Form the differential equation having y = ( ^ -1 x)^ 2 + A ^ -1 +… - Vidyadip

Question

Form the differential equation having $\text{y} = (\sin^{-1}\text{x})^{2} + \text{A}\cos^{-1} + \text{ B},$ where A and B are arbitrary constants, as its general solution.

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Answer

We have, $\text{y}=(\sin^{-1}\text{x})^2+\text{A}\cos^{-1}\text{x}+\text{B}$
On differentiating w.r.t.x, we get
$\frac{\text{dy}}{\text{dx}}=\frac{2\sin^{-1}\text{x}}{\sqrt{1-\text{x}^2}}+\frac{(-\text{A})}{\sqrt{1-\text{x}^2}}$
$\Rightarrow\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2\sin^{-1}\text{x}-\text{A}$
Again, differentiating w.r.t.x, we get
$\sqrt{1-\text{x}^2}\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}+\frac{\text{dy}}{\text{dx}}.\frac{-2\text{x}}{2\sqrt{1+\text{x}^2}}=\frac{2}{{\sqrt{1-\text{x}}^2}}$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\frac{\text{x}}{\sqrt{1-\text{x}^2}}.\sqrt{1-\text{x}^2}\frac{\text{dy}}{\text{dx}}=2$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{x}\frac{\text{dy}}{\text{dx}}=2$
$\Rightarrow(1-\text{x}^2)\frac{\text{d}^2\text{y}}{\text{d}\text{x}^2}-\text{x}\frac{\text{dy}}{\text{dx}}-2=0$
which is the required differential equation.

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