Question

Differential equations of the first order and first degree.
Q. ydx + $\sqrt{1-x^2}\sin {}^{-1}xdy=0$

Answer 1

Dear student
$$\begin{gathered} \mathrm{ydx}+\sqrt{1-{\mathrm{x}}^2}{\sin }^{-1}\mathrm{xdy}=0 \\ \sqrt{1-{\mathrm{x}}^2}{\sin }^{-1}\mathrm{xdy}=-\mathrm{ydx} \\ \frac{-\mathrm{dy}}{\mathrm{y}}=\frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}{\sin }^{-1}\mathrm{x}} \\ \mathrm{Integrating}\mathrm{both}\mathrm{sides},\mathrm{we}\mathrm{get} \\ -\int \frac{\mathrm{dy}}{\mathrm{y}}=\int \frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}{\sin }^{-1}\mathrm{x}} \\ \mathrm{Consider},\int \frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}{\sin }^{-1}\mathrm{x}}...\left(1\right) \\ \mathrm{Put}{\sin }^{-1}\mathrm{x}=\mathrm{t} \\ \mathrm{differentiate}\mathrm{wrt}\mathrm{x},\mathrm{we}\mathrm{get} \\ \frac{1}{\sqrt{1-{\mathrm{x}}^2}}=\frac{\mathrm{dt}}{\mathrm{dx}} \\ \Rightarrow \frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}}=\mathrm{dt} \\ \mathrm{So},\left(1\right)\mathrm{becomes} \\ \int \frac{\mathrm{dt}}{\mathrm{t}} \\ =\log \left|\mathrm{t}\right|+\mathrm{c} \\ =\log \left|{\sin }^{-1}\mathrm{x}\right|+\mathrm{c} \\ \mathrm{Now}\mathrm{consider},-\int \frac{\mathrm{dy}}{\mathrm{y}}=\int \frac{\mathrm{dx}}{\sqrt{1-{\mathrm{x}}^2}{\sin }^{-1}\mathrm{x}} \\ -\log \left|\mathrm{y}\right|=\log \left|{\sin }^{-1}\mathrm{x}\right|+\mathrm{c} \\ {\mathrm{c}}_1=\log \left|{\sin }^{-1}\mathrm{x}\right|+\log \left|\mathrm{y}\right|,\mathrm{where}{\mathrm{c}}_1=-\mathrm{c} \\ {\mathrm{e}}^{{\mathrm{c}}_1}={\mathrm{e}}^{\log \left|{\mathrm{ysin}}^{-1}\mathrm{x}\right|} \\ \mathrm{C}=\left|{\mathrm{ysin}}^{-1}\mathrm{x}\right|,\mathrm{where}\mathrm{C}={\mathrm{e}}^{{\mathrm{c}}_1}\left[{\mathrm{e}}^{\mathrm{log\theta }}=\mathrm{\theta }\right] \end{gathered}$$
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