MCQ
The differential equation whose solution is $y = A\sin x + B\cos x,$ is
  • $\frac{{{d^2}y}}{{d{x^2}}} + y = 0$
  • B
    $\frac{{{d^2}y}}{{d{x^2}}} - y = 0$
  • C
    $\frac{{dy}}{{dx}} + y = 0$
  • D
    None of these

Answer

Correct option: A.
$\frac{{{d^2}y}}{{d{x^2}}} + y = 0$
a
(a) $y = A\sin x + B\cos x$==> $\frac{{dy}}{{dx}} = A\cos x - B\sin x$

==> $\frac{{{d^2}y}}{{d{x^2}}} = - A\sin x - B\cos x$$ = - (A\sin x + B\cos x) = - y$

==> $\frac{{{d^2}y}}{{d{x^2}}} + y = 0$  is the required differential equation.

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The solution of the differential equation $\frac{\text{dy}}{\text{dx}}+\frac{2\text{xy}}{1+\text{x}^2}=\frac{1}{(1+\text{x}^2)^2}$ is:

  1. $\text{y}(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$

  2. $\frac{\text{y}}{1+\text{x}^2}=\text{C}+\tan^{-1}\text{x}$

  3. $\text{y}\log(1+\text{x}^2)=\text{C}+\tan^{-1}\text{x}$

  4. $\text{y}(1+\text{x}^2)=\text{C}+\sin^{-1}\text{x}$