^2e^{Cx} = 0$. This is incorrect.

Let's examine again. $y dx + (x+x^2y)dy = 0$ $y dx + x dy + x^2 y dy = 0$ $d(xy) + x^2 y dy = 0$ Dividing by $xy$: $\frac{d(xy)}{xy} + x dy = 0$ $\frac{d(xy)}{xy} = -x dy$ Integrating is difficult.

Let $M = y$, $N = x+x^2y$, then $\frac{\partial M}{\partial y} = 1$, $\frac{\partial N}{\partial x} = 1+2xy$. It is not exact.

Let's try dividing the original equation by $x^2$. $\frac{y}{x^2} dx + (\frac{1}{x} + y)dy = 0$ $\frac{y}{x^2} dx + \frac{1}{x} dy + y dy = 0$ $d(-\frac{y}{x}) + y dy = 0$ $-\frac{y}{x} + \frac{y^2}{2} = C$ This doesn't work either.

Try $y = Cx$ $dy = C dx$ $Cx dx + (x+x^2 Cx) C dx = 0$ $Cx + Cx + C^2x^3 = 0$ $2Cx + C^2x^3 = 0$ $2 + Cx^2 = 0$, which is incorrect.

Given $\ln y = Cx$, $y = e^{Cx}$, $\frac{dy}{dx} = C e^{Cx} = Cy$ $ydx + xdy + x^2 y dy = 0$ $ydx + x(Cy dx) + x^2y(Cy dx) = 0$ $y + Cxy + Cx^2 y^2 = 0$ $e^{Cx} + Cx e^{Cx} + Cx^2 e^{2Cx} = 0$ $1 + Cx + Cx^2 e^{Cx} = 0$

If $\ln y = Cx$, $y = e^{Cx}$. Then $y dx + (x+x^2y)dy = 0$ implies $y dx + x dy + x^2 y dy = 0$ $d(xy) + x^2 y dy = 0$ $\frac{d(xy)}{xy} + x dy = 0$ Let $z = xy$. $\frac{dz}{z} + x dy = 0$ $\int \frac{dz}{z} = -\int x dy$ $\ln z = -\int x dy$ $\ln(xy) = -\int x dy$

The only way I see to obtain the answer in option A from this equation is if $x$ is a constant. Let's rewrite the original equation:

$y dx + x dy + x^2 y dy = 0$ $d(xy) + x^2 y dy = 0$

Divide by $y$: $\frac{d(xy)}{y} + x^2 dy = 0$ Integrate: $\int \frac{d(xy)}{y} + \int x^2 dy = 0$

Consider $\frac{dy}{y} + \frac{dx}{x} + x y dy = 0$, integrate each term. $\ln y + \int (\frac{1}{y} + x) dx = C$, then $\ln y = Cx$.

This is where I realize my mistake. The correct answer given is (A) which is incorrect. This question is flawed.

Common Mistakes & Tips

• Double-check the algebra, especially when manipulating the differential equation.
• Be careful when applying the integrating factor method.
• Recognize the type of differential equation correctly.

Summary

After carefully reviewing the steps and different solution methods, the provided answer (A) is incorrect. The correct answer is (B).

Final Answer

The question is incorrect. The correct option should be (B), $-\frac{1}{xy} + \log y = C$. There is no correct option.