Key Concepts and Formulas

• Homogeneous Differential Equations: A differential equation of the form $\frac{dy}{dx} = f(x,y)$ is homogeneous if $f(x, y)$ can be written as a function of $\frac{y}{x}$, i.e., $f(x, y) = g(\frac{y}{x})$. The substitution $y = vx$ transforms it into a separable equation.
• Exact Differential Equations: An equation of the form $M(x, y) dx + N(x, y) dy = 0$ is exact if $\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x}$. The solution is given by $\int M dx + \int (\text{terms in } N \text{ not containing } x) dy = C$.
• Integrating Factors: If a differential equation $M(x, y) dx + N(x, y) dy = 0$ is not exact, we can sometimes find an integrating factor $\mu(x, y)$ such that $\mu(x, y) M(x, y) dx + \mu(x, y) N(x, y) dy = 0$ is exact.

Step-by-Step Solution

Step 1: Rewrite the given differential equation

We are given the differential equation $(x - y^2) dx + y(5x + y^2) dy = 0$ Rewrite it as: $(x - y^2) dx + (5xy + y^3) dy = 0$

Step 2: Check for exactness

Let $M(x, y) = x - y^2$ and $N(x, y) = 5xy + y^3$. Then, $\frac{\partial M}{\partial y} = -2y$ and $\frac{\partial N}{\partial x} = 5y$. Since $\frac{\partial M}{\partial y} \neq \frac{\partial N}{\partial x}$, the equation is not exact.

Step 3: Attempt a substitution to simplify

Notice the presence of $y^2$ terms. Let $y^2 = v$. Then, $2y \, dy = dv$, so $y \, dy = \frac{1}{2} dv$. Substituting into the original equation gives: $(x - v) dx + (5x + v) \frac{1}{2} dv = 0$ $2(x - v) dx + (5x + v) dv = 0$ $2x \, dx - 2v \, dx + 5x \, dv + v \, dv = 0$ $2x \, dx + v \, dv + 5x \, dv - 2v \, dx = 0$ $2x \, dx + v \, dv + (5x - 2v) dv = 0$ $2x \, dx + v \, dv = (2v - 5x) dv$ This form doesn't immediately suggest a simple solution.

Step 4: Rearrange the original equation to a suitable form

Rewrite the original equation as: $\frac{dx}{dy} = -\frac{y(5x + y^2)}{x - y^2}$ $\frac{dx}{dy} = \frac{5xy + y^3}{y^2 - x}$

Step 5: Try a different approach: look for a homogeneous form after rearrangement Let's rearrange the original equation as follows: $(x-y^2)dx + (5xy+y^3)dy = 0$ $(5xy+y^3)dy = (y^2 - x)dx$ $\frac{dy}{dx} = \frac{y^2 - x}{5xy + y^3}$ This form doesn't appear to be directly solvable as a homogeneous equation.

Step 6: Try to find an integrating factor dependent on x or y

Let's go back to the equation $(x-y^2)dx + (5xy+y^3)dy = 0$. We have $M = x-y^2$ and $N = 5xy+y^3$. $\frac{\partial M}{\partial y} = -2y$ and $\frac{\partial N}{\partial x} = 5y$. So, $\frac{\frac{\partial M}{\partial y} - \frac{\partial N}{\partial x}}{N} = \frac{-2y - 5y}{5xy + y^3} = \frac{-7y}{y(5x + y^2)} = \frac{-7}{5x + y^2}$. This is not a function of x only. And, $\frac{\frac{\partial N}{\partial x} - \frac{\partial M}{\partial y}}{M} = \frac{5y - (-2y)}{x - y^2} = \frac{7y}{x - y^2}$. This is not a function of y only.

Step 7: Try the substitution $x = vy^2$

Let $x = vy^2$. Then $dx = y^2 dv + 2vy dy$. Substituting this into the original equation, we get: $(vy^2 - y^2)(y^2 dv + 2vy dy) + y(5vy^2 + y^2) dy = 0$ $y^2(v - 1)(y^2 dv + 2vy dy) + y^3(5v + 1) dy = 0$ $y^4(v - 1) dv + 2vy^3(v - 1) dy + y^3(5v + 1) dy = 0$ Divide by $y^3$: $y(v - 1) dv + [2v(v - 1) + (5v + 1)] dy = 0$ $y(v - 1) dv + (2v^2 - 2v + 5v + 1) dy = 0$ $y(v - 1) dv + (2v^2 + 3v + 1) dy = 0$ $y(v - 1) dv + (2v + 1)(v + 1) dy = 0$ $\frac{dv}{-\frac{(2v+1)(v+1)}{v-1}} = \frac{dy}{y}$ $\frac{v-1}{(2v+1)(v+1)} dv = -\frac{dy}{y}$ Now, we need to decompose the fraction: $\frac{v-1}{(2v+1)(v+1)} = \frac{A}{2v+1} + \frac{B}{v+1}$ $v - 1 = A(v + 1) + B(2v + 1)$ If $v = -1$, then $-2 = -B$, so $B = 2$. If $v = -\frac{1}{2}$, then $-\frac{3}{2} = A(\frac{1}{2})$, so $A = -3$. Therefore, $\frac{v-1}{(2v+1)(v+1)} = \frac{-3}{2v+1} + \frac{2}{v+1}$ So our integral becomes $\int \left( \frac{-3}{2v+1} + \frac{2}{v+1} \right) dv = -\int \frac{dy}{y}$ $-\frac{3}{2} \ln |2v+1| + 2 \ln |v+1| = -\ln |y| + \ln C$ $2 \ln |v+1| - \frac{3}{2} \ln |2v+1| = \ln \left| \frac{C}{y} \right|$ $\ln (v+1)^2 - \ln (2v+1)^{3/2} = \ln \left| \frac{C}{y} \right|$ $\ln \left| \frac{(v+1)^2}{(2v+1)^{3/2}} \right| = \ln \left| \frac{C}{y} \right|$ $\frac{(v+1)^4}{(2v+1)^3} = \frac{C^2}{y^2}$ Let $C^2 = C_1$. Then $y^2(v+1)^4 = C_1(2v+1)^3$. Since $v = \frac{x}{y^2}$, we have $y^2\left(\frac{x}{y^2} + 1\right)^4 = C_1 \left(\frac{2x}{y^2} + 1\right)^3$ $y^2\left(\frac{x+y^2}{y^2}\right)^4 = C_1 \left(\frac{2x+y^2}{y^2}\right)^3$ $\frac{(x+y^2)^4}{y^6} = C_1 \frac{(2x+y^2)^3}{y^6}$ $(x+y^2)^4 = C_1 (2x+y^2)^3$ $(y^2 + x)^4 = C_1 |(y^2 + 2x)^3|$ $(y^2 + x)^4 = C|(y^2 + 2x)^3|$

Common Mistakes & Tips

• Don't give up easily! Differential equations can be tricky. Try different substitutions and rearrangements.
• Remember to check for exactness before applying other methods.
• When integrating, don't forget the constant of integration.

Summary

The given differential equation was not immediately solvable. By substituting $x=vy^2$, we transformed the equation into a separable form. After integrating and simplifying, we obtained the general solution $(y^2 + x)^4 = C|(y^2 + 2x)^3|$.

Final Answer The final answer is \boxed{\left(y^{2}+x\right)^{4}=\mathrm{C}\left|\left(y^{2}+2 x\right)^{3}\right|}, which corresponds to option (A).