The solution of the equation sqrt{a + x} frac | vedclass

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Question

(A) Given the differential equation: $\sqrt{a + x} \frac{dy}{dx} + x = 0$Separating the variables,we get: $dy = -\frac{x}{\sqrt{a + x}} dx$Integrating

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$3y + 2\sqrt{a + x}(x - 2a) = 3c$

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(A) Given the differential equation: $\sqrt{a + x} \frac{dy}{dx} + x = 0$Separating the variables,we get: $dy = -\frac{x}{\sqrt{a + x}} dx$Integrating both sides: $\int dy = -\int \frac{x}{\sqrt{a + x}} dx$To solve the integral,rewrite the numerator: $\int \frac{x}{\sqrt{a + x}} dx = \int \frac{x + a - a}{\sqrt{a + x}} dx = \int (\sqrt{a + x} - \frac{a}{\sqrt{a + x}}) dx$Thus,$y = -[\int (a + x)^{1/2} dx - a \int (a + x)^{-1/2} dx]$$y = -[\frac{2}{3}(a + x)^{3/2} - 2a(a + x)^{1/2}] + c$$y = -\frac{2}{3}(a + x)^{3/2} + 2a(a + x)^{1/2} + c$Multiply by $3$: $3y = -2(a + x)^{3/2} + 6a(a + x)^{1/2} + 3c$Factor out $-2\sqrt{a + x}$: $3y = -2\sqrt{a + x} [(a + x) - 3a] + 3c$$3y = -2\sqrt{a + x} (x - 2a) + 3c$Rearranging gives: $3y + 2\sqrt{a + x}(x - 2a) = 3c$.

The solution of the equation $\sqrt{a + x} \frac{dy}{dx} + x = 0$ is

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