y = ke^{sin ^{-1} x} + 3 is a solution of whi | vedclass

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(A) Given the equation: $y = ke^{\sin^{-1} x} + 3$. Subtracting $3$ from both sides,we get: $y - 3 = ke^{\sin^{-1} x}$. Differentiating both sides wit

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$\sqrt{1 - x^2} \frac{dy}{dx} = y - 3$

Vedclass · Answer

(A) Given the equation: $y = ke^{\sin^{-1} x} + 3$. Subtracting $3$ from both sides,we get: $y - 3 = ke^{\sin^{-1} x}$. Differentiating both sides with respect to $x$: $\frac{d}{dx}(y - 3) = \frac{d}{dx}(ke^{\sin^{-1} x})$. $\frac{dy}{dx} = k \cdot e^{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}}$. Since $ke^{\sin^{-1} x} = y - 3$,we substitute this into the derivative: $\frac{dy}{dx} = (y - 3) \cdot \frac{1}{\sqrt{1 - x^2}}$. Multiplying both sides by $\sqrt{1 - x^2}$,we obtain: $\sqrt{1 - x^2} \frac{dy}{dx} = y - 3$.

$y = ke^{\sin ^{-1} x} + 3$ is a solution of which differential equation?

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