The solution of the differential equation `y"dy"/"dx"=x[y^2/x^2 + (phi(y^2/x^2))/((phi')(y^2/x^2))]` is (where c is a constant )

Options

A. `phi(y^2/x^2)=cx `
B. `xphi(y^2/x^2)=c`
C. `phi(y^2/x^2)=cx^2`
D. `x^2phi(y^2/x^2)=c`

Answer

To solve the given differential equation \( y \frac{dy}{dx} = x \left( \frac{y^2}{x^2} + \frac{\phi\left(\frac{y^2}{x^2}\right)}{\phi'\left(\frac{y^2}{x^2}\right)} \right) \), we will follow these steps:

### Step 1: Substitute \( y = vx \)
Let \( y = vx \), where \( v \) is a function of \( x \). Then, we can differentiate \( y \) with respect to \( x \):

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

### Step 2: Substitute into the differential equation
Substituting \( y \) and \( \frac{dy}{dx} \) into the original equation gives:

\[ y \frac{dy}{dx} = vx(v + x \frac{dv}{dx}) = x \left( \frac{(vx)^2}{x^2} + \frac{\phi\left(\frac{(vx)^2}{x^2}\right)}{\phi'\left(\frac{(vx)^2}{x^2}\right)} \right) \]

### Step 3: Simplify the equation
This simplifies to:

\[ vx \left( v + x \frac{dv}{dx} \right) = x \left( v^2 + \frac{\phi(v^2)}{\phi'(v^2)} \right) \]

Dividing both sides by \( x \) (assuming \( x \neq 0 \)):

\[ v \left( v + x \frac{dv}{dx} \right) = v^2 + \frac{\phi(v^2)}{\phi'(v^2)} \]

### Step 4: Rearranging the equation
Rearranging gives:

\[ v + x \frac{dv}{dx} = 1 + \frac{\phi(v^2)}{v^2 \phi'(v^2)} \]

### Step 5: Isolate \( x \frac{dv}{dx} \)
Now, isolate \( x \frac{dv}{dx} \):

\[ x \frac{dv}{dx} = \frac{\phi(v^2)}{v^2 \phi'(v^2)} \]

### Step 6: Separate variables
Rearranging gives:

\[ v^2 \phi'(v^2) dv = \frac{\phi(v^2)}{x} dx \]

### Step 7: Integrate both sides
Integrate both sides:

\[ \int v^2 \phi'(v^2) dv = \int \frac{\phi(v^2)}{x} dx \]

Let \( t = v^2 \), then \( dv = \frac{1}{2\sqrt{t}} dt \):

\[ \frac{1}{2} \int \phi'(t) dt = \int \frac{\phi(t)}{x} dx \]

### Step 8: Solve the integrals
The left side integrates to \( \frac{1}{2} \phi(t) + C_1 \) and the right side integrates to \( \log x + C_2 \):

\[ \frac{1}{2} \phi(v^2) = \log x + C \]

### Step 9: Solve for \( y \)
Substituting back \( v = \frac{y}{x} \):

\[ \frac{1}{2} \phi\left(\frac{y^2}{x^2}\right) = \log x + C \]

### Step 10: Final solution form
This gives us the relationship between \( y \) and \( x \):

\[ \phi\left(\frac{y^2}{x^2}\right) = 2 \log x + C' \]

Thus, the solution of the differential equation is:

\[ \frac{y^2}{x^2} = \phi^{-1}(2 \log x + C') \]