Show that each of the following differential equations is homogeneous and solve each of them :
`2xy dx+(x^(2)+2y^(2))dy=0`

Answer

To solve the differential equation \(2xy \, dx + (x^2 + 2y^2) \, dy = 0\), we will follow these steps:

### Step 1: Rewrite the equation
We can rewrite the given differential equation in the standard form:
\[ 2xy \, dx + (x^2 + 2y^2) \, dy = 0 \]
This can be rearranged to:
\[ 2xy \, dx = - (x^2 + 2y^2) \, dy \]
Dividing both sides by \(dx\) gives:
\[ \frac{dy}{dx} = -\frac{2xy}{x^2 + 2y^2} \]

### Step 2: Check for homogeneity
To check if the equation is homogeneous, we substitute \(x = \lambda x\) and \(y = \lambda y\):
\[ \frac{dy}{dx} = -\frac{2(\lambda x)(\lambda y)}{(\lambda x)^2 + 2(\lambda y)^2} = -\frac{2\lambda^2 xy}{\lambda^2(x^2 + 2y^2)} = -\frac{2xy}{x^2 + 2y^2} \]
Since the equation remains unchanged, we conclude that the differential equation is homogeneous.

### Step 3: Substitute \(y = tx\)
Let \(y = tx\), where \(t = \frac{y}{x}\). Then, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = t + x\frac{dt}{dx} \]
Substituting \(y\) and \(\frac{dy}{dx}\) into the equation gives:
\[ t + x\frac{dt}{dx} = -\frac{2x(tx)}{x^2 + 2(tx)^2} \]
Simplifying the right side:
\[ -\frac{2tx^2}{x^2 + 2t^2x^2} = -\frac{2t}{1 + 2t^2} \]
Thus, we have:
\[ t + x\frac{dt}{dx} = -\frac{2t}{1 + 2t^2} \]

### Step 4: Rearranging the equation
Rearranging gives:
\[ x\frac{dt}{dx} = -\frac{2t}{1 + 2t^2} - t \]
Combining terms:
\[ x\frac{dt}{dx} = -\frac{2t + t(1 + 2t^2)}{1 + 2t^2} = -\frac{t(3 + 2t^2)}{1 + 2t^2} \]

### Step 5: Separate variables
Separating variables gives:
\[ \frac{1 + 2t^2}{t(3 + 2t^2)} dt = -\frac{1}{x} dx \]

### Step 6: Integrate both sides
Now we integrate both sides:
\[ \int \frac{1 + 2t^2}{t(3 + 2t^2)} dt = -\int \frac{1}{x} dx \]
The left side can be simplified using partial fractions:
\[ \frac{1 + 2t^2}{t(3 + 2t^2)} = \frac{A}{t} + \frac{B}{3 + 2t^2} \]
Solving for \(A\) and \(B\) gives:
\[ A = \frac{1}{3}, \quad B = \frac{1}{2} \]
Thus, we have:
\[ \int \left( \frac{1/3}{t} + \frac{1/2}{3 + 2t^2} \right) dt = -\ln |x| + C \]
Integrating gives:
\[ \frac{1}{3} \ln |t| + \frac{1}{2} \cdot \frac{1}{\sqrt{2}} \tan^{-1}\left(\frac{t\sqrt{2}}{\sqrt{3}}\right) = -\ln |x| + C \]

### Step 7: Substitute back for \(t\)
Substituting back \(t = \frac{y}{x}\) leads to:
\[ \frac{1}{3} \ln \left|\frac{y}{x}\right| + \frac{1}{2\sqrt{2}} \tan^{-1}\left(\frac{y\sqrt{2}}{x\sqrt{3}}\right) = -\ln |x| + C \]

### Final Solution
This gives the implicit solution of the differential equation.

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