Question

Show that the differential equation \(y^{3} d y+\left(x+y^{2}\right) d x=0\) can be reduced to a homogeneous equation. Hence, solve it.

Short answer

Question: Show that the given differential equation can be reduced to a homogeneous equation and find its solution: $$y^3 dy + (x+y^2) dx = 0.$$ Answer: After rearranging the equation and making the substitution \(v = y^2\), we obtain a homogeneous differential equation: $$\frac{dv}{dx} = -\frac{2(x+v)}{v^2}.$$ Integrating both sides, we find the general solution: $$\frac{1}{3}y^6 = -x^2 - xy^2 + C.$$

Step-by-step solution

Write Down the Given Equation

The given differential equation is $$y^3 dy + (x+y^2) dx = 0.$$

Rearrange the Equation to Calculate \(\frac{dy}{dx}\)

We can write the equation in the form $$\frac{dy}{dx} = - \frac{x + y^2}{y^3}.$$

Substituting \(v = y^2\) and Calculating \(\frac{dy}{dx}\)

We'll make a substitution to reduce the equation to a homogeneous equation. Let \(v = y^2\). Then, \(\frac{dv}{dy} = 2y\). Now, we can find \(\frac{dy}{dx}\) using the chain rule: $$\frac{dy}{dx} = \frac{dy}{dv} \frac{dv}{dx} \Rightarrow \frac{dv}{dx} = \frac{dy}{dx} \cdot \frac{1}{\frac{dy}{dv}}.$$ We know that \(\frac{dy}{dv} = \frac{1}{2y} = \frac{1}{2\sqrt{v}}\), and \(\frac{dy}{dx} = - \frac{x + y^2}{y^3} = -\frac{x + v}{v\sqrt{v}}\). So, we have: $$\frac{dv}{dx} = - \frac{x + v}{v\sqrt{v}} \cdot 2\sqrt{v} = -\frac{2(x+v)}{v^2}.$$

Separating Variables and Integrating

Now, we have a first-order differential equation in the form of \(\frac{dv}{dx} = -\frac{2(x+v)}{v^2}\). We'll now separate variables and integrate both sides to find the general solution: $$\int v^2 dv = -2 \int (x + v) dx.$$ Integrating both sides, we get: $$\frac{1}{3}v^3 = -2\left(\frac{1}{2}x^2+xv\right) + C.$$

Substitute Back and Obtain the Solution

Finally, we substitute back \(y^2\) for \(v\) and obtain the general solution: $$\frac{1}{3}y^6 = -x^2 - xy^2 + C.$$ This is the solution to the given differential equation.