Question

Solution of the differential equation $y' = \frac{x^2+y^2}{xy}$, where $y(1) = -2$ is given by

• A. $y^2 = 4x^2 \log x^2 + x^2$
• B. $y^2 = x^2 \log x - x^2$
• C. $y^2 = x \log x^2 + 4x^2$
• D. $y^2 = x^2 \log x^2 + 4x^2$

Hint:

You can easily check your answer by verifying the initial condition. Plug $x = 1$ into the options to see which one yields $y^2 = 4$: For (D): $y^2 = 1^2 \log(1) + 4(1)^2 = 0 + 4 = 4 \Rightarrow y = \pm 2$. This step immediately helps eliminate options that do not satisfy the initial state boundary condition!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given a first-order homogeneous differential equation with an initial condition $y = -2$ when $x = 1$. We need to find its particular solution.

Step 2: Key Formula or Approach:
The given equation is homogeneous because the total degree of every term in the numerator and denominator is 2. We solve homogeneous differential equations by substituting: $$y = vx \Rightarrow \frac{dy}{dx} = v + x\frac{dv}{dx}$$ This substitution will transform our equation into a variable separable format.

Step 3: Detailed Explanation:
Rewrite the given equation: $$\frac{dy}{dx} = \frac{x^2 + y^2}{xy}$$ Substitute $y = vx$ and $\frac{dy}{dx} = v + x\frac{dv}{dx}$: $$v + x\frac{dv}{dx} = \frac{x^2 + (vx)^2}{x(vx)}$$ $$v + x\frac{dv}{dx} = \frac{x^2(1 + v^2)}{x^2v}$$ $$v + x\frac{dv}{dx} = \frac{1 + v^2}{v}$$ Subtract $v$ from both sides: $$x\frac{dv}{dx} = \frac{1 + v^2}{v} - v$$ $$x\frac{dv}{dx} = \frac{1 + v^2 - v^2}{v}$$ $$x\frac{dv}{dx} = \frac{1}{v}$$ Separate the variables $v$ and $x$: $$v \, dv = \frac{1}{x} dx$$ Integrating both sides: $$\int v \, dv = \int \frac{1}{x} dx$$ $$\frac{v^2}{2} = \log|x| + c$$ Substitute back $v = \frac{y}{x}$: $$\frac{y^2}{2x^2} = \log|x| + c \Rightarrow y^2 = 2x^2 \log|x| + 2cx^2$$ Using the property $2\log|x| = \log x^2$: $$y^2 = x^2 \log x^2 + 2cx^2$$ Now, apply the initial condition $y(1) = -2$: $$(-2)^2 = (1)^2 \log(1^2) + 2c(1)^2$$ $$4 = 0 + 2c \Rightarrow 2c = 4$$ Substitute $2c = 4$ back into the general solution equation: $$y^2 = x^2 \log x^2 + 4x^2$$

Step 4: Final Answer:
The particular solution is $y^2 = x^2 \log x^2 + 4x^2$, which corresponds exactly to option (D).