Q119 / 160 · AP EAMCET 2024 (23 May Shift 1) · Mathematics
The differential equation for which $a x+b y=1$ is general solution is
- A$\frac{d y}{d x}=x+c$
- B$y \frac{d^2 y}{d x^2}+x=1$
- Correct answerC$\frac{d y}{d x}$
- D$\frac{d^3 y}{d x^3}=0$
Solution
$a x+b y=1$
Differentiating w.r.t. $x$
$a+b \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-a}{b}$
Differentiating w.r.t. $x$ $\Rightarrow \frac{d^2 y}{d x^2}=0$ is the required differential equation.
Differentiating w.r.t. $x$ $\Rightarrow \frac{d^2 y}{d x^2}=0$ is the required differential equation.
Hence, the correct answer is option: 'C'.