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If the differential equation obtained by eliminating A,B from $y=\left(\sin ^{-1} x\right)^2+A \cos ^{-1} x+B$ is $\left(a-x^2\right) y^{\prime \prime}-x y^{\prime}=b$, then $\frac{b+a}{b-a}=$
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$3$
we are given that
$y=\left(\sin ^{-1} x\right)^2+A \cos ^{-1} x+B$...(i)
$\mathrm{y}^1=2 \sin ^{-1}(\mathrm{x}) \times \frac{1}{\sqrt{1-\mathrm{x}^2}}-\frac{\mathrm{A}}{\sqrt{1-\mathrm{x}^2}}$...(ii)
$=\mathrm{A}=2 \sin ^{-1} \mathrm{z}-\sqrt{1-\mathrm{x}^2}, \mathrm{y}^1$
Now $y^1=\frac{2 \sin ^{-1}(x)-A}{\sqrt{1-x^2}}$
$\therefore \mathrm{y}^4=\frac{2 \sqrt{1-\mathrm{x}^2+2 \mathrm{x} \sin ^{-1} \mathrm{x}-\mathrm{Ax}}}{\left(1+\mathrm{x}^2\right)^{3 / 2}}=2 \sqrt{1-\mathrm{x}^2}+2 \mathrm{x} \sin ^{-1} \mathrm{x}$
$\frac{-2 x \sin ^{-1} x+x y^1 \sqrt{1-x^2}}{\left(1-x^2\right)^{3 / 2}}$
$\begin{aligned} & y^{\prime \prime}=\frac{2 \sqrt{1-x^2}+x y^{\prime} \sqrt{1-x^2}}{\left(1-x^2\right)^{3 / 2}} \\ & \Rightarrow y^{\prime \prime}=\frac{2+x y^{\prime}}{\left(1-x^2\right)} \\ & \Rightarrow y^{\prime \prime}=\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}=2\end{aligned}$
Now, $\frac{\mathrm{b}+\mathrm{a}}{\mathrm{b}-\mathrm{a}}=\frac{2+1}{2-1}=3$
$y=\left(\sin ^{-1} x\right)^2+A \cos ^{-1} x+B$...(i)
$\mathrm{y}^1=2 \sin ^{-1}(\mathrm{x}) \times \frac{1}{\sqrt{1-\mathrm{x}^2}}-\frac{\mathrm{A}}{\sqrt{1-\mathrm{x}^2}}$...(ii)
$=\mathrm{A}=2 \sin ^{-1} \mathrm{z}-\sqrt{1-\mathrm{x}^2}, \mathrm{y}^1$
Now $y^1=\frac{2 \sin ^{-1}(x)-A}{\sqrt{1-x^2}}$
$\therefore \mathrm{y}^4=\frac{2 \sqrt{1-\mathrm{x}^2+2 \mathrm{x} \sin ^{-1} \mathrm{x}-\mathrm{Ax}}}{\left(1+\mathrm{x}^2\right)^{3 / 2}}=2 \sqrt{1-\mathrm{x}^2}+2 \mathrm{x} \sin ^{-1} \mathrm{x}$
$\frac{-2 x \sin ^{-1} x+x y^1 \sqrt{1-x^2}}{\left(1-x^2\right)^{3 / 2}}$
$\begin{aligned} & y^{\prime \prime}=\frac{2 \sqrt{1-x^2}+x y^{\prime} \sqrt{1-x^2}}{\left(1-x^2\right)^{3 / 2}} \\ & \Rightarrow y^{\prime \prime}=\frac{2+x y^{\prime}}{\left(1-x^2\right)} \\ & \Rightarrow y^{\prime \prime}=\left(1-x^2\right) y^{\prime \prime}-x y^{\prime}=2\end{aligned}$
Now, $\frac{\mathrm{b}+\mathrm{a}}{\mathrm{b}-\mathrm{a}}=\frac{2+1}{2-1}=3$
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