MCQ
Which of the following differential equations has $y=c_{1} e^{x}+c_{2} e^{-x}$ as the general solution?
- ✓$\frac{d^{2} y}{d x^{2}}-y=0$
- B$\frac{d^{2} y}{d x^{2}}+y=0$
- C$\frac{d^{2} y}{d x^{2}}+1=0$
- D$\frac{d^{2} y}{d x^{2}}-1=0$
$y=c_{1} e^{x}+c_{2} e^{x}$ .........$(1)$
Differentiating with respect to $\mathrm{x}$, we get:
$\frac{d y}{d x}=c_{1} e^{x}-c_{2} e^{-x}$
Again, differentiating with respect to $\mathrm{x}$, we get:
$\frac{d^{2} y}{d x^{2}}=c_{1} e^{x}+c_{2} e^{-x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=y$
$\Rightarrow \frac{d^{2} y}{d x^{2}}-y=0$
This is the required differential equation of the given equation of curve.
Hence, the correct answer is $\mathrm{A}$.
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$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
$P(x) = 2x^3 + x^2 + 3x - 2? $
$(i)$ It has exactly one positive real root.
$(ii)$ It has either one or three negative roots.
$(iii)$It has a root between $0$ and $1.$
$(iv)$ It must have exactly two real roots.
$(v)$ It has a negative root between $- 2$ and $-1.$
$(vi)$ It has no complex roots.