The general solution of the differential equation `(x^2+xy)y'=y^2` is

Options

A. `e^(y/x)=cx`
B. `e^(-y/x)=cy`
C. `e^(-y/x)=cxy`
D. `e^((-2y)/x)=cy`

Answer

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

### Step 1: Rewrite the equation
We start by rewriting the given differential equation in a more manageable form. We can express \(y'\) as \(\frac{dy}{dx}\):

\[ (x^2 + xy) \frac{dy}{dx} = y^2 \]

### Step 2: Separate variables
Next, we can isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{y^2}{x^2 + xy} \]

### Step 3: Identify the homogeneity
Notice that the equation is homogeneous. We can use the substitution \(y = vx\), where \(v\) is a function of \(x\). Thus, we have:

\[ \frac{dy}{dx} = x \frac{dv}{dx} + v \]

Substituting \(y = vx\) into the equation gives:

\[ \frac{dy}{dx} = \frac{(vx)^2}{x^2 + x(vx)} = \frac{v^2 x^2}{x^2(1 + v)} = \frac{v^2}{1 + v} \]

### Step 4: Substitute and simplify
Now substituting \(\frac{dy}{dx}\) into our equation:

\[ x \frac{dv}{dx} + v = \frac{v^2}{1 + v} \]

Rearranging gives:

\[ x \frac{dv}{dx} = \frac{v^2}{1 + v} - v \]

### Step 5: Simplify the right-hand side
We can simplify the right-hand side:

\[ \frac{v^2}{1 + v} - v = \frac{v^2 - v(1 + v)}{1 + v} = \frac{v^2 - v - v^2}{1 + v} = \frac{-v}{1 + v} \]

Thus, we have:

\[ x \frac{dv}{dx} = \frac{-v}{1 + v} \]

### Step 6: Separate variables again
Now we can separate the variables:

\[ \frac{1 + v}{v} dv = -\frac{1}{x} dx \]

### Step 7: Integrate both sides
Integrating both sides gives:

\[ \int \left(\frac{1}{v} + \frac{1}{1 + v}\right) dv = -\int \frac{1}{x} dx \]

This results in:

\[ \ln |v| + \ln |1 + v| = -\ln |x| + C \]

### Step 8: Combine logarithms
We can combine the logarithms on the left-hand side:

\[ \ln |v(1 + v)| = -\ln |x| + C \]

### Step 9: Exponentiate to solve for \(v\)
Exponentiating both sides gives:

\[ |v(1 + v)| = \frac{C}{|x|} \]

### Step 10: Substitute back for \(y\)
Recall that \(v = \frac{y}{x}\), so we substitute back:

\[ \left|\frac{y}{x}(1 + \frac{y}{x})\right| = \frac{C}{|x|} \]

Multiplying through by \(x^2\) gives:

\[ |y(1 + \frac{y}{x})| = C \]

### Final Step: General solution
Thus, the general solution of the differential equation is:

\[ y(1 + \frac{y}{x}) = C \]

or

\[ y^2 + xy - Cx = 0 \]