If \( y \frac{dy}{dx} = x \left[ \frac{\phi(y^2/x^2)}{\phi'(y^2/x^2)} + \frac{y^2}{x^2} \right], x>0, \phi>0, \) and \( y(1) = -1 \), then \( \phi\left(\frac{y^2}{4}\right) \) is equal to :
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Whenever a differential equation contains terms like \( \phi(y/x) \) or \( \phi(y^2/x^2) \), it is likely homogeneous. Substitution \( v = y/x \) or \( v = y^2/x^2 \) should be your first step.
Step 1: Understanding the Concept:
This is a homogeneous differential equation because the expression depends on the ratio \( y^2/x^2 \). Step 2: Key Formula or Approach:
Let \( v = \frac{y^2}{x^2} \). Then \( y^2 = v x^2 \).
Differentiate with respect to \( x \): \( 2y \frac{dy}{dx} = x^2 \frac{dv}{dx} + 2vx \). Step 3: Detailed Explanation:
The given equation is:
\[ 2y \frac{dy}{dx} = 2x \left[ \frac{\phi(v)}{\phi'(v)} + v \right] \]
Substitute \( 2y \frac{dy}{dx} = x^2 \frac{dv}{dx} + 2vx \):
\[ x^2 \frac{dv}{dx} + 2vx = 2x \frac{\phi(v)}{\phi'(v)} + 2vx \]
\[ x^2 \frac{dv}{dx} = 2x \frac{\phi(v)}{\phi'(v)} \implies \frac{\phi'(v)}{\phi(v)} dv = \frac{2}{x} dx \]
Integrating both sides:
\[ \ln \phi(v) = 2 \ln x + \ln C = \ln(C x^2) \]
\[ \phi(v) = C x^2 \implies \phi(y^2/x^2) = C x^2 \]
Use \( y(1) = -1 \). At \( x = 1 \), \( y^2 = 1 \):
\[ \phi(1/1) = C \cdot 1^2 \implies C = \phi(1) \]
So, the general solution is \( \phi(y^2/x^2) = \phi(1) x^2 \).
We need to find \( \phi(y^2/4) \). This corresponds to taking \( x = 2 \).
\[ \phi(y^2/2^2) = \phi(1) \cdot 2^2 = 4\phi(1) \] Step 4: Final Answer:
The value is \( 4\phi(1) \).
