\( x^2 + 2y^2 = x^4 \)
\( x^2 + y^2 = x^4 \)
Given the differential equation:
\( x y \frac{dy}{dx} = x^2 + 2y^2 \)
Divide by \( x y \):
\( \frac{dy}{dx} = \frac{x^2 + 2y^2}{x y} \)
Separate terms on the RHS:
\( \frac{dy}{dx} = \frac{x}{y} + \frac{2y}{x} \)
Multiply by \( y \):
\( y \frac{dy}{dx} = x + \frac{2y^2}{x} \)
To simplify, use the substitution \( z = y^2 \). Then \( \frac{dz}{dx} = 2y \frac{dy}{dx} \).
From the original equation, substitute \( z \):
\( x y \frac{dy}{dx} = x^2 + 2y^2 \Rightarrow x \cdot \frac{1}{2} \cdot \frac{dz}{dx} = x^2 + 2z \)
This yields:
\( \frac{1}{2} x \frac{dz}{dx} = x^2 + 2z \)
Multiply by 2:
\( x \frac{dz}{dx} = 2x^2 + 4z \)
Rearrange into standard linear form:
\( \frac{dz}{dx} - \frac{4z}{x} = 2x \)
This is a linear differential equation. The integrating factor (IF) is:
\( \text{IF} = e^{\int -\frac{4}{x} dx} = x^{-4} \)
Multiply the equation by the IF:
\( x^{-4} \frac{dz}{dx} - \frac{4z}{x^5} = 2x^{-3} \)
The left side is the derivative of \( z x^{-4} \):
\( \frac{d}{dx}(z x^{-4}) = 2x^{-3} \)
Integrate both sides:
\( z x^{-4} = \int 2x^{-3} dx = -x^{-2} + C \)
Solve for \( z \):
\( z = x^4 (-x^{-2} + C) = -x^2 + Cx^4 \)
Substitute back \( z = y^2 \):
\( y^2 = -x^2 + Cx^4 \Rightarrow x^2 + y^2 = Cx^4 \)
Apply the initial condition \( y(1) = 0 \):
\( x = 1, y = 0 \Rightarrow 1^2 + 0^2 = C(1)^4 \Rightarrow C = 1 \)
Final solution:
\( \boxed{x^2 + y^2 = x^4} \)