MCQ
$x^2+y^2=a^2$ is a solution of
- A$\frac{d^2 y}{d x^2}+\frac{d y}{d x}-y=0$
- B$y=x \sqrt{1+\left(\frac{d y}{d x}\right)^2}+a^2 y$
- ✓$y=x \frac{d y}{d x}+a \sqrt{1+\left(\frac{d y}{d x}\right)^2}$
- D$\frac{d^2 y}{d x^2}=(x+1) \frac{d y}{d x}$
Hint : $x^2+y^2=a^2 \quad \therefore 2 x+2 y \frac{d y}{d x}=0$
$\therefore \frac{d y}{d x}=-\frac{x}{y}$
$=x\left(-\frac{x}{y}\right)+a \sqrt{1+\frac{x^2}{y^2}}=-\frac{x^2}{y}+a \times \frac{a}{y}$
$=\frac{a^2-x^2}{y}=\frac{y^2}{y}=y$.
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