The solution of the differential equation sqr | vedclass

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(A) Given differential equation is $\sqrt{1-y^2} dx + (x - \sin^{-1} y) dy = 0$. Dividing by $dy$ and $\sqrt{1-y^2}$, we get $\frac{dx}{dy} + \frac{x}

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$x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}$

Vedclass · Answer

(A) Given differential equation is $\sqrt{1-y^2} dx + (x - \sin^{-1} y) dy = 0$. Dividing by $dy$ and $\sqrt{1-y^2}$, we get $\frac{dx}{dy} + \frac{x}{\sqrt{1-y^2}} = \frac{\sin^{-1} y}{\sqrt{1-y^2}}$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \frac{1}{\sqrt{1-y^2}}$ and $Q(y) = \frac{\sin^{-1} y}{\sqrt{1-y^2}}$. The integrating factor is $IF = e^{\int P(y) dy} = e^{\int \frac{1}{\sqrt{1-y^2}} dy} = e^{\sin^{-1} y}$. The solution is $x \cdot IF = \int Q(y) \cdot IF dy + c$. $x e^{\sin^{-1} y} = \int \frac{\sin^{-1} y}{\sqrt{1-y^2}} e^{\sin^{-1} y} dy + c$. Let $t = \sin^{-1} y$, then $dt = \frac{1}{\sqrt{1-y^2}} dy$. $x e^{\sin^{-1} y} = \int t e^t dt + c = (t e^t - e^t) + c = e^t(t - 1) + c$. Substituting $t = \sin^{-1} y$, we get $x e^{\sin^{-1} y} = e^{\sin^{-1} y}(\sin^{-1} y - 1) + c$. Dividing by $e^{\sin^{-1} y}$, we get $x = \sin^{-1} y - 1 + c e^{-\sin^{-1} y}$.

The solution of the differential equation $\sqrt{1-y^2} dx + x dy - \sin^{-1} y dy = 0$ is

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