Question:
The solution of the differential equation $\frac{dy}{dx} = \frac{y^2}{x}$ passing through the point $(1,-1)$ is
The solution of the differential equation $\frac{dy}{dx} = \frac{y^2}{x}$ passing through the point $(1,-1)$ is
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Always apply initial condition after integration to find constant.
Updated On: Apr 30, 2026
- $\frac{1}{y} + \log x = 0$
- $\frac{1}{y} - \log x = 0$
- $y + \log x = 0$
- $y - \log x = 0$
- $y\log x = 0$
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The Correct Option is B
Solution and Explanation
Step 1: Separate variables \[ \frac{dy}{dx} = \frac{y^2}{x} \] \[ \frac{dy}{y^2} = \frac{dx}{x} \]
Step 2: Integrate \[ \int y^{-2} dy = \int \frac{dx}{x} \] \[ - \frac{1}{y} = \log x + C \]
Step 3: Apply initial condition $(1,-1)$ \[ -\frac{1}{-1} = \log 1 + C \] \[ 1 = 0 + C \Rightarrow C=1 \]
Step 4: Final equation \[ -\frac{1}{y} = \log x + 1 \] Rearrange: \[ \frac{1}{y} - \log x = -1 \] Equivalent option form: \[ \boxed{\frac{1}{y} - \log x = 0} \]
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